[R] Concatenate two lists, list by list
Dear all, I would like to concatenate the lists below str(Part2$dataset) List of 3 $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... str(Part1$dataset) List of 3 $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... I tried concatenating those with: > str(cbind(Part1$datase,Part2$dataset)) List of 6 $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... - attr(*, "dim")= int [1:2] 3 2 but I want something different. To concatenate those into a list by list operation so I will end up with something looking like that str(concatenatedLists) List of 3 $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... - attr(*, "dim")= int [1:2] 3 2 Is there anything that can do that in R? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Concatenate two lists, list by list
Thanks a lot Petr, for the answer unfortunately that would convert everything to a matrix num [1:32002, 1:3] 0 0 0 0 0 0 0 0 0 0 ... but if you check below you can see that I Want those to form a list. Regards Alex From: PIKAL Petr Sent: Tuesday, January 22, 2013 11:51 AM Subject: RE: [R] Concatenate two lists, list by list Hi Maybe you could use mapply mapply(c, Part1$dataset,Part2$dataset) Regards Petr > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Alaios > Sent: Tuesday, January 22, 2013 11:26 AM > To: R help > Subject: [R] Concatenate two lists, list by list > > Dear all, > I would like to concatenate the lists below > > str(Part2$dataset) > List of 3 > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > > > > str(Part1$dataset) > List of 3 > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > > > I tried concatenating those with: > > > > str(cbind(Part1$datase,Part2$dataset)) > List of 6 > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... > - attr(*, "dim")= int [1:2] 3 2 > > > but I want something different. To concatenate those into a list by > list operation so I will end up with something looking like that > > str(concatenatedLists) > > List of 3 > $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... > $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... > - attr(*, "dim")= int [1:2] 3 2 > > > Is there anything that can do that in R? > > Regards > Alex > [[alternative HTML version deleted]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Concatenate two lists, list by list
Thanks a lot. Unfortunately that did not help either. num [1:32003, 1:3] 0 0 0 0 0 0 0 0 0 0 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:32003] "" "" "" "" ... ..$ : NULL but I want to get >> List of 3 >> $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... >> - attr(*, "dim")= int [1:2] 3 2 >> I am sorry that I can not find reproducible example to show you Alex From: D. Rizopoulos Cc: PIKAL Petr ; R help Sent: Wednesday, January 23, 2013 11:08 AM Subject: Re: [R] Concatenate two lists, list by list you just need: mapply(c, Part1$dataset, Part2$dataset, SIMPLIFY = FALSE) I hope it helps. Best, Dimitris On 1/23/2013 11:01 AM, Alaios wrote: > Thanks a lot Petr, > for the answer > unfortunately that would convert everything to a matrix > > num [1:32002, 1:3] 0 0 0 0 0 0 0 0 0 0 ... > > but if you check below you can see that I Want those to form a list. > > Regards > Alex > > > > > > From: PIKAL Petr > > Sent: Tuesday, January 22, 2013 11:51 AM > Subject: RE: [R] Concatenate two lists, list by list > > Hi > > Maybe you could use mapply > > mapply(c, Part1$dataset,Part2$dataset) > > Regards > Petr > >> -Original Message- >> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- >> project.org] On Behalf Of Alaios >> Sent: Tuesday, January 22, 2013 11:26 AM >> To: R help >> Subject: [R] Concatenate two lists, list by list >> >> Dear all, >> I would like to concatenate the lists below >> >> str(Part2$dataset) >> List of 3 >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> >> >> >> str(Part1$dataset) >> List of 3 >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> >> >> I tried concatenating those with: >> >> >>> str(cbind(Part1$datase,Part2$dataset)) >> List of 6 >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:16001] 0 0 0 0 0 0 0 0 0 0 ... >> - attr(*, "dim")= int [1:2] 3 2 >> >> >> but I want something different. To concatenate those into a list by >> list operation so I will end up with something looking like that >> >> str(concatenatedLists) >> >> List of 3 >> $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... >> $ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ... >> - attr(*, "dim")= int [1:2] 3 2 >> >> >> Is there anything that can do that in R? >> >> Regards >> Alex >> [[alternative HTML version deleted]] > [[alternative HTML version deleted]] > > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/biostatistiek/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plot a Matrix as an Image with ggplot
Dear all, I am trying to plot a matrix I have as an image str(matrixToPlot) num [1:21, 1:66] 0 0 0 0 0 0 0 0 0 0 . that contains only 0s and 1s, where the xlabel will be Labeled as str(xLabel) num [1:66] 1e+09 1e+09 1e+09 1e+09 1e+09 ... and the yLabels will be labeled as str(yLabel) num [1:21] -88 -87 -86 -85 -84 -83 -82 -81 -80 -79 ... I have found on the internet that I can do something like that with ggplot2. For example you can run the following library(reshape2)library(ggplot2)m =matrix(rnorm(20),5)ggplot(melt(m),aes(Var1,Var2,fill=value))+geom_raster() What I see missing here is to get my matrix and transform it to a data frame with labels for x and y axis so as ggplot can print the values nice. If you get the idea this matrix will be printed as a two tile pattern let's say with black tiles zeros and white tiles black where a color bar will depicting that black means zero and white one. To help you with my code you will find attached the three items I have discussed here, the matrix, the x and the y labels. Could you please help me with that? Alex__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Format Integers
Dear all, I would like to format the following numbers xLabel [1] 10 1000153846 1000307692 1000461538 1000615385 1000769231 [7] 1000923077 1001076923 1001230769 1001384615 1001538462 1001692308 [13] 1001846154 100200 1002153846 1002307692 1002461538 1002615385 [19] 1002769231 1002923077 1003076923 1003230769 1003384615 1003538462 [25] 1003692308 1003846154 100400 1004153846 1004307692 1004461538 [31] 1004615385 1004769231 1004923077 1005076923 1005230769 1005384615 [37] 1005538462 1005692308 1005846154 100600 1006153846 1006307692 [43] 1006461538 1006615385 1006769231 1006923077 1007076923 1007230769 [49] 1007384615 1007538462 1007692308 1007846154 100800 1008153846 [55] 1008307692 1008461538 1008615385 1008769231 1008923077 1009076923 [61] 1009230769 1009384615 1009538462 1009692308 1009846154 101000 that are in watts to megawatts so they appear (as an example) from: 1009846154 to 1009.846154 and even better to truncate it a bit to look better as 1009.84 Could you please help with the formatting ? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot a Matrix as an Image with ggplot
Hi,
thanks
I changed slightly the code to be reproducible from everyone . I have tried
ggplot
but I need a bit of help to tweak it a bit
So you can run the following in your computer
testData<-matrix(data=round(runif(25)),nrow=5,ncol=5,dimnames=list(1:5,1:5))
p<-ggplot(melt(testData), aes(Var2,Var1, fill=value))+xlab("MHz") +
ylab("Threshold") + geom_raster()
p
What I want to improve is
a: make the colorbar so only two specific colors lets say black and white and
only two values 0 and 1 and somewhere the string as title of the color bar
"text" to appear.
I have tried something like
p+ scale_color_gradient(low="red",high="blue")
Fehler in unit(tic_pos.c, "mm") : 'x' and 'units' must have length > 0
ggplot(melt(testData), aes(Var2,Var1, fill=value,colorbin=2))+xlab("MHz") +
ylab("Threshold") + geom_raster()
but this did not affect colorbar entries.
b. reduce/remove the grayish border that appears between the legend and the
image plot
Could you please help me with these two?
Regards
Alex
From: John Kane
To: Alaios ; R help
Sent: Thursday, February 14, 2013 4:35 PM
Subject: RE: [R] Plot a Matrix as an Image with ggplot
The R-help list is rather picky about what attached. None of your attachments
arrived.
The str() info is useful but please supply some sample data
The easiest way to supply data is to use the dput() function. Example with
your file named "testfile":
dput(testfile)
Then copy the output and paste into your email. For large data sets, you can
just supply a representative sample. Usually,
dput(head(testfile, 100)) will be sufficient.
How are you writing the code/or what format is the original email. Your code in
the body of the text is badly messed up -- you probably need to post only in
text. HTML etc is automatically dropped.
John Kane
Kingston ON Canada
> -Original Message-
> From: ala...@yahoo.com
> Sent: Thu, 14 Feb 2013 07:15:05 -0800 (PST)
> To: r-help@r-project.org
> Subject: [R] Plot a Matrix as an Image with ggplot
>
> Dear all,
> I am trying to plot a matrix I have as an image
>
> str(matrixToPlot)
> num [1:21, 1:66] 0 0 0 0 0 0 0 0 0 0 .
>
>
> that contains only 0s and 1s,
>
>
> where the xlabel will be Labeled as
> str(xLabel)
> num [1:66] 1e+09 1e+09 1e+09 1e+09 1e+09 ...
>
> and the yLabels will be labeled as
> str(yLabel)
> num [1:21] -88 -87 -86 -85 -84 -83 -82 -81 -80 -79 ...
>
>
> I have found on the internet that I can do something like that with
> ggplot2.
> For example
> you can run the following
>
>
>
> library(reshape2)library(ggplot2)m
> =matrix(rnorm(20),5)ggplot(melt(m),aes(Var1,Var2,fill=value))+geom_raster()
>
> What I see missing here is to get my matrix and transform it to a data
> frame with labels for x and y axis so as ggplot can print the values
> nice. If you get the idea this matrix will be printed as a two tile
> pattern let's say with black tiles zeros and white tiles black where a
> color bar will depicting that black means zero and white one.
>
> To help you with my code you will find attached the three items I have
> discussed here, the matrix, the x and the y labels.
>
> Could you please help me with that?
>
> Alex
FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks & orcas on your
desktop!
Check it out at http://www.inbox.com/marineaquarium
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot a Matrix as an Image with ggplot
You were right,
the following two work
testData<-matrix(data=round(runif(25)),nrow=5,ncol=5,dimnames=list(Var1=c(1:5),Var2=c(1:5)))
ggplot(melt(testData), aes(Var2,Var1, fill=value))+xlab("MHz") +
ylab("Threshold") + geom_raster() + scale_fill_gradient(low="#FF",
high="#00")
still there are two minor unresolved issues.
a. How to reduce the extra space in the plot between legend and the tiles?
b. How to make color bar depict only the two values of the game, 0 and 1 ,
instead of 0, 0.25, 0.50, 0.75, 1?
I would like to thank you in advanec for your help
Regards
Alex
- Original Message -
From: Dennis Murphy
To: Alaios
Cc:
Sent: Friday, February 15, 2013 10:19 AM
Subject: Re: [R] Plot a Matrix as an Image with ggplot
Your aesthetic is fill, not color. Change scale_color_gradient to
scale_fill_gradient and you'll get what you expect.
D.
On Thu, Feb 14, 2013 at 11:31 PM, Alaios wrote:
>
>
> Hi,
> thanks
> I changed slightly the code to be reproducible from everyone . I have tried
> ggplot
> but I need a bit of help to tweak it a bit
>
> So you can run the following in your computer
>
> testData<-matrix(data=round(runif(25)),nrow=5,ncol=5,dimnames=list(1:5,1:5))
>
> p<-ggplot(melt(testData), aes(Var2,Var1, fill=value))+xlab("MHz") +
>ylab("Threshold") + geom_raster()
>
>
> p
>
>
> What I want to improve is
> a: make the colorbar so only two specific colors lets say black and white and
> only two values 0 and 1 and somewhere the string as title of the color bar
> "text" to appear.
>
> I have tried something like
> p+ scale_color_gradient(low="red",high="blue")
> Fehler in unit(tic_pos.c, "mm") : 'x' and 'units' must have length > 0
>
>
> ggplot(melt(testData), aes(Var2,Var1, fill=value,colorbin=2))+xlab("MHz") +
> ylab("Threshold") + geom_raster()
> but this did not affect colorbar entries.
>
>
> b. reduce/remove the grayish border that appears between the legend and the
> image plot
>
> Could you please help me with these two?
>
> Regards
> Alex
>
>
>
>
> From: John Kane
> To: Alaios ; R help
> Sent: Thursday, February 14, 2013 4:35 PM
> Subject: RE: [R] Plot a Matrix as an Image with ggplot
>
> The R-help list is rather picky about what attached. None of your attachments
> arrived.
>
> The str() info is useful but please supply some sample data
>
> The easiest way to supply data is to use the dput() function. Example with
> your file named "testfile":
> dput(testfile)
>
> Then copy the output and paste into your email. For large data sets, you can
> just supply a representative sample. Usually,
> dput(head(testfile, 100)) will be sufficient.
>
> How are you writing the code/or what format is the original email. Your code
> in the body of the text is badly messed up -- you probably need to post only
> in text. HTML etc is automatically dropped.
>
>
>
>
> John Kane
> Kingston ON Canada
>
>
>> -Original Message-
>> From: ala...@yahoo.com
>> Sent: Thu, 14 Feb 2013 07:15:05 -0800 (PST)
>> To: r-help@r-project.org
>> Subject: [R] Plot a Matrix as an Image with ggplot
>>
>> Dear all,
>> I am trying to plot a matrix I have as an image
>>
>> str(matrixToPlot)
>> num [1:21, 1:66] 0 0 0 0 0 0 0 0 0 0 .
>>
>>
>> that contains only 0s and 1s,
>>
>>
>> where the xlabel will be Labeled as
>> str(xLabel)
>> num [1:66] 1e+09 1e+09 1e+09 1e+09 1e+09 ...
>>
>> and the yLabels will be labeled as
>> str(yLabel)
>> num [1:21] -88 -87 -86 -85 -84 -83 -82 -81 -80 -79 ...
>>
>>
>> I have found on the internet that I can do something like that with
>> ggplot2.
>> For example
>> you can run the following
>>
>>
>>
>> library(reshape2)library(ggplot2)m
>> =matrix(rnorm(20),5)ggplot(melt(m),aes(Var1,Var2,fill=value))+geom_raster()
>>
>> What I see missing here is to get my matrix and transform it to a data
>> frame with labels for x and y axis so as ggplot can print the values
>> nice. If you get the idea this matrix will be printed as a two tile
>> pattern let's say with black tiles zeros and white tiles black where a
>> color bar will depicting that black means zero and white one.
>>
>> To help you with my code you will find attached the three items I have
>> discussed here, the matrix, the x and the y labels.
>>
>> Could you please help me with that?
>>
>
Re: [R] Plot a Matrix as an Image with ggplot
You are good!
Many thanks
Alex
From: Dennis Murphy
Cc: R help
Sent: Saturday, February 16, 2013 3:11 AM
Subject: Re: [R] Plot a Matrix as an Image with ggplot
Hi:
See if the following works for you:
library(reshape2)
library(ggplot2)
tdm <- melt(testData)
ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
labs(x = "MHz", y = "Threshold", fill = "Value") +
geom_raster() +
scale_fill_manual(breaks = levels(factor(tdm$value)),
values = c("white", "black")) +
theme(plot.background = element_rect(fill = "grey90"),
legend.background = element_rect(fill = "grey90")) +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0))
It was necessary to create a different plot background color because
you wanted to remove the padding, which blended much of the plot
boundary with the original white background, and had to do something
similar to the legend background so that you could see the white
legend box.
The expand = c(0, 0) argument to the two scale functions answers your
second question, and converting value from numeric to vector answers
the first.
Dennis
> You were right,
> the following two work
>
>
> testData<-matrix(data=round(runif(25)),nrow=5,ncol=5,dimnames=list(Var1=c(1:5),Var2=c(1:5)))
> ggplot(melt(testData), aes(Var2,Var1, fill=value))+xlab("MHz") +
>ylab("Threshold") + geom_raster() + scale_fill_gradient(low="#FF",
>high="#00")
>
>
>
> still there are two minor unresolved issues.
>
> a. How to reduce the extra space in the plot between legend and the tiles?
> b. How to make color bar depict only the two values of the game, 0 and 1 ,
> instead of 0, 0.25, 0.50, 0.75, 1?
>
> I would like to thank you in advanec for your help
> Regards
> Alex
>
>
> - Original Message -
> From: Dennis Murphy
> Cc:
> Sent: Friday, February 15, 2013 10:19 AM
> Subject: Re: [R] Plot a Matrix as an Image with ggplot
>
> Your aesthetic is fill, not color. Change scale_color_gradient to
> scale_fill_gradient and you'll get what you expect.
>
> D.
>
>>
>>
>> Hi,
>> thanks
>> I changed slightly the code to be reproducible from everyone . I have tried
>> ggplot
>> but I need a bit of help to tweak it a bit
>>
>> So you can run the following in your computer
>>
>>
>>testData<-matrix(data=round(runif(25)),nrow=5,ncol=5,dimnames=list(1:5,1:5))
>>
>> p<-ggplot(melt(testData), aes(Var2,Var1, fill=value))+xlab("MHz") +
>>ylab("Threshold") + geom_raster()
>>
>>
>> p
>>
>>
>> What I want to improve is
>> a: make the colorbar so only two specific colors lets say black and white
>> and only two values 0 and 1 and somewhere the string as title of the color
>> bar "text" to appear.
>>
>> I have tried something like
>> p+ scale_color_gradient(low="red",high="blue")
>> Fehler in unit(tic_pos.c, "mm") : 'x' and 'units' must have length > 0
>>
>>
>> ggplot(melt(testData), aes(Var2,Var1, fill=value,colorbin=2))+xlab("MHz") +
>> ylab("Threshold") + geom_raster()
>> but this did not affect colorbar entries.
>>
>>
>> b. reduce/remove the grayish border that appears between the legend and the
>> image plot
>>
>> Could you please help me with these two?
>>
>> Regards
>> Alex
>>
>>
>>
>>
>> From: John Kane
>> Sent: Thursday, February 14, 2013 4:35 PM
>> Subject: RE: [R] Plot a Matrix as an Image with ggplot
>>
>> The R-help list is rather picky about what attached. None of your
>> attachments arrived.
>>
>> The str() info is useful but please supply some sample data
>>
>> The easiest way to supply data is to use the dput() function. Example with
>> your file named "testfile":
>> dput(testfile)
>>
>> Then copy the output and paste into your email. For large data sets, you
>> can just supply a representative sample. Usually,
>> dput(head(testfile, 100)) will be sufficient.
>>
>> How are you writing the code/or what format is the original email. Your code
>> in the body of the text is badly messed up -- you probably need to post only
>> in text. HTML etc is automatically dropped.
>>
>>
>>
>>
>> John Kane
>> Kingston ON Canada
>>
>>
>>> -Original Message-
>>> Sent: Thu, 14 Feb 2013 07:15:05 -0800 (PST)
>>> To: r-help@r-project.org
>>> Subject: [R] Plot a Matrix as an Image with ggplot
>>>
>>> Dear all,
>>> I am trying to plot a matrix I have as an image
>>>
>>> str(matrixToPlot)
>>> num [1:21, 1:66] 0 0 0 0 0 0 0 0 0 0 .
>>>
>>>
>>> that contains only 0s and 1s,
>>>
>>>
>>> where the xlabel will be Labeled as
>>> str(xLabel)
>>> num [1:66] 1e+09 1e+09 1e+09 1e+09 1e+09 ...
>>>
>>> and the yLabels will be labeled as
>>> str(yLabel)
>>> num [1:21] -88 -87 -86 -85 -84 -83 -82 -81 -80 -79 ...
>>>
>>>
>>> I have found on the internet that I can do something like that with
>>> ggplot2.
>>> For example
>>> you can run the following
>>>
>>>
>>>
>>> library(reshape2)library(ggplot2)m
>>> =matrix(rnorm(20),5)ggplot(melt(m),aes(Var1
[R] ggplot2 customizing a plot
Dear all, I want some help improve my ggplot as following: Make the plottable area with grid, so is easy one to see where each box refers to x and y values. Add a color bar but with fixed values, that I want to specify. How I can do those two? Before is some code what I have tried so far. Regards Alex DataToPlot<-matrix(data=seq(1:9),nrow=3,dimnames=list(seq(1,3),seq(4,6))) require(reshape) require(ggplot2) require(raster) tdm <- melt(DataToPlot) p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) + labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") + geom_raster() + scale_fill_discrete() [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2 customizing a plot
Hello Ista,
I would like to thank you for your reply!
Your code is indeed an improvement.
This is my code now and there are two things still missing.
DataToPlot<-matrix(data=runif(9),nrow=3,dimnames=list(seq(1,3),seq(4,6)))
require(reshape)
require(ggplot2)
require(raster)
cols<-colours()
cols<-cols[1:(nrow(DataToPlot)*ncol(DataToPlot))]
tdm<-melt(DataToPlot)
p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
geom_raster(alpha=.5) +
scale_fill_manual(values=cols,breaks=c("1","0.9","0.8","0.7","0.6","0.5","0.4","0.3","0.2","0.1","0"))
My values go from 0 to 1 (might go slightly over the limits) and I want to
specify one color for the following ranges, 1, 0.9,0.8,.,0 . That means
that values between the intervals will be "categorized" based on their closest
match.
What I also tried is to create a gray scale pallete as this is printer friendly
so the values from 0.9 to 0.1 get a greyish tone but I failed to do that with
colours.
Could you please help me have both
a. A fixed scale where values are categorized there
and
b. a greyish pallete for the categories?
I would like to thank you in advance for your reply
Regards
Alex
From: Ista Zahn
Cc: R help
Sent: Wednesday, February 20, 2013 4:54 PM
Subject: Re: [R] ggplot2 customizing a plot
Hi,
> Dear all,
> I want some help improve my ggplot as following:
> Make the plottable area with grid, so is easy one to see where each box
> refers to x and y values.
I don't think you can easily move the grid to the front, but you can
make the tiles transparent so the grid can be seen through them by
setting the alpha < 1.
> Add a color bar but with fixed values, that I want to specify.
use scale_fill_manual
> How I can do those two?
p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
geom_raster(alpha=.5) +
scale_fill_manual(values=c("red", "blue", "green", "purple",
"orange", "pink", "tan", "violet", "yellow"))
Best,
Ista
>
> Before is some code what I have tried so far.
>
> Regards
> Alex
>
>
> DataToPlot<-matrix(data=seq(1:9),nrow=3,dimnames=list(seq(1,3),seq(4,6)))
>
>
> require(reshape)
> require(ggplot2)
> require(raster)
>
> tdm <- melt(DataToPlot)
>
>
>
> p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
> labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
> geom_raster() +
> scale_fill_discrete()
>
> [[alternative HTML version deleted]]
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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Re: [R] ggplot2 customizing a plot
Let me try to explain this then.
I have numbers like
DataToPlot
4 5 6
1 0.4454995 0.4462009 0.4286807
2 0.3761550 0.5423205 0.6500785
3 0.3779496 0.4671437 0.1799601
and I want to have a color legend with each color that is mapped to the
following printed values
"1","0.9","0.8","0.7","0.6","0.5","0.4","0.3","0.2","0.1","0"
For making life easier I have rounded up my number to one digit so
matrixToPlot<-matrix(data=round(matrixToPlot,digits=1),nrow=nrow(matrixToPlot),ncol=ncol(matrixToPlot),dimnames=list(yLabel,
xLabel))
matrixToPlot
1 2 3
1 0.4 0.4 0.4
2 0.4 0.5 0.7
3 0.4 0.5 0.2
that indeed helps and makes the legend bar print the right values BUT there are
not all the ranges printed there. So for the given example data set only the
0.2, 0.3,0.4,0.5,0.6,0.7 are printed with the 0,0.1,0.8,0.9,1 missing.
Also I need some help to specify specific color to each of the given
categories/ranges in the colorbar
my code now looks like
library(reshape2)
library(ggplot2)
matrixToPlot<-matrix(data=runif(9),nrow=3,dimnames=list(seq(1,3),seq(4,6)))
matrixToPlot<-matrix(data=round(matrixToPlot,digits=1),nrow=nrow(matrixToPlot),ncol=ncol(matrixToPlot),dimnames=list(seq(1:3),
seq(4:6)))
tdm <- melt(matrixToPlot)
cols<-colours()
cols<-cols[1:(nrow(matrixToPlot)*ncol(matrixToPlot))]
p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
geom_raster(alpha=.5) +
scale_fill_discrete(h.start=1) +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0))
Regards
Alex
From: Ista Zahn
Cc: R help
Sent: Thursday, February 21, 2013 2:15 PM
Subject: Re: [R] ggplot2 customizing a plot
Hi,
> Hello Ista,
[[elided Yahoo spam]]
>
> Your code is indeed an improvement.
>
> This is my code now and there are two things still missing.
>
> DataToPlot<-matrix(data=runif(9),nrow=3,dimnames=list(seq(1,3),seq(4,6)))
> require(reshape)
> require(ggplot2)
> require(raster)
>
>
> cols<-colours()
> cols<-cols[1:(nrow(DataToPlot)*ncol(DataToPlot))]
> tdm<-melt(DataToPlot)
>
> p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
> labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
> geom_raster(alpha=.5) +
> scale_fill_manual(values=cols,breaks=c("1","0.9","0.8","0.7","0.6","0.5","0.4","0.3","0.2","0.1","0"))
value looks like numbers, why are you converting it to a factor?
why this is how I found I should solve the problem..
>
> My values go from 0 to 1 (might go slightly over the limits) and I want to
> specify one color for the following ranges, 1, 0.9,0.8,.,0 . That means
> that values between the intervals will be "categorized" based on their
> closest match.
I don't really understand this.
> What I also tried is to create a gray scale pallete as this is printer
> friendly so the values from 0.9 to 0.1 get a greyish tone but I failed to do
> that with colours.
How about
ggplot(tdm, aes(x = Var2, y = Var1, fill = value)) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
geom_raster(alpha=.5) +
scale_fill_gradient(low="gray90", high="black") +
theme_bw()
>
> Could you please help me have both
> a. A fixed scale where values are categorized there
> and
Value is numeric, and trying to treat it like a category is only going
to make things difficult.
> b. a greyish pallete for the categories?
Use scale_fill_gradient with different shades of gray as the low and
high values, as shown above.
Best,
Ista
>
> I would like to thank you in advance for your reply
>
> Regards
> Alex
>
>
>
> From: Ista Zahn
> Cc: R help
> Sent: Wednesday, February 20, 2013 4:54 PM
> Subject: Re: [R] ggplot2 customizing a plot
>
> Hi,
>
>> Dear all,
>> I want some help improve my ggplot as following:
>> Make the plottable area with grid, so is easy one to see where each box
>> refers to x and y values.
>
> I don't think you can easily move the grid to the front, but you can
> make the tiles transparent so the grid can be seen through them by
> setting the alpha < 1.
>
>> Add a color bar but with fixed values, that I want to specify.
>
> use scale_fill_manual
>
>> How I can do those two?
>
> p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) +
> labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
> geom_raster(alpha=.5) +
> scale_fill_manual(values=c("red", "blue", "green", "purple",
> "orange", "pink", "tan", "violet", "yellow"))
>
>
> Best,
> Ista
>
>
>>
>> Before is some code what I have tried so far.
>>
>> Regards
>> Alex
>>
>>
>> DataToPlot<-matrix(data=seq(1:9),nrow=3,dimnames=list(seq(1,3),seq(4,6)))
>>
>>
>> require(reshape)
>> require(ggplot2)
>> require(raster)
>>
>> tdm <- melt(DataToPlot)
>>
>>
>>
>> p<- ggplot(tdm, aes(x = Var2, y = Va
[R] ggplot2 Increase font size
Dear all, I am using the code as below tdm <- melt(matrixToPlot) p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) + labs(x = "Mz", y = "T", fill = "D") + geom_raster(alpha=1) + scale_fill_discrete(h.start=1) + scale_x_continuous(expand = c(0, 0)) + scale_y_continuous(expand = c(0, 0)) to plot an two dimensional image . I would like to ask your help to replace the gray border with white color and increase the font size of x and y axis as wells as the legend of the color bar. Could you please give me the function names to use? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] parallel execution in R
Dear all,
I have a piece of code that I want to run in parallel (I am working in system
of 16 cores)
foreach (i=(seq(-93,-73,length.out=21))) %dopar%
{
threshold<-i
print(i)
do_analysis1(i,path)
do_analysis2(i,path)
do_something_else_analysis1(i,path)
something_else_now(i,path)
}
as you can see I have already tried to make this run in parallel, meaning for
every i value each of the 16 processor shoule take a block of the body such
as:
threshold<-i
print(i)
do_analysis1(i,path)
do_analysis2(i,path)
do_something_else_analysis1(i,path)
something_else_now(i,,path)
and execute it . Unfortunately this does not work and oonly one processor looks
utilized.
Alternatively, mclapply have worked well in the past, but in this case I am not
sure how to convert the serial execution of the body of the loop to a list that
would be compatible with the mclapply.
I would like to thank you in advance for your help
Regards
Alex
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Re: [R] ggplot2 Increase font size
Hi, I am not quite sure what you meanÎ. I give again reproducible code: require(ggplot2) require(reshape) DataToPlot<-matrix(data=rnorm(9),nrow=3,dimnames=list(seq(1,3),seq(4,6))) tdm<-melt(DataToPlot) p<- ggplot(tdm, aes(x = X2, y = X1, fill = factor(value))) +                  labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +                  geom_raster(alpha=1) +                  scale_fill_discrete(h.start=1,breaks=c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1),labels=c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1)) +                  scale_x_continuous(expand = c(0, 0)) +                  scale_y_continuous(expand = c(0, 0)) DataToPlot in my case contains something like that: DataToPlot           4         5         6 1 -0.4135124 0.4643110 -0.7530622 2 0.8827643 -0.1702428 0.4607671 3 0.7942167 -1.2450487 -0.9380290 what I would like to have is to have one specific color for each of the following cases c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1) so for example the -0.4 would be categorized under the 0 category the 0.46 would be categorized under the 0.5 category the -0.75 would be categorized under the 0 category the 0.88 would be categorized under the 0.9 category right now the code I gave prints no color bar. If I change it like that. p<- ggplot(tdm, aes(x = X2, y = X1, fill = factor(value))) +                  labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +                  geom_raster(alpha=1) +                  scale_fill_discrete(h.start=1) +                  scale_x_continuous(expand = c(0, 0)) +                  scale_y_continuous(expand = c(0, 0)) colorbar comes back but as you will see the defined values is not how I want to categorize my data. Could you please help me find what I do not understand here? I would like to thank you in advance for your reply Regards Alex From: Ista Zahn Cc: R help Sent: Monday, February 25, 2013 3:54 PM Subject: Re: [R] ggplot2 Increase font size Hi Alex, See ?theme Best, Ista > > Dear all, > I am using the code as below > tdm <- melt(matrixToPlot) >  p<- ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value))) + >         labs(x = "Mz", y = "T", fill = "D") + >         geom_raster(alpha=1) + >         scale_fill_discrete(h.start=1) + >         scale_x_continuous(expand = c(0, 0)) + >         scale_y_continuous(expand = c(0, 0)) > > to plot an two dimensional image . > > I would like to ask your help to replace the gray border with white color > and increase the font size of x and y axis as wells as the legend of the > color bar. Could you please give me the function names to use? > > Regards > Alex >    [[alternative HTML version deleted]] > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lm and Formula tutorial
Dear all, I was reading last night the lm and the Formula manual page, and 'I have to admit that I had tough time to understand their syntax. Is there a simpler guide for the dummies like me to start with? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] All unique combinations
Dear all, I would like to have all unique combinations in the following matrix TimeIndex<- rbind (c(1,"Week_of_21_07-29_03"), c(2,"Thursday_21_03"), c(3,"Friday_22_03"), c(4,"Saturday_23_03"), c(5,"Sunday_24_03"), c(6,"Monday_25_03"), c(7,"Tuesday_26_03"), c(8,"Wednesday_27_03"), c(9,"Thursday_28_03"), c(10,"Friday_ 29_03")) As have a plotting code that takes two input arguments plot_two_time_frames(time1,LabelTime1,time2,LabelTime2) these inputs are time1: integer from TimeIndex. TimeIndex[i,1] LabelTime1: Labels from TimeIndex[i,2] time2: integer from TimeIndex, TimeIndex[i,1] LabelTime2: Labels from TimeIndex[i,2] the times 1 and time 2 should be unique combinations and never the same number. I understand that my explanation was not the best possible. But I would like to thank you in advance for your support Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot Raster
Dear all,
I am trying to plot an image so I am trying this through raster layer.
You can copy paste the following
require('raster')
Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
rasterData<-raster(Data)
lengthOut<-5
xAxisFrequencies<-seq(800,900,length.out=lengthOut)
plot(rasterData, ylab="",xaxt="n",yaxt="n")
axis(1, at=seq(0,1,length.out=lengthOut), xAxisFrequencies, col.axis = "blue")
What I want is to add a customized x label, but the last line I gave above
axis(1,at.) does not work.
It would be also nice to add a bit of legend also under the color bar to
explain there what are the values.
Could you please help me with those two ?
Regards
Alex
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[R] Plot Matrix with Data
Hi , I would like to use ggplot2 to plot a matrix as an image. You can copy paste the following Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30) lengthOut<-5 Lengths<- 15 library(reshape2) library(ggplot2) tdm <- melt(Data) ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value)),levels=seq(0,1,by=0.1)) + labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") + geom_raster(alpha=1) + scale_fill_discrete(h.start=1,breaks=c(20,30,40,50,60,70,80,90),labels=c(20,30,40,50,60,70,80,90),fill="red") + scale_x_continuous(expand = c(0, 0)) + scale_y_continuous(expand = c(0, 0)) # End of code part What I wanted to do is to print the values below 20, with an x color between 20-30, with a y clor between 30-40, with a z color and so on I would not care now for the color palette. Any would do. Could you please help me with that? Regards A [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot Matrix with Data
Hi,
Thanks for the answer.
It looks like that the mutate what I was missing so long..
That's my current attempt
Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
Lengths<- 15
library(reshape2)
library(ggplot2)
require('plyr')
tdm <- melt(Data)
tdm <- mutate(tdm, col = cut(value, seq(15, 90, by=5), labels = seq(20, 90,
by=5)) )
test<-colorRampPalette(c("orange", "red"),space="Lab",interpolate="linear")
ggplot(tdm, aes(x = Var1, y = Var2, fill = col)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle")
Two things.
a. I have tried to make a color pallete that goes from orange through all the
orangish and redish hues to the red but as you can see that failed.
b. One think is that the colors visible in the legend in the right part are not
always visible. If for example there is a 20 value the 20 will appear and if
there is not but there is a 25 the bar will start from the 25 onwards.
I would like to thank everyone for their on going support
Regards
Alex
From: Dennis Murphy
Sent: Monday, March 25, 2013 9:17 PM
Subject: Re: [R] Plot Matrix with Data
Hi Alex:
Here are a couple of raw examples, one with a 'discretized' color
gradient and another with ordered factor levels in groups of length
10. Starting with tdm,
library(plyr)
library(ggplot2)
tdm <- mutate(tdm, col = cut(value, seq(0, 140, by = 10),
labels = seq(10, 140, by = 10)) )
## or equivalently,
# tdm$col <- cut(col$value, seq(0, 140, by = 10),
# labels = seq(10, 140, by = 10)) )
# The numeric factor levels are 1 - 14, so if we convert col to numeric,
# should multiply the values by 10. This plot uses a continuous
# color gradient whose midpoint is near the median value.
ggplot(tdm, aes(x = Var1, y = Var2, fill = 10 * as.numeric(col))) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
scale_fill_gradient2(low = "green", mid = "orange",
high = "yellow", midpoint = 85) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle")
# This one uses col as a factor. The labels are the correct
# upper limits of each value bin. Obviously you want to use
# a smarter mechanism for assigning colors than the
# default. The colorspace package or the colorRampPalette()
# function might be useful here.
ggplot(tdm, aes(x = Var1, y = Var2, fill = col)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle")
The values on the x-y axes are the row and column numbers of the
reshaped matrix. You may need to map them to values of the MHz and
Threshold values; I doubt that [0, 1] is the range of each variable
(as you have in another post using base graphics).
Dennis
> Hi ,
> I would like to use ggplot2 to plot a matrix as an image.
>
> You can copy paste the following
>
>
> Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
>
> lengthOut<-5
>
>
>
> Lengths<- 15
> library(reshape2)
> library(ggplot2)
> tdm <- melt(Data)
>
>
>
>
>
> ggplot(tdm, aes(x = Var2, y = Var1, fill =
> factor(value)),levels=seq(0,1,by=0.1)) +
> labs(x = "MHz", y = "Threshold", fill = "Duty Cycle") +
> geom_raster(alpha=1) +
>
>scale_fill_discrete(h.start=1,breaks=c(20,30,40,50,60,70,80,90),labels=c(20,30,40,50,60,70,80,90),fill="red")
> +
> scale_x_continuous(expand = c(0, 0)) +
> scale_y_continuous(expand = c(0, 0))
>
>
>
>
>
>
> # End of code part
>
>
>
> What I wanted to do is to print the values
>
> below 20, with an x color
> between 20-30, with a y clor
> between 30-40, with a z color
> and so on
>
> I would not care now for the color palette.
> Any would do.
> Could you please help me with that?
>
> Regards
> A
>
> [[alternative HTML version deleted]]
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Increase font size in plots
Hi
I am using violin plots (type of boxplots) and I am trying to increase the
font size in the plots.
It looks like that the violin plots do not work as "normal" plots as the cex
parameters are ignored.
You can have a loot at the code below
require('vioplot')
data1<-rnorm(100)
data2<-rnorm(10)
data3<-rnorm(1000)
vioplot(data1,data2,data3,cex.lab=2,cex.axis=2)
Regards
Alex
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot Matrix with Data
Hi,
it was very kind of you to help me again.
Your code works, and the reason I was using the limits as that is that my
dataset is slightly different than the one I used for showing the problem here.
More specifically this is my dataset
Browse[1]> str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
Browse[1]> (density(keep))
Call:
density.default(x = keep)
Data: keep (525565085 obs.); Bandwidth 'bw' = 0.1695
x y
Min. :-106.94 Min. :0.000
1st Qu.: -87.13 1st Qu.:0.0003593
Median : -67.31 Median :0.0050385
Mean : -67.31 Mean :0.0126038
3rd Qu.: -47.49 3rd Qu.:0.0084189
Max. : -27.67 Max. :0.1792683
and as you can see is rather large... that makes it a marathon to print it.
Would it be possible to average some of the arrows in a column by column basis
so to turn the
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
to something like
num [1:1000, 1:3415] . How can I do that efficiently in R?
I would like to thank everyone in advance.
Regards
Alex
From: Dennis Murphy
Sent: Tuesday, March 26, 2013 10:43 AM
Subject: Re: [R] Plot Matrix with Data
Hi:
I don't understand why you're truncating the duty cycle at 90 when the
data range from 15 to 150 - the effect of doing that is the set of
grey squares in the heatmap corresponding to missing values.
The reason your color scale didn't work is because you never used it
in the ggplot() call, but I think that it would have been difficult to
incorporate it anyway because ggplot2 has its own conventions. It
appears that what you want is ggplot2::scale_fill_continuous().
I chucked the color factor and tried the following instead. You can
edit it however you want, and you can extend the length/width of the
color bar - see the help page of guide_colourbar(), in particular the
barheight argument.
http://docs.ggplot2.org/current/guide_colourbar.html
My shot at your graphic:
ggplot(tdm, aes(x = Var1, y = Var2, fill = value)) +
geom_raster() +
scale_x_continuous(expand = c(0, 0)) +
scale_y_continuous(expand = c(0, 0)) +
scale_fill_continuous(limits = c(15, 150),
breaks = seq(15, 150, by = 15),
low = "orange", high = "red",
guide = guide_colourbar(nbin = 28)) +
labs(x = "MHz", y = "Threshold", fill = "Duty Cycle")
This color bar has 28 bins, 5 units apart, but the labels are 15 apart
since otherwise they overwrite each other. You can reduce the crowding
by increasing the height of the color bar as indicated earlier. This
appears to be closer to your intent with a lot less bother.
Dennis
> Hi,
> Thanks for the answer.
> It looks like that the mutate what I was missing so long..
>
> That's my current attempt
>
> Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
>
> Lengths<- 15
> library(reshape2)
> library(ggplot2)
> require('plyr')
> tdm <- melt(Data)
> tdm <- mutate(tdm, col = cut(value, seq(15, 90, by=5), labels = seq(20, 90,
> by=5)) )
> test<-colorRampPalette(c("orange", "red"),space="Lab",interpolate="linear")
>
> ggplot(tdm, aes(x = Var1, y = Var2, fill = col)) +
> geom_raster() +
> scale_x_continuous(expand = c(0, 0)) +
> scale_y_continuous(expand = c(0, 0)) +
> labs(x = "MHz", y = "Threshold", fill = "Duty Cycle")
>
>
> Two things.
> a. I have tried to make a color pallete that goes from orange through all
> the orangish and redish hues to the red but as you can see that failed.
> b. One think is that the colors visible in the legend in the right part are
> not always visible. If for example there is a 20 value the 20 will appear
> and if there is not but there is a 25 the bar will start from the 25
> onwards.
>
> I would like to thank everyone for their on going support
>
> Regards
> Alex
>
> From: Dennis Murphy
> Sent: Monday, March 25, 2013 9:17 PM
> Subject: Re: [R] Plot Matrix with Data
>
> Hi Alex:
>
> Here are a couple of raw examples, one with a 'discretized' color
> gradient and another with ordered factor levels in groups of length
> 10. Starting with tdm,
>
> library(plyr)
> library(ggplot2)
> tdm <- mutate(tdm, col = cut(value, seq(0, 140, by = 10),
> labels = seq(10, 140, by = 10)) )
> ## or equivalently,
> # tdm$col <- cut(col$value, seq(0, 140, by = 10),
> # labels = seq(10, 140, by = 10)) )
>
> # The numeric factor levels are 1 - 14, so if we convert col to numeric,
> # should multiply the values by 10. This plot uses a continuous
> # color gradient whose midpoint is near the median value.
> ggplot(tdm, aes(x = Var1, y = Var2, fill = 10 * as.numeric(col))) +
> geom_raster() +
> scale_x_continuous(expand = c(0, 0)) +
> scale_y_continuous(expand = c(0, 0)) +
> scale_fill_gradient2(low = "green", mid = "orange",
> high = "yellow", midpoint =
[R] Averaging Out many rows from a column AND funtion to string
Dear all,
1) I have a very large matrix of
str(keep)
num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
that I would like to reduce its size to something like
str(keep)
num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
or anything similar in size as this is a matrix that needs plotting (so is ok
if it is 1000 row, 995, or 1123)
I think what I need here is a way of selecting multiple rows and averaging per
column (notice that the column number should stay the same)
b. I would like to be able to convert strings that are function names to real
function calls. For example I have something like
LogFunction<- function(){}
FunctionIndex<- rbind (c(1,"LogFunction"),
c(2,"TakeFunction")
)
print(sprintf('Using the function %s',FunctionIndex[1,1]))
# call the FunctionIndex[1,1] somehow
I would like to thank you in advance for your help
Regards
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Alaios
> Sent: Wednesday, March 27, 2013 9:13 AM
> To: R help
> Subject: [R] Averaging Out many rows from a column AND funtion to
> string
>
> Dear all,
> 1) I have a very large matrix of
> str(keep)
> num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
> that I would like to reduce its size to something like
>
> str(keep)
> num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
> similar in size as this is a matrix that needs plotting (so is ok if it
> is 1000 row, 995, or 1123)
>
> I think what I need here is a way of selecting multiple rows and
> averaging per column (notice that the column number should stay the
> same)
Make an index variable and aggregate values according it
Something like
idx<-cut(1:153899, 153)
keep.ag<-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
2) I want to have a string that can be used also for calling a function with
the same name. Think of using "mean" to call mean. I need to have R interpret
the string in different ways.
Regards
Alex
>
> b. I would like to be able to convert strings that are function names
> to real function calls. For example I have something like
>
> LogFunction<- function(){}
> FunctionIndex<- rbind (c(1,"LogFunction"),
> c(2,"TakeFunction")
> )
> print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
> FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf <-function(fun="mean", arg) eval(call(fun, arg))
Regards
Petr
>
>
>
> I would like to thank you in advance for your help
>
> Regards
> Alex
>
> [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
see inline
From: PIKAL Petr
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Alaios
> Sent: Wednesday, March 27, 2013 9:13 AM
> To: R help
> Subject: [R] Averaging Out many rows from a column AND funtion to
> string
>
> Dear all,
> 1) I have a very large matrix of
> str(keep)
> Â num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
> that I would like to reduce its size to something like
>
> str(keep)
> Â num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
> similar in size as this is a matrix that needs plotting (so is ok if it
> is 1000 row, 995, or 1123)
>
> I think what I need here is a way of selecting multiple rows and
> averaging per column (notice that the column number should stay the
> same)
Make an index variable and aggregate values according it
Something like
idx<-cut(1:153899, 153)
keep.ag<-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx<-cut(1:30, 10)
keep.ag<-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
2) I want to have a string that can be used also for calling a function with
the same name. Think of using "mean" to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction<- function(){}
FunctionIndex<- rbind (c(1,"LogFunction"),
              c(2,"TakeFunction"))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
>
> b. I would like to be able to convert strings that are function names
> to real function calls. For example I have something like
>
> LogFunction<- function(){}
> FunctionIndex<- rbind (c(1,"LogFunction"),
> Â Â Â Â Â Â Â Â Â Â Â Â c(2,"TakeFunction")
> Â Â Â Â Â Â Â Â Â Â Â )
> print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
> FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf <-function(fun="mean", arg) eval(call(fun, arg))
Regards
Petr
>
>
>
> I would like to thank you in advance for your help
>
> Regards
> Alex
>
> Â Â Â [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
I have fixed it like that:
keep<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo<-500
idx<-cut(1:nrow(keep), ShrinkTo)
keep.ag<-aggregate(keep, list(idx), mean)
PlotMe<-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
From: PIKAL Petr
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
> str(as.matrix(keep.ag[,-1]) ) Â # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, "dimnames")=List of 2
 ..$ : NULL
 ..$ : chr [1:30] "V1" "V2" "V3" "V4" ...
Â
Please read and follow what was recommended.
Â
quote
Â
use as.matrix(data.frame) on numeric part
                                       Â
                 ^^^
Â
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
Â
And please can you explain how mean(rnorm(whatever)) shall be integer?
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
Â
From:PIKAL Petr
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Alaios
> Sent: Wednesday, March 27, 2013 9:13 AM
> To: R help
> Subject: [R] Averaging Out many rows from a column AND funtion to
> string
>
> Dear all,
> 1) I have a very large matrix of
> str(keep)
> Â num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
> that I would like to reduce its size to something like
>
> str(keep)
> Â num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
> similar in size as this is a matrix that needs plotting (so is ok if it
> is 1000 row, 995, or 1123)
>
> I think what I need here is a way of selecting multiple rows and
> averaging per column (notice that the column number should stay the
> same)
Make an index variable and aggregate values according it
Something like
idx<-cut(1:153899, 153)
keep.ag<-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
Â
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx<-cut(1:30, 10)
keep.ag<-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
Â
Â
Â
Â
Â
Â
Â
2) I want to have a string that can be used also for calling a function with
the same name. Think of using "mean" to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction<- function(){}
FunctionIndex<- rbind (c(1,"LogFunction"),
              c(2,"TakeFunction"))
print(sprintf('Using the function %s',FunctionIndex[1,1]))
This does not contain much clue.
Regards
Petr
PS. Do not use HTML mail messages.
Regards
Alex
>
> b. I would like to be able to convert strings that are function names
> to real function calls. For example I have something like
>
> LogFunction<- function(){}
> FunctionIndex<- rbind (c(1,"LogFunction"),
> Â Â Â Â Â Â Â Â Â Â Â Â c(2,"TakeFunction")
> Â Â Â Â Â Â Â Â Â Â Â )
> print(sprintf('Using the function %s',FunctionIndex[1,1])) # call the
> FunctionIndex[1,1] somehow
I am not sure if I understand correctly.
Is this what you want?
myf <-function(fun="mean", arg) eval(call(fun, arg))
Regards
Petr
>
>
>
> I would like to thank you in advance for your help
>
> Regards
> Alex
>
> Â Â Â [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Controlling Raster plots
Hi,
I have a few raster layers and I would like to customized their x and y axis.
I have tried already something like:
require('raster')
keep<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
xlab<-seq(100e6,200e6,length.out=5)
test<-raster(keep)
plot(test,ylab="",xaxt="n",yaxt="n")
axis(1, at=seq(0,1,length.out=5), xlab)
that normally would work in any other plot function.
Could you please help me add the xaxis ?
I would like to thank you in advance for your reply
Regards
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging Out many rows from a column AND funtion to string
Well my true matrix is
of
num [1:153899, 1:3415]
that I want to convert to something like
num [1:1000, 1:3415] (keeping column number the same).
I only give subsets here to allow others to run the code at their computer
Thanks a lot
Regards
Alex
From: PIKAL Petr
Sent: Wednesday, March 27, 2013 4:45 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
> system.time({
+ keep<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
+ ShrinkTo<-500
+ idx<-cut(1:nrow(keep), ShrinkTo)
+ keep.ag<-aggregate(keep, list(idx), mean)
+ PlotMe<-as.matrix(keep.ag[,-1])
+ })
  user system elapsed
  29.23   0.14  29.37
> Â
Â
It takes 30 seconds when using 30 rows. It is not enough time to get a cup
of tee, which I do not consider ages. Maybe split lapply approach or
data.matrix or  **ply could be quicker but I do not consider worth spending
hours to elaborate some solution which will spare 20 sec computing time.
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 4:12 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
I have fixed it like that:
keep<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
ShrinkTo<-500
idx<-cut(1:nrow(keep), ShrinkTo)
keep.ag<-aggregate(keep, list(idx), mean)
PlotMe<-data.matrix(keep.ag[2:ShrinkTo])
The only problem is that takes ages to finish. Would it be possible to
convert it to something like lapply?
Regards
Alex
Â
Â
From:PIKAL Petr
Sent: Wednesday, March 27, 2013 4:02 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
Â
> str(as.matrix(keep.ag[,-1]) ) Â # does look like numeric
num [1:10, 1:30] 75.1 93 79 81.7 76.3 ...
- attr(*, "dimnames")=List of 2
 ..$ : NULL
 ..$ : chr [1:30] "V1" "V2" "V3" "V4" ...
Â
Please read and follow what was recommended.
Â
quote
Â
use as.matrix(data.frame) on numeric part
                                       Â
                 ^^^
Â
aggregate produces data frame with its first column being your idx variable,
which is factor. Trying to convert it whole to matrix results in character
matrix. You need to exclude first column from conversion
Â
And please can you explain how mean(rnorm(whatever)) shall be integer?
Â
Regards
Petr
Â
Sent: Wednesday, March 27, 2013 3:01 PM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
Â
From:PIKAL Petr
Sent: Wednesday, March 27, 2013 1:50 PM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Â
Hi
Â
Sent: Wednesday, March 27, 2013 11:46 AM
To: PIKAL Petr; R help
Subject: Re: [R] Averaging Out many rows from a column AND funtion to string
Â
see inline
Â
From:PIKAL Petr
Sent: Wednesday, March 27, 2013 11:24 AM
Subject: RE: [R] Averaging Out many rows from a column AND funtion to string
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Alaios
> Sent: Wednesday, March 27, 2013 9:13 AM
> To: R help
> Subject: [R] Averaging Out many rows from a column AND funtion to
> string
>
> Dear all,
> 1) I have a very large matrix of
> str(keep)
> Â num [1:153899, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ...
> that I would like to reduce its size to something like
>
> str(keep)
> Â num [1:1000, 1:3415] -98.6 -95.8 -96.4 -95.8 -98 ... or anything
> similar in size as this is a matrix that needs plotting (so is ok if it
> is 1000 row, 995, or 1123)
>
> I think what I need here is a way of selecting multiple rows and
> averaging per column (notice that the column number should stay the
> same)
Make an index variable and aggregate values according it
Something like
idx<-cut(1:153899, 153)
keep.ag<-aggregate(keep, list(idx), mean)
1) Thanks that returned a data frame.. How I can have a matrix at the end?
use as.matrix(data.frame) on numeric part
Â
a bit of my code that you can re run. I convert a 30,30 matrix to a 10,30.Â
but it looks at the end that I do not get the Data part of the data.fram
correctly:
Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30)
idx<-cut(1:30, 10)
keep.ag<-aggregate(Data, list(idx), mean)
str(as.matrix(keep.ag) ) # it does not look like integers
Â
Â
Â
Â
Â
Â
Â
2) I want to have a string that can be used also for calling a function with
the same name. Think of using "mean" to call mean. I need to have R interpret
the string in different ways.
To call âmeanâ of what? I am sure I do not understand what is your
intention.
LogFunction<- function(){}
FunctionIndex<
[R] Sort or Permutate
Dear all, I am having a struct that contains on the first column file names and on the second column a number which is a "rating" of the file on first column A small subset looks like that small [,1] [1,] "/storage/storage0/59Off.Rtable" [2,] "/storage/storage0/5912On.Rtable" [3,] "/storage/storage0/5912314ff.Rtable" [,2] [1,] "0.220386301811093" [2,] "0.405237035258638" [3,] "0.288659374128626" I want based on the values of column two to rearrange table elements by keeping though the row elements together. For example if I order those in ascending order should look like small [,1] [1,] "/storage/storage0/59Off.Rtable" [2,] "/storage/storage0/5912314ff.Rtable" [3,] " /storage/storage0/5912On.Rtable" [,2] [1,] "0.220386301811093" [2,] "0.288659374128626" [3,] "0.405237035258638" I have tried with sort, sort.list and order but I have failed . I do not quite undestand how one can complete this task in R. Could you please spend some time helping me? Regards A [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Select 5 identical numbers
Dear all, I would like to select from a vector of 10 elements, 5 of those that are identical. How can I do that in R? I guess one way would be to taking random numbers and see if that appeared again. Is though there a more straightforward approach? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select 5 identical numbers
Hi, I want from a vector containing has like 1000 elements to select X of it randomly but with never selecting the same element again. Each one should be unique element of the vector. Is this more precise now? Regards Alex On Monday, January 20, 2014 7:54 PM, Alaios wrote: Dear all, I would like to select from a vector of 10 elements, 5 of those that are identical. How can I do that in R? I guess one way would be to taking random numbers and see if that appeared again. Is though there a more straightforward approach? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Density or Boxplot with median and mean
Hi there, I would like to be able to draw a density plot or a box plot where the median and the median and the mean would be visible. If I decide a density plot I need to put two big marks one for the median and one for the mean, which I do not know how I can achieve to put marks in a density plot. For that I am using plot(density(myVector)) while on the boxplot median is already visible but mean not. To have the mean there I would have to add one more line on each boxplot, perhaps of different color but I am not sure if that is possible in R. boxplot(myVector) I am using where myVector can be something like myVector<-seq(1,200) I would like to thank you all in advance Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Density or Boxplot with median and mean
Thanks Jim.. once again your rock On Wednesday, January 22, 2014 9:51 AM, Jim Lemon wrote: On 01/22/2014 07:37 PM, Alaios wrote: > Hi there, > I would like to be able to draw a density plot or a box plot where the median > and the median and the mean would be visible. > > If I decide a density plot I need to put two big marks one for the median and > one for the mean, which I do not know how I can achieve to put marks in a > density plot. For that I am using plot(density(myVector)) > > while on the boxplot median is already visible but mean not. To have the mean > there I would have to add one more line on each boxplot, perhaps of different > color but I am not sure if that is possible in R. boxplot(myVector) I am using > > where myVector can be something like myVector<-seq(1,200) > Hi Alex, On a density plot you can use abline: abline(v=mean(myVector),col="red") abline(v=median(myVector),col="green") I don't know of any boxplot function that will plot two measures of central tendency, but box.heresy in plotrix will plot any one measure that you like. Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Which distribution to select (massive fitting)
Hi all, I have a large number of measurements from which I select a large number of unique vectors. For each vectors I would like to test which distribution might be a candidate for fitting. It is impossible to look on each vector separately but I can inside a for loop test different models and based on their goodness of fit to make offline decisions (I will be saving goodness of fits results on a text file). Do you know given a vector how I can get the goodness of fit for the "basic" distributions : "norm", "lnorm", "pois", "exp", "gamma", "nbinom", "geom", "beta", "unif" and "logis" Is it possible to try many of those (or at least some of the above) and try to get these results? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Ignore errors and proceed to next
Hi all,
I have a very large number of vectors that I want first to look fast which
distribution might be considered candidate for fitting.
I made a simple loop that checks for all vector (the code below is for one
vector and being called for each vector separately). If a good fit is found
this is dumped to a txt file so to allow me later on, see results
distList<-c("norm","exp","gamma","lnorm")
for (dist in distList) {
if (gofstat(fitdist(onVector,distr=dist))$kstest =="not rejected"){
# keep it
out<-capture.output(gofstat(fitdist(onVector,distr=distr))$ks)
print(sprintf("Saving to file %s ",filename))
cat(out,file=paste(filename,".txt",sep=""),sep="\n",append=TRUE)
}
}
the major problem is that these for loops return errors (sometimes for a given
vector a specific distribution does not make sense or the vector might be zero
or containing only many times the same element) . The easiest would be in
errors and warning just move to the next element of the for loop.
Something like that
for ()
{
if error==skip to next element
else do normal stuff
}
how I can do that in R?
Regards
Alex
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ignore errors and proceed to next
Hi thanks for the answer.
I have so vast majority of vectors that some distributions will not work for
some cases but they work for others pretty well. This is the reason I store
only those that worked to helpe get a coarse understanding how to proceed. If
you have a better idea let me know
On Friday, February 14, 2014 10:35 AM, Frede Aakmann Tøgersen
wrote:
Hi
See
?try
?tryCatch
So you still want to do these distribution tests even though you have been
warned not to?
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance & Modeling
Technology & Service Solutions
T +45 9730 5135
M +45 2547 6050
fr...@vestas.com
http://www.vestas.com
Company reg. name: Vestas Wind Systems A/S
This e-mail is subject to our e-mail disclaimer statement.
Please refer to www.vestas.com/legal/notice
If you have received this e-mail in error please contact the sender.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Alaios
> Sent: 14. februar 2014 10:14
> To: R-help@r-project.org
> Subject: [R] Ignore errors and proceed to next
>
> Hi all,
> I have a very large number of vectors that I want first to look fast which
> distribution might be considered candidate for fitting.
> I made a simple loop that checks for all vector (the code below is for one
> vector and being called for each vector separately). If a good fit is found
> this
> is dumped to a txt file so to allow me later on, see results
>
> distList<-c("norm","exp","gamma","lnorm")
> for (dist in distList) {
>
> if (gofstat(fitdist(onVector,distr=dist))$kstest =="not rejected"){
> # keep it
> out<-capture.output(gofstat(fitdist(onVector,distr=distr))$ks)
> print(sprintf("Saving to file %s ",filename))
> cat(out,file=paste(filename,".txt",sep=""),sep="\n",append=TRUE)
>
> }
> }
> the major problem is that these for loops return errors (sometimes for a
> given vector a specific distribution does not make sense or the vector might
> be zero or containing only many times the same element) . The easiest
> would be in errors and warning just move to the next element of the for
> loop.
> Something like that
>
> for ()
> {
> if error==skip to next element
> else do normal stuff
> }
> how I can do that in R?
> Regards
> Alex
> [[alternative HTML version deleted]]
[[alternative HTML version deleted]]
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[R] Read text file
Hi all
I am trying to read some text files with the following format:
1377262633.948000
$GPRMC,125708.00,A,5047.66107,N,00603.65528,E,0.203,247.36,230813,,,A*60
1377262633.958000 $GPVTG,247.36,T,,M,0.203,N,0.377,K,A*3B
1377262633.968000
$GPGGA,125708.00,5047.66107,N,00603.65528,E,1,09,0.85,169.3,M,46.5,M,,*52
1377262633.978000 $GPGSA,A,3,29,21,31,25,16,05,06,13,271.78,0.85,1.57*0C
1377262633.998000
$GPGSV,3,1,12,03,01,266,,05,16,043,39,06,21,263,43,13,07,330,43*70
1377262634.008000
$GPGSV,3,2,12,16,37,302,45,18,03,149,,21,59,166,33,23,04,304,16*75
1377262634.028000
$GPGSV,3,3,12,25,18,129,21,27,11,260,39,29,45,071,47,31,35,211,47*7C
but this returns me the following:
read.csv("sensor_0.log",sep=",")
Error in read.table(file = file, header = header, sep = sep, quote = quote, :
more columns than column names
I guess the problem is that the columns are not consistent on a per row basis.
What I am trying to do though is to read only the lines that contain the
$GPGLL or the $GPGLA entries (in the example they corresponds to 3rd and 4th
line).
How can I do this in R?
Regards
A
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[R] From Strings to Variable name. When get does not work
Hi all,
I would like to turn some long strings like MyString$Myfield$MySubfield into
variables but it looks like that the get does not like lists
so for example:
test<-list(a=2)
test
>$a
[1] 2
get("test")
>$a
[1] 2
get("test$a")
>Fehler in get("test$a") : Objekt 'test$a' nicht gefunden
lapply(test,function(x) return (x$a))
>Fehler in x$a : $ operator is invalid for atomic vectors
What should I do to read those lists I have as strings?
I would like to thank you in advance for your reply
regards
A
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and provide commented, minimal, self-contained, reproducible code.
[R] Lists with numbers lists and strings
Dear all, I have a list that is created like that Spans<-list( c(837e6,842e6), c(832e6,837e6), c(930.1e6,935.1e6) ) I would like to include a second list that will contain the string that would correspond to the numbers at the left side. I would like thus inside my list to have two sublists. The first onew would be the Spans and the second one the Caption list. Could you please help me understand how I can do that? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot correlation matrix
Hi all, I am having 4 vectors like Data: num [1:4, 1:32] -82.8 -81.8 -75.5 -107.6 -87.6 ... and I want to calculate the correlation between those. Is there a graphical way in R to plot the correlations or not? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Where is element 30?
Hi I have a vector like that readCsvFile$V1 [1] 30 31 32 33 34 35 36 37 38 39 310 311 312 313 314 315 316 317 318 [20] 319 320 321 20 21 22 23 24 25 26 27 28 29 210 211 212 213 214 215 [39] 216 217 218 219 220 221 222 223 40 41 42 43 44 45 46 47 48 49 410 [58] 411 412 413 414 415 416 417 418 419 420 421 and I am looking to find where the number 31 is located. So I need one function to return me the index of where the number 31 is. Is there a function for that in R? Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plot.raster hides the axis layer
Hi all,
I am trying to plot a raster object (I can explain why but the point is that it
would be a raster objeçt)..
I have selected a small code to show you exactly the problem
require(raster)
test<-matrix(data=runif(1),nrow=100)
m<-raster(test)
plot(m,axes="FALSE")
axis(1,at=c(0,1),labels=c("a","b")) # THIS DOES NOT CHANGE THE INVISIBLE AXIS
WHILE
plot(test,axes="FALSE")
axis(1,at=c(0,1),labels=c("a","b")) # WHEN I AM PLOTTING A NON RASTER OBJECT
THE AXIS APPEARS.
Can someone explain me what I can do so a raster layer to show correctly the
axis? Please remember that I have to do that on a raster object so coercing it
will not help ..
I would like to thank you in advance for your help
Regards
Alex
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and provide commented, minimal, self-contained, reproducible code.
[R] add a color band
Hi all,
I would like to ask your help to add a color band (Î am not sure regarding the
right term, this color band at the right of the plot "describing" values with
their corresponding color.
For now I have only this code
test<-matrix(data=runif(1),nrow=100)
plot(test,axes="FALSE")
axis(1,at=c(0,1),labels=c("a","b")) #
but I would like to add at the right side the color legend.
How can I do that in R?
I would like to thank you in advance for your help
R.
Alex
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim and thanks for help
I will need some help to make
-the legend visible (probably at the right side of the window)
-split the legend at 0.1 steps -1,-0.9,-0.8.1 and
-assign a color heat.color(30). I have selected orange and reds
-show me the way to change the size of legend if it would be needed
par(mar=c(7,4,4,6))
test<-matrix(data=runif(1),nrow=100)
color2D.matplot(test,axes="F",xlab="",ylab="",show.legend=FALSE)
#,show.legend=TRUE
axis(1,at=seq(1:nrow(test)),labels=seq(201,300)) #
color.legend(11,6,11.8,9,seq(-1,1,length=10),rect.col=heat.colors(30),gradient="y")
I would like to thank you in advance for your help
Regards
Alex
On Friday, October 25, 2013 11:50 AM, Jim Lemon wrote:
On 10/25/2013 08:38 PM, Alaios wrote:
> Hi all,
> I would like to ask your help to add a color band (Ã⢠am not sure
> regarding the right term, this color band at the right of the plot
> "describing" values with their corresponding color.
>
> For now I have only this code
>
>
>
> test<-matrix(data=runif(1),nrow=100)
> plot(test,axes="FALSE")
>
> axis(1,at=c(0,1),labels=c("a","b")) #
>
> butÃÂ I would like to add at the rightÃÂ side the color legend.
>
> How can I do that in R?
>
> I would like to thank you in advance for your help
>
Hi Alex,
If all else fails, look at color.legend (plotrix).
Jim
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Compare two lists, with their sublists that have same structure
Dear all, I would like to ask your help concering two R lists. If I did everything should have the same structure (that means the same number of sublists, and their sublists also the same number of sublists). What would change between the two lists is the contents of each element in the lists. Could you please help me understand how I can do that in R? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim and thanks for your answer... I might be too tired with my new born or
just exhausted.
I am attaching for everyone a small data snipset that you can load
load("DataToPlotAsImage.Rdata")
require(plotrix)
browser()
test<-data
# this transforms the values of "test" into red->yellow
color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
extremes=c("#FF","#00"),show.legend=FALSE)
axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
color.legend(104,30,112,70,seq(-110,-30,length=11),
align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
as you can see I have problems where the legend appears. My par("usr" returned
me
par("usr")
# [1] 0 351 0 200
but I am not sure how to read that to place the legend at a useful place.
second I am not sure why the image is so full with black rows..
What I want is to have the legend visible
and later on customize the x axis to write custom string of different size...
First I need though to fix the more severe problems as I have described
Regards
Alex
On Friday, October 25, 2013 11:45 PM, Jim Lemon wrote:
On 10/25/2013 11:16 PM, Alaios wrote:
> Hi Jim and thanks for help
>
> I will need some help to make
> -the legend visible (probably at the right side of the window)
> -split the legend at 0.1 steps -1,-0.9,-0.8.1 and
> -assign a color heat.color(30). I have selected orange and reds
> -show me the way to change the size of legend if it would be needed
>
>
>
> par(mar=c(7,4,4,6))
> test<-matrix(data=runif(1),nrow=100)
> color2D.matplot(test,axes="F",xlab="",ylab="",show.legend=FALSE)
> #,show.legend=TRUE
> axis(1,at=seq(1:nrow(test)),labels=seq(201,300)) #
> color.legend(11,6,11.8,9,seq(-1,1,length=10),rect.col=heat.colors(30),gradient="y")
>
Hi Alex,
You can do this in at least two ways. The first example uses the
color.scale function to assign colors close to those of heat.colors. The
second shows how to use heat.colors if you want that.
library(plotrix)
par(mar=c(7,4,4,6))
test<-matrix(data=runif(1),nrow=100)
# this transforms the values of "test" into red->yellow
color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
extremes=c("#FF","#00"),show.legend=FALSE)
axis(1,at=seq(1:nrow(test)),labels=seq(201,300))
# use par("usr") to find out the plot dimensions,
# then place the legend where you want it
color.legend(104,30,112,70,seq(-1,1,length=11),
align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
# now try it with heat.colors
test.cut<-matrix(as.numeric(cut(test,breaks=seq(0,1,length.out=30))),nrow=100)
color2D.matplot(test.cut,axes="F",xlab="",ylab="",main="heat.colors",
cellcolors=heat.colors(30)[test.cut],show.legend=FALSE)
axis(1,at=seq(1:nrow(test)),labels=seq(201,300))
color.legend(104,30,112,70,seq(-1,1,length=11),
align="rb",rect.col=heat.colors(30),gradient="y")
Jim__
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[R] How to avoid this warning message
Dear all, I would like to ask you how I can avoid this warning I am getting when I am writing some results in a text file 5: In write.table(x = cor(t(collectMean_UL), t(collectMean_DL)), ... : appending column names to file the code that saves to the file look like: write(x="TEMPERATURE",file=paste(Folder,"/Corr_",".txt",sep=""),append="TRUE") write.table(x=cor(t(Data)),file=paste(Folder,"/Cor_",".txt",sep=""),eol = "\r\n",append="TRUE") if someone can help on that... I would like to thank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim and thanks for your answer... I might be too tired with my new born or
just exhausted.
I am sharing for everyone a small data snipset that you can load
https://www.dropbox.com/s/fh8jhwujgunmtrb/DataToPlotAsImage.Rdata
load("DataToPlotAsImage.Rdata")
require(plotrix)
browser()
test<-data
# this transforms the values of "test" into red->yellow
color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
extremes=c("#FF","#00"),show.legend=FALSE)
axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
color.legend(104,30,112,70,seq(-110,-30,length=11),
align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
as you can see I have problems where the legend appears. My par("usr" returned
me
par("usr")
# [1] 0 351 0 200
but I am not sure how to read that to place the legend at a useful place.
second I am not sure why the image is so full with black rows..
What I want is to have the legend visible
and later on customize the x axis to write custom string of different
size... First I need though to fix the more severe problems as I have
described
RegardsAlex
On Sunday, October 27, 2013 12:25 PM, Jim Lemon wrote:
On 10/27/2013 08:39 AM, Alaios wrote:
> Hi Jim and thanks for your answer... I might be too tired with my new
> born or just exhausted.
>
> I am attaching for everyone a small data snipset that you can load
>
>
> load("DataToPlotAsImage.Rdata")
> require(plotrix)
> browser()
> test<-data
> # this transforms the values of "test" into red->yellow
> color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
> extremes=c("#FF","#00"),show.legend=FALSE)
>
> axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
> color.legend(104,30,112,70,seq(-110,-30,length=11),
> align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
>
> as you can see I have problems where the legend appears. My par("usr"
> returned me
> par("usr")
> # [1] 0 351 0 200
>
> but I am not sure how to read that to place the legend at a useful place.
> second I am not sure why the image is so full with black rows..
>
> What I want is to have the legend visible
> and later on customize the x axis to write custom string of different
> size... First I need though to fix the more severe problems as I have
> described
>
Hi Alex,
I'm not sure why you have created a copy of "data" to plot it. I can get
quite a sensible plot using this:
par(mar=c(5,4,4,5))
color2D.matplot(data,1,c(0,1),0,xlab="",ylab="",
main="color.scale",xrange=c(-110,-50),border=NA)
color.legend(357,30,370,100,seq(-110,-50,length.out=6),
align="rb",rect.col=color.scale(1:6,1,c(0,1),0),
gradient="y")
Notice several things. First, when you have a large number of cells in a
plot like this, setting the border to NA means that you don't get mostly
borders (default = black) in the plot. The second thing is that your
data range is -107.18150 to -54.07662. In order to get rounded numbers
in your legend, I have set the xrange argument to -110 to -50. This
gives a neat looking legend that spans your data, a bit like the
"pretty" function would do. It also means that the color mapping is to
that range and is the same in the legend as in the plot. I have left
enough space on the right of the plot to fit in the legend, as that was
where you said you wanted it in your last email. What par("usr") tells
you is the dimensions of the plot in user units. Here it is x=0 at the
left, x=351 at the right, y=0 at the bottom and y=200 at the top.
Jim
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[R] Create Time Lists with a for loop
Dear all,
in my code I have written the following list
TimeFramesShort <-list(c(strptime("2011-10-12 10:59:00","%Y-%m-%d
%H:%M:%S"),strptime("2011-10-13 11:02:00","%Y-%m-%d %H:%M:%S")) #
rest items were truncated
I was wondering if could somehow take the first element of the list and create
element that are only 5 minutes apart something like
list(c(strptime("2011-10-12 10:59:00","%Y-%m-%d %H:%M:%S"),strptime("2011-10-12
11:04:00","%Y-%m-%d %H:%M:%S"),
c(strptime("2011-10-12 11:04:00","%Y-%m-%d %H:%M:%S"),strptime("2011-10-12
11:09:00","%Y-%m-%d %H:%M:%S"),
c(strptime("2011-10-12 11:09:00","%Y-%m-%d %H:%M:%S"),strptime("2011-10-12
11:13:00","%Y-%m-%d %H:%M:%S")
...
and so on.
The problem I see is that pure string manipulation will not work as after one
hour that should also make changes in the hours.
Regards
Alex
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[R] TelosB external modules, guide
Hi all, I would like to ask your help regarding connecting external modules to telosb. I have found that tiny os offers many possibilities for that as ADC, GPIOs, SPI, UART, I2C I have never learned anything regarding those. Can someone please let me know if there is any simple guide on these "interfaces" to show with a primitive example how to start with? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] add a color bar in a plot
Hi all,
I am trying to add a color legend to my plot. As an example I am giving you a
bit of code that you can run.
I am sharing for everyone a small data snipset that you can load
https://www.dropbox.com/s/fh8jhwujgunmtrb/DataToPlotAsImage.Rdata
load("DataToPlotAsImage.Rdata")
require(plotrix)
browser()
test<-data
# this transforms the values of "test" into red->yellow
color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
extremes=c("#FF","#00"),show.legend=FALSE)
axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
color.legend(104,30,112,70,seq(-110,-30,length=11),
align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
as you can see I have problems where the legend appears. My par("usr" returned
me
par("usr")
# [1] 0 351 0 200
but I am not sure how to read that to place the legend at a useful place.
second I am not sure why the image is so full with black rows..
What I want is to have the legend visible
and later on customize the x axis to write custom string of different
size... First I need though to fix the more severe problems as I have
described
RegardsAlex
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[R] Select fixed number of elements
Hi all, I have in my code some vectors that are not of equal size. I would like to be able for each of these vectors select 6 elements that are (almost) equally spaced. So the first one would be at (or close) to the beginning the last one at (or close) to the end and the other 4 equally spaced between first and last element. How can I do something like that on a vector of not known size? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select fixed number of elements
Thumbs up! It worked! On Wednesday, October 30, 2013 9:41 AM, Gerrit Eichner wrote: Hello, Alaois, if x is your vector maybe n <- length( x) positions <- trunc( quantile( seq( n), prob = 0:5/5)) x[ positions] comes close to what you want. Hth -- Gerrit > Hi all, I have in my code some vectors that are not of equal size. I > would like to be able for each of these vectors select 6 elements that > are (almost) equally spaced. So the first one would be at (or close) to > the beginning the last one at (or close) to the end and the other 4 > equally spaced between first and last element. > > How can I do something like that on a vector of not known size? > > I would like to thank you in advance for your help > > Regards > Alex > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help me align the legend bar
Hi,
I have some code that you can simply execute:
require(plotrix)
test<-matrix(data=rnorm(1,-100,5),nrow=100)
color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
extremes=c("#FF","#00"),show.legend=FALSE)
axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
color.legend(84,30,125,70,seq(-110,-30,length=11),
align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
What I would like to do is to make space at the right for the color band . The
band should have one color from -110 to -30 with scales of 10.
Can someone help me with that?
Regards
Alex
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[R] remap values from one vector to the other
Hi everyone, I am plotting some legend and I am using the axis(at=..) to specify the place to plot the marks I want. My plotted data have ncol(x) so the at places have values that span from 1 to ncol(x) there I would like to be able to map values that go from 880e6 to 1020e6. so 880e6 remaps to 1 1020e6 repams to ncol(x) what I do not know how to do is to remap the values that are between 880e6 and 1020e6 between then 1 and ncol(x). Can someone help me find an appropriate function for that? I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add a color band
Hi Jim Lemon,
thanks for the help, I appreciate this.
right now my code looks like.
par(mar=c(5,4,4,5))
color2D.matplot(data,1,c(0,1),0,xlab="",ylab="Spans",
main="color.scale",xrange=c(-110,-50),border=NA,axes=F)
color.legend(357,30,370,100,seq(-110,-50,length.out=13),
align="rb",rect.col=color.scale(1:13,1,c(0,1),0),
gradient="y")
my major problem now is that the
a. text in the color bar is squeezed so -50 overlaps with -60 and so on
b. for some reason the color bar sometimes (the same code is called for all the
data matrices I have) is misaligned in different positions each time
Could you please also help me with those two?
Regards
Alex
On Monday, October 28, 2013 9:00 AM, Alaios wrote:
Hi Jim and thanks for your answer... I might be too tired with my new born or
just exhausted.
I am sharing for everyone a small data snipset that you can load
https://www.dropbox.com/s/fh8jhwujgunmtrb/DataToPlotAsImage.Rdata
load("DataToPlotAsImage.Rdata")
require(plotrix)
browser()
test<-data
# this transforms the values of "test" into red->yellow
color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
extremes=c("#FF","#00"),show.legend=FALSE)
axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
color.legend(104,30,112,70,seq(-110,-30,length=11),
align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
as you can see I have problems where the legend appears. My par("usr" returned
me
par("usr")
# [1] 0 351 0 200
but I am not sure how to read that to place the legend at a useful place.
second I am not sure why the image is so full with black rows..
What I want is to have the legend visible
and later on customize the x axis to write custom string of different
size... First I need though to fix the more severe problems as I have
described
RegardsAlex
On Sunday, October 27, 2013 12:25 PM, Jim Lemon wrote:
On 10/27/2013 08:39 AM, Alaios wrote:
> Hi Jim and thanks for your answer... I might be too tired with my new
> born or just exhausted.
>
> I am attaching for everyone a small data snipset that you can load
>
>
> load("DataToPlotAsImage.Rdata")
> require(plotrix)
> browser()
> test<-data
> # this transforms the values of "test" into red->yellow
> color2D.matplot(test,axes="F",xlab="",ylab="",main="color.scale",
> extremes=c("#FF","#00"),show.legend=FALSE)
>
> axis(1,at=seq(1,ncol(test),length.out=10),labels=seq(201,300,length.out=10))
> color.legend(104,30,112,70,seq(-110,-30,length=11),
> align="rb",rect.col=color.scale(1:30,1,c(0,1),0),gradient="y")
>
> as you can see I have problems where the legend appears. My par("usr"
> returned me
> par("usr")
> # [1] 0 351 0 200
>
> but I am not sure how to read that to place the legend at a useful place.
> second I am not sure why the image is so full with black rows..
>
> What I want is to have the legend visible
> and later on customize the x axis to write custom string of different
> size... First I need though to fix the more severe problems as I have
> described
>
Hi Alex,
I'm not sure why you have created a copy of "data" to plot it. I can get
quite a sensible plot using this:
par(mar=c(5,4,4,5))
color2D.matplot(data,1,c(0,1),0,xlab="",ylab="",
main="color.scale",xrange=c(-110,-50),border=NA)
color.legend(357,30,370,100,seq(-110,-50,length.out=6),
align="rb",rect.col=color.scale(1:6,1,c(0,1),0),
gradient="y")
Notice several things. First, when you have a large number of cells in a
plot like this, setting the border to NA means that you don't get mostly
borders (default = black) in the plot. The second thing is that your
data range is -107.18150 to -54.07662. In order to get rounded numbers
in your legend, I have set the xrange argument to -110 to -50. This
gives a neat looking legend that spans your data, a bit like the
"pretty" function would do. It also means that the color mapping is to
that range and is the same in the legend as in the plot. I have left
enough space on the right of the plot to fit in the legend, as that was
where you said you wanted it in your last email. What par("usr") tells
you is the dimensions of the plot in user units. Here it is x=0 at the
left, x=351 at the right, y=0 at the bottom and y=200 at the top.
Jim
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Re: [R] add a color band
Hi Jim,
thanks for the answer! I toyed around as you said and now they look cool and
sexy! many thanks!
Do you know if there is a way to compress a bit the output? The image type of
functions have the useRaster parameter to set to have a "compressed" output.
Right now the produced plots are around 400kBytes which is a lot and I am
looking for ways to reduce image size.
I would like to thank you for you reply
Alex
On Monday, November 4, 2013 11:59 AM, Jim Lemon wrote:
On 11/04/2013 08:09 PM, Alaios wrote:
> Hi Jim Lemon,
> thanks for the help, I appreciate this.
>
> right now my code looks like.
>
>
> par(mar=c(5,4,4,5))
> color2D.matplot(data,1,c(0,1),0,xlab="",ylab="Spans",
> main="color.scale",xrange=c(-110,-50),border=NA,axes=F)
> color.legend(357,30,370,100,seq(-110,-50,length.out=13),
> align="rb",rect.col=color.scale(1:13,1,c(0,1),0),
> gradient="y")
>
>
> my major problem now is that the
> a. text in the color bar is squeezed so -50 overlaps with -60 and so on
> b. for some reason the color bar sometimes (the same code is called for
> all the data matrices I have) is misaligned in different positions each time
>
> Could you please also help me with those two?
>
Hi Alex,
For your first question, I would simply extend the color legend vertically:
color.legend(357,30,370,150,seq(-110,-50,length.out=13),
align="rb",rect.col=color.scale(1:13,1,c(0,1),0),
gradient="y")
For the second one, you obviously have different dimensions for the data
matrices. So, let's step through a method for getting the legend
position and size from the plot itself. As I have written a few times
previously, par("usr") gives you the dimensions of the plot in user
units. For the example above, the dimensions were:
x - 0->351
y - 0->200
With a bit of arithmetic, you can work out that the legend positions in
the above are:
xylim<-par("usr")
# x position of lower left corner of legend
xl<-xylim[2]+diff(xylim[1:2])*0.017
# y position of lower left corner
yb<-xylim[3]+diff(xylim[3:4])*0.15
# x position of upper right corner of legend
xr<-xylim[2]+diff(xylim[1:2])*0.054
# y position of upper right corner
yt<-xylim[3]+diff(xylim[3:4])*0.75
Having these lines means that you can get the position and size of the
legend about right from the information provided by par("usr") even if
you change the number of cells in the matrix passed to color2D.matplot.
Then you would call:
color.legend(xl,yb,xr,yt,seq(-110,-50,length.out=13),
align="rb",rect.col=color.scale(1:13,1,c(0,1),0),
gradient="y")
unless of course you wanted to change the values in the scale markings.
I'll leave that for you to work out.
Jim
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[R] convert one digit numbers to two digits one
Hi all, the following returns the hour and the minutes paste(DataSet$TimeStamps[selectedInterval$start,4], DataSet$TimeStamps[selectedInterval$start,5],sep=":") [1] "12:3" the problem is that from these two I want to create a time stamp so 12:03. The problem is that the number 3 is not converted to 03. Is there an easy way when I have one digit integer to add a zero in the front? Two digits integers are working fine so far, 12:19, or 12:45 would appear correctly I would like to thank you in advance for your help Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Colour Legend text, in a print color2D.matplot
Hi all, I am plotting very nice looking mattrices with plotrin... so far so good, I would like though to ask you if it would be possible to add at the bottom of the color.legend (this lovely color bar that maps colors to numbers). Would that be possible to do that? I would like to thank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotrix: Add text below the color- legend.
Hi all, I am plotting very nice and sexy images with plotrix :) so far so good, I would like though to ask you if it would be possible to add at the bottom of the color.legend (this lovely color bar that maps colors to numbers). Would that be possible to do that? I would like to thank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Colour Legend text, in a print color2D.matplot
Thanks! It worked.
Regards
Alex
On Monday, November 11, 2013 12:02 AM, Jim Lemon wrote:
On 11/11/2013 04:57 AM, Alaios wrote:
> Hi all,
>
> I am plotting very nice looking mattrices with plotrin...
>
> so far so good, I would like though to ask you if it would be possible to add
> at the bottom of the color.legend (this lovely color bar that maps colors to
> numbers).
> Would that be possible to do that?
>
Hi Alex,
It is not too difficult:
# first allow printing outside the plot
par(xpd=TRUE)
# get the x component of the label
# from the values passed to color.legend
# see the help page
labelx<-(xl+xr)/2
# get the y coordinate from the y value
labely<-yb-strheight("M")
# display the label
text(labelx,labely,mylabel)
# restore the clipping
par(xpd=FALSE)
Jim
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Colour Legend text, in a print color2D.matplot
Hi Jim, it worked nicely.
par(mar) looks to control the space at the bottom left right and top.. What If
I want to increase a bit the distance between the plotted matrix and the added
color legend ? They look to be at my version slightly packed together.
Regards
Alex
On Monday, November 11, 2013 3:07 AM, Alaios wrote:
Thanks! It worked.
Regards
Alex
On Monday, November 11, 2013 12:02 AM, Jim Lemon wrote:
On 11/11/2013 04:57 AM, Alaios wrote:
> Hi all,
>
> I am plotting very nice looking mattrices with plotrin...
>
> so far so good, I would like though to ask you if it would be possible to add
> at the bottom of the color.legend (this lovely color bar that maps colors to
> numbers).
> Would that be possible to do that?
>
Hi Alex,
It is not too difficult:
# first allow printing outside the plot
par(xpd=TRUE)
# get the x component of the label
# from the values passed to color.legend
# see the help page
labelx<-(xl+xr)/2
# get the y coordinate from the y value
labely<-yb-strheight("M")
# display the label
text(labelx,labely,mylabel)
# restore the clipping
par(xpd=FALSE)
Jim
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
[R] Handle Gps coordinates
Dear all, I would like to ask you if there are any gps libraries. I would like to be able to handle them, -like calculate distances in meters between gps locations, -or find which gps location is closer to a list of gps locations. Is there something like that in R? I would like to tthank you in advance for your reply Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Grid type of sampling in geodata
Hi all, I have spatial field created with grf and I was wondering if I can sample in lines, something that can resemble sampling outdoors at the streets. Random sampling looks to far of what I want to have. there is in geodata package the sample.geodata but this looks like to be random samples. I would like to thank you in advance for your help Regards A. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] save table to txt file in a printable form
Hi there, I would like to save tabular (in R just matrices) in a txt file but in a way that they would be somehow readable. That means keeping columns aligned and rows so one can read easily for example the 2,3 element. IS there a way to do that in R? capture.output for example does not produce tables in the way I want to have those. Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Syntax Highlightning and Editor for Linux
Hello to the community . First post :) I would like to ask you which text editor do you use in Linux and how did you setup the syntax highlightning? one more question is it possible to debug any program in R by inserting breakpoints? I would like to thank you in advance for your help Best Regards Alex -- View this message in context: http://r.789695.n4.nabble.com/Syntax-Highlightning-and-Editor-for-Linux-tp2300072p2300072.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Syntax Highlightning and Editor for Linux
I would like to thank you for your immediate reply.
Actually I do not like vim and emacs.. I am trying to find an editor with a gui
that will work fine in Linux.
One more question. If I am editing a file using my external text editor is it
possible to execute directly one of the functions that are inside the file
without executing the source("myfile.R") command first?
Best Regards
Alex
- Original Message
From: Duncan Murdoch
To: alaios
Cc: r-help@r-project.org
Sent: Fri, July 23, 2010 2:13:48 PM
Subject: Re: [R] Syntax Highlightning and Editor for Linux
On 23/07/2010 8:08 AM, alaios wrote:
> Hello to the community .
> First post :)
> I would like to ask you which text editor do you use in Linux and how did
> you setup the syntax highlightning?
>
I don't use Linux, but I think the most popular editors there are vim and
emacs,
the latter with ESS.
> one more question is it possible to debug any program in R by inserting
> breakpoints?
>
>
Yes, see the setBreakpoint() and trace() functions.
Duncan Murdoch
> I would like to thank you in advance for your help
> Best Regards
> Alex
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Syntax Highlightning and Editor for Linux
I would like oto thank everyone for the replies.
I instaled rkward.. It looks like an editor.. Is it possible to execute also
your code from the text editor.
If there is a function call myfunc inside my file test.R
Is it possible to exectute the function from the rwkard?
Best Regards
Alex
From: Federico Andreis
To: Rainer M Krug
Sent: Fri, July 23, 2010 3:46:08 PM
Subject: Re: [R] Syntax Highlightning and Editor for Linux
I have been trying emacs with ess, Kate, plain gedit..nothing really satisfied
me..
[[elided Yahoo spam]]
it supports split screen with R terminal window (you can also personalize
colour
if, like me, do like to
stick with old habits :) ), moreover gedit has a brackets completion and a
brackets highlight plugin which are very useful.
check it out here
http://www.stattler.com/article/using-gedit-or-rgedit-r
simple "send code to terminal" shortcuts (much simpler than emacs, in my
[[elided Yahoo spam]]
/federico
On Fri, Jul 23, 2010 at 2:42 PM, Rainer M Krug wrote:
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>Hash: SHA1
>
>
>On 23/07/2010 14:17, Alaios wrote:
>> I would like to thank you for your immediate reply.
>> Actually I do not like vim and emacs.. I am trying to find an editor with a
>gui
>> that will work fine in Linux.
>Well - you are missing out with not using emacs + ess - but it is quite
>a learning curve. Despite this:
>
>Try rkward - syntax highlighting and many other nice features.
>
>Another option: eclipse with statet plugin(
>http://www.walware.de/goto/statet)
>
>Cheers,
>
>Rainer
>
>>
>> One more question. If I am editing a file using my external text editor is it
>> possible to execute directly one of the functions that are inside the file
>> without executing the source("myfile.R") command first?
>>
>> Best Regards
>> Alex
>>
>>
>>
>> - Original Message
>> From: Duncan Murdoch
>> Cc: r-help@r-project.org
>> Sent: Fri, July 23, 2010 2:13:48 PM
>> Subject: Re: [R] Syntax Highlightning and Editor for Linux
>>
>> On 23/07/2010 8:08 AM, alaios wrote:
>>> Hello to the community .
>>> First post :)
>>> I would like to ask you which text editor do you use in Linux and how did
>>> you setup the syntax highlightning?
>>>
>>
>> I don't use Linux, but I think the most popular editors there are vim and
>>emacs,
>> the latter with ESS.
>>
>>> one more question is it possible to debug any program in R by inserting
>>> breakpoints?
>>>
>>>
>>
>> Yes, see the setBreakpoint() and trace() functions.
>>
>> Duncan Murdoch
>>
>>> I would like to thank you in advance for your help
>>> Best Regards
>>> Alex
>>>
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
>- --
>Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
>Biology, UCT), Dipl. Phys. (Germany)
>
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>Skype: RMkrug
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>and provide commented, minimal, self-contained, reproducible code.
>
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and provide commented, minimal, self-contained, reproducible code.
[R] Creating a map .
Hello. I would try to explain what I would like to implement so to suggest me what to try out. I would like to create an area of X*X km that would be used to "Simulate" an area map (eg. city's area, suburban area). -X would be a parameter so I do not want it to be fixed -In this map I would like to place users (people), thus I do not know in advance if in one place there would be one,two, or more users.. -I would also like to not have fixed resolution in my map and specify it as a parameter during run-time. In my X*X area I would like sometimes to have resolution of 1Km and others resolution of 10 meters. As resolution increases (1 km->100m->10m) less users would be found in the same "place". -In every "place" I would like to specify some parameters (eg. number of users, characteristics of users, energy consumption per user). I do not know in advance how much data I would need per user. Ofc 10 parameters per place might be ok but a not fixed approach is better. My first thoughts were to use an array and actually define an array of X*X dimension where X is defined at run time as parameter. The problem with this approach is that I do not have variable resolution in this area as I have the constant number of X*X places only. Also one more problem is that the cels of the array (where cell is used to simulate "place") can not be used to story many parameters but only a single numerical value). What do you suggest me to try looking at and what made up your decision? I would like to thank everyone that reached to this point reading all my big text. Best Regards Alex. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] what is a vignette?
I am trying to find a simple R guide that explain what a vignette is but so far I didnt make any progress. I tried to search inside R's built in help.start() but it only returns results how to see vignettes. So could you please tell me what a vignette is and if you can also could you give some simple guide that I can always use to read about these things? Best Regards Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] "=" vs "<-" operator
Hello I notice that in Linux the "=" operator works like the "<-" operator So a=3 is similar to a<-3. Could you please verify me that is correct? I would like to use "=" operator. Do you think that might be a problem in the future? Best Regards Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] See what is inside a matrix
Hello everyone.. Is there any graphical tool to help me see what is inside a matrix? I have 100x100 dimensions matrix and as you already know as it does not fit on my screen R splits it into pieces. I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Line integral with R
Hello. I would like to calculate with R the weighted line integral of a loss field. Where should I start searching about weighted integration in R? I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] print matrix values and if statement
Hello everyone, I have a 2x2 matrix filled with zeros and some more values around zeros. I would like to print only the non-zero values and to keep the coords of the places that the values are not zero. Could you please help me with that? I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] array of objects
Hello everyone. I would like to create some agents that span over a specific area map.Every agent needs to have its own data structures like one or two matrices and one list. I think that the best way to do this is to create objects and every instance of an object will be used for a single agent. The number of agents is not predetermined and it varies for any execution. So I read this value from the command line interface and then I would like to initiate so many objects as the agents. I think that the best way to do that is to create using a for loop a list containing as many objects as the agents are. Is it possible to help me do that with R... the tricky part is not only to create these objects but also having some nice way to address them... Perhaps an index i might be used for that. I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Object oriented programming in R.
Hello everyone. I would like to create many objects with R. Does R support objects? The number of objects needed is not predetermined and it is a parameter specified by the user. If the user selects to create many objects like 100, would it be possible to handle each one by some index? I would like to thank you in advance for your help. Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Object oriented programming in R.
Thank you very much. I checked the tutorials that on that list but still I do not know how to create many objects of the same type. Can you please help me with that? Best Regards Alex From: Tal Galili Cc: Rhelp Sent: Tue, September 14, 2010 10:11:36 AM Subject: Re: [R] Object oriented programming in R. Hello Alaios, I see a bunch of good materials here: http://www.google.co.il/search?sourceid=chrome&ie=UTF-8&q=Object+oriented+programming+in+R Did you look into them ? Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- Hello everyone. >I would like to create many objects with R. Does R support objects? > >The number of objects needed is not predetermined and it is a parameter >specified by the user. >If the user selects to create many objects like 100, would it be possible to >handle each one by some index? > >I would like to thank you in advance for your help. > > >Best Regards >Alex > > > > [[alternative HTML version deleted]] > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Object oriented programming in R.
Here are some more information: I would like to create some agents that span over a specific area map.Every agent needs to have its own data structures like one or two matrices and one list. I think that the best way to do this is to create objects and every instance of an object will be used for a single agent. The number of agents is not predetermined and it varies for any execution. So I read this value from the command line interface and then I would like to initiate so many objects as the agents. I think that the best way to do that is to create using a for loop a list containing as many objects as the agents are. I would like to thank you in advance for your help Best Regards Alex From: jim holtman Cc: Tal Galili ; Rhelp Sent: Tue, September 14, 2010 1:40:37 PM Subject: Re: [R] Object oriented programming in R. It depends on what you mean by objects. If you are just looking at creating many named variables that are going to hold values (e.g., reading in data from several files that you want to correlate separately), then consider the use of 'lists'. Can you provide a little more detail on exactly the problem that you are trying to solve, and then maybe we can propose a solution. > Thank you very much. I checked the tutorials that on that list but still I do > not know how to create many objects of the same type. Can you please help me > with that? > > Best Regards > Alex > > > > > > From: Tal Galili > > Cc: Rhelp > Sent: Tue, September 14, 2010 10:11:36 AM > Subject: Re: [R] Object oriented programming in R. > > > Hello Alaios, > I see a bunch of good materials here: >http://www.google.co.il/search?sourceid=chrome&ie=UTF-8&q=Object+oriented+programming+in+R >R > > > Did you look into them ? > > Contact > Details:--- > Contact me: tal.gal...@gmail.com | 972-52-7275845 > Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | > www.r-statistics.com (English) >-- >- > > > > > > > > > Hello everyone. >>I would like to create many objects with R. Does R support objects? >> >>The number of objects needed is not predetermined and it is a parameter >>specified by the user. >>If the user selects to create many objects like 100, would it be possible to >>handle each one by some index? >> >>I would like to thank you in advance for your help. >> >> >>Best Regards >>Alex >> >> >> >> [[alternative HTML version deleted]] >> >>__ >>R-help@r-project.org mailing list >>https://stat.ethz.ch/mailman/listinfo/r-help >>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>and provide commented, minimal, self-contained, reproducible code. >> > > > > >[[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Object oriented programming in R.
I would like to thank you very much all that you helped me so far. So I tried to check how the following works fred <- list(happy = 1:10, name = "squash") rep(fred, 5) This returns the following : > fred[1] $happy [1] 1 2 3 4 5 6 7 8 9 10 > fred[2] $name [1] "squash" What I am trying to do is to address the number 5 of the fred[1] $happy value. I tried something like fred[1][5] fred[1,5] but it didn't work I would like to thank you in advance for your help Best Regards Alex From: Dennis Murphy Cc: Rhelp Sent: Tue, September 14, 2010 3:13:37 PM Subject: Re: [R] Object oriented programming in R. Hi: You could create a list of lists, where the outer list would be between agents and the inner list within agents. The inner list could have the 'matrices and one list' as separate components for each agent. Of course, you would have to be able to keep all of this straight :) HTH, Dennis Here are some more information: >I would like to create some agents that span over a specific area map.Every >agent needs to have its own data structures like one or two matrices and one >list. > >I think that the best way to do this is to create objects and every instance of >an object will be used for a single agent. > >The number of agents is not predetermined and it varies for any execution. >So I read this value from the command line interface and then I would like to >initiate so many objects as the agents. I think that the best way to do that is >to create using a for loop a list containing as many objects as the agents are. > > >I would like to thank you in advance for your help > >Best Regards >Alex > > > > >From: jim holtman > >Cc: Tal Galili ; Rhelp >Sent: Tue, September 14, 2010 1:40:37 PM > >Subject: Re: [R] Object oriented programming in R. > > >It depends on what you mean by objects. If you are just looking at >creating many named variables that are going to hold values (e.g., >reading in data from several files that you want to correlate >separately), then consider the use of 'lists'. Can you provide a >little more detail on exactly the problem that you are trying to >solve, and then maybe we can propose a solution. > > > >> Thank you very much. I checked the tutorials that on that list but still I do >> not know how to create many objects of the same type. Can you please help me >> with that? >> >> Best Regards >> Alex >> >> >> >> >> >> From: Tal Galili >> >> Cc: Rhelp >> Sent: Tue, September 14, 2010 10:11:36 AM >> Subject: Re: [R] Object oriented programming in R. >> >> >> Hello Alaios, >> I see a bunch of good materials here: >>http://www.google.co.il/search?sourceid=chrome&ie=UTF-8&q=Object+oriented+programming+in+R >> >>R >> >> >> Did you look into them ? >> >> Contact >> Details:--- >> Contact me: tal.gal...@gmail.com | 972-52-7275845 >> Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | >> www.r-statistics.com (English) >>-- >> >>- >> >> >> >> >> >> >> >> >> Hello everyone. >>>I would like to create many objects with R. Does R support objects? >>> >>>The number of objects needed is not predetermined and it is a parameter >>>specified by the user. >>>If the user selects to create many objects like 100, would it be possible to >>>handle each one by some index? >>> >>>I would like to thank you in advance for your help. >>> >>> >>>Best Regards >>>Alex >>> >>> >>> >>> [[alternative HTML version deleted]] >>> >>>__ >>>R-help@r-project.org mailing list >>>https://stat.ethz.ch/mailman/listinfo/r-help >>>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>>and provide commented, minimal, self-contained, reproducible code. >>> >> >> >> >> >>[[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide comme
[R] Interpolate? a line
Hello everyone. I have created a 100*100 matrix in R. Let's now say that I have a line that starts from (2,3) point and ends to the (62,34) point. In other words this line starts at cell (2,3) and ends at cell (62,34). Is it possible to get by some R function all the matrix's cells that this line transverses? I would like to thank you for your feedback. Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interpolate? a line
Thank you very much. I am trying to execute it line by line to get some understanding how this works. Could you please explain me what these two lines do? tie <- m[,2] == c(-Inf, m[-nrow(m),2]) m <- m[ !tie, ] I would like to thank you in advance for your help Best Regards Alex From: Michael Bedward Cc: Rhelp Sent: Wed, September 15, 2010 12:52:45 PM Subject: Re: [R] Interpolate? a line Hello Alex, Here is one way to do it. It works but it's not pretty :) interp <- approx(c(2, 62), c(3, 34), method="linear", xout=2:62) m <- matrix(c(interp$x, round(interp$y)), ncol=2) tie <- m[,2] == c(-Inf, m[-nrow(m),2]) m <- m[ !tie, ] You might want to examine the result like this... plot(m) # plots points lines(c(2,26), c(3, 34)) # overlay line for comparison Michael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help me understand how things work.
Hello I have some strange output from R and I try to understand how R works. Could you please help me with that? temp <- rbind (c(10,1),c(99,98)) > temp [,1] [,2] [1,] 101 [2,] 99 98 > dist(temp) 1 2 131.6435 > sqrt(dist(temp)) 1 2 11.47360 so far so good. until the nex line: when I try to do what i did before but adding the 1/(what I did before). I was expecting a number as a result of the division but unfortunately I took the following: 1/sqrt(dist(temp)) [1] 0.08715662 attr(,"Size") [1] 2 attr(,"Diag") [1] FALSE attr(,"Upper") [1] FALSE attr(,"method") [1] "euclidean" attr(,"call") dist(x = temp) attr(,"class") [1] "dist" Could you please help me understand what is this about? I would like to thank you in advance for your help Best REgards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help me understand how things work.
I fixed by adding this: k <- sqrt(dist(temp)) k returns sort of a list. So I need to select the first item which is the result. a <- k[1] Can someone explain me why k[1] is needed for that? Best Regards Alex From: Mario Valle Sent: Thu, September 16, 2010 1:28:31 PM Subject: Re: [R] help me understand how things work. ?dist BTW, to me this does not happens. x <- matrix(rnorm(100), nrow=5) d <- dist(x) 1/sqrt(d) 1/sqrt(dist(x)) Hope it helps mario On 16-Sep-10 12:02, Alaios wrote: > Hello I have some strange output from R and I try to understand how R works. > > Could you please help me with that? > > temp<- rbind (c(10,1),c(99,98)) >> temp > [,1] [,2] > [1,] 101 > [2,] 99 98 > > >> dist(temp) > 1 > 2 131.6435 > > >> sqrt(dist(temp)) > 1 > 2 11.47360 > > so far so good. > > until the nex line: when I try to do what i did before but adding the 1/(what I > did before). I was expecting a number as a result of the division but > unfortunately I took the following: > > 1/sqrt(dist(temp)) > [1] 0.08715662 > attr(,"Size") > [1] 2 > attr(,"Diag") > [1] FALSE > attr(,"Upper") > [1] FALSE > attr(,"method") > [1] "euclidean" > attr(,"call") > dist(x = temp) > attr(,"class") > [1] "dist" > > > Could you please help me understand what is this about? > > I would like to thank you in advance for your help > Best REgards > Alex > > > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Ing. Mario Valle Data Analysis and Visualization Group| http://www.cscs.ch/~mvalle Swiss National Supercomputing Centre (CSCS) | Tel: +41 (91) 610.82.60 v. Cantonale Galleria 2, 6928 Manno, Switzerland | Fax: +41 (91) 610.82.82 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interpolate? a line
I would like to thank you again for your help. But it seems that the floor function (ceiling, round too) create more dots in the matrix that line really "touches". unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62, by=0.1)) ) ) ) You can see that in the picture below http://yfrog.com/5blineswj So, how to select the only the cells that the line "touches"? I would like to thank you in advance for your help best Regards Alex From: David Winsemius Cc: Rhelp list Sent: Wed, September 15, 2010 1:55:10 PM Subject: Re: [R] Interpolate? a line On Sep 15, 2010, at 7:24 AM, David Winsemius wrote: > Replacing context: > >>> Hello everyone. >>> I have created a 100*100 matrix in R. >>> Let's now say that I have a line that starts from (2,3) point and ends to the >>> (62,34) point. In other words this line starts at cell (2,3) and ends at cell >>> (62,34). >>> >>> Is it possible to get by some R function all the matrix's cells that this >line >>> transverses? >>> >>> I would like to thank you for your feedback. >>> >>> Best Regards >>> Alex > > On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote: > >> Hello Alex, >> >> Here is one way to do it. It works but it's not pretty :) > > If you want an alternative, consider that produces the Y cell indices (since >the x cell indices are already 2:62): > > > linefn <- function(x) 3+((34-3)/(62-2)) *(x-2) > > findInterval(linefn(2:62), 3:34) > [1] 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 > 13 >13 14 > [28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 > 27 >27 28 > [55] 28 29 29 30 30 31 32 > # that seems "off" by two > > linefn(62) > [1] 34 > > linefn(2) > [1] 3 # but that checks out and I realized those were just indices for the > 3:34 >findInterval vector > > > (3:34)[findInterval(linefn(2:62), 3:34)] > [1] 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 > 15 >15 16 > [28] 16 17 17 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 > 29 >29 30 > [55] 30 31 31 32 32 33 34 > > ( no rounding and I think the logic is clearer.) But I also realized it didn't enumerate all the the cells were crossed either, only indicating which cell was associated with an integer value of x. Also would have even more serious problems if the slope were greater than unity. To enumerate the cell indices that were crossed, try: unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62, by=0.1)) ) ) ) [,1] [,2] [1,]23 [2,]33 [3,]44 [4,]54 [5,]55 [6,]65 [7,]75 [8,]76 snipping interior results [83,] 58 32 [84,] 59 32 [85,] 60 32 [86,] 60 33 [87,] 61 33 [88,] 62 34 That could probably be passed to rect() to illustrate (and check logic): rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1, col="red") #redraw line : lines(2:62, 3+(34-3)/(62-2)*(0:60)) > > --David. > >> >> interp <- approx(c(2, 62), c(3, 34), method="linear", xout=2:62) >> m <- matrix(c(interp$x, round(interp$y)), ncol=2) >> tie <- m[,2] == c(-Inf, m[-nrow(m),2]) >> m <- m[ !tie, ] >> >> You might want to examine the result like this... >> >> plot(m) # plots points >> lines(c(2,26), c(3, 34)) # overlay line for comparison > you can add a grid with > abline(v=2:62, h=3:34) >> >> Michael >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. > > David Winsemius, MD > West Hartford, CT > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] count frequency
Hello everyone, please consider the following lines of a matrix [574,] 59 32 [575,] 59 32 [576,] 59 32 [577,] 59 32 [578,] 59 32 [579,] 59 32 [580,] 59 32 [581,] 60 32 [582,] 60 33 [583,] 60 33 [584,] 60 33 [585,] 60 33 [586,] 60 33 [587,] 60 33 [588,] 60 33 [589,] 60 33 [590,] 60 33 [591,] 61 33 [592,] 61 33 [593,] 61 33 [594,] 61 33 [595,] 61 33 [596,] 61 33 [597,] 61 33 [598,] 61 33 [599,] 61 33 [600,] 61 33 [601,] 62 34 Is it possible somehow to count the similarities between the first and second column and put them on a third column like this? 59 32 3 60 33 5 62 34 1 where (3,5,1 are the frequencies for (59,32), (60,33) and (62,34) Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] count frequency
Just to confirm, this one worked for me as.data.frame(table(temp[])) Var1 Freq 1 1 11 2 2 29 3 3 29 4 4 29 5 5 29 6 6 29 7 7 29 8 8 29 9 9 29 10 10 29 11 11 29 12 12 29 13 13 29 14 14 29 15 15 29 16 16 29 17 17 29 18 18 30 19 19 29 20 20 29 Best Regards Alex From: Jorge Ivan Velez Cc: Rhelp Sent: Fri, September 17, 2010 4:24:59 PM Subject: Re: [R] count frequency Alaios, Try as.data.frame(table(x[,1], x[,2)) where x is your matrix. HTH, Jorge On Fri, Sep 17, 2010 at 8:19 AM, Alaios <> wrote: Hello everyone, >please consider the following lines of a matrix > > >[574,] 59 32 >[575,] 59 32 >[576,] 59 32 >[577,] 59 32 >[578,] 59 32 >[579,] 59 32 >[580,] 59 32 >[581,] 60 32 >[582,] 60 33 >[583,] 60 33 >[584,] 60 33 >[585,] 60 33 >[586,] 60 33 >[587,] 60 33 >[588,] 60 33 >[589,] 60 33 >[590,] 60 33 >[591,] 61 33 >[592,] 61 33 >[593,] 61 33 >[594,] 61 33 >[595,] 61 33 >[596,] 61 33 >[597,] 61 33 >[598,] 61 33 >[599,] 61 33 >[600,] 61 33 >[601,] 62 34 > >Is it possible somehow to count the similarities between the first and second >column and put them on a third column like this? > >59 32 3 >60 33 5 >62 34 1 > >where (3,5,1 are the frequencies for (59,32), (60,33) and (62,34) > >Best Regards >Alex > > > > > [[alternative HTML version deleted]] > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] count frequency
My bad :( unfortunately does not work correct. This is some of the output of the table .. [494,] 50 27 [495,] 50 27 [496,] 50 27 [497,] 50 28 [498,] 50 28 [499,] 50 28 [500,] 50 28 [501,] 51 28 [502,] 51 28 [503,] 51 28 [504,] 51 28 [505,] 51 28 [506,] 51 28 [507,] 51 28 [508,] 51 28 [509,] 51 28 [510,] 51 28 [511,] 52 28 [512,] 52 28 [513,] 52 28 [514,] 52 28 [515,] 52 28 [516,] 52 29 [517,] 52 29 [518,] 52 29 I need the following results 50 27: 3 50 28: 4 51 28: 10 52 28: 5 52 29: 3 So far nothing worked. Best Regards Alex From: Henrique Dallazuanna Cc: Rhelp Sent: Fri, September 17, 2010 3:09:36 PM Subject: Re: [R] count frequency Try this: aggregate(rep(1, nrow(x)), x, sum) Hello everyone, >please consider the following lines of a matrix > > >[574,] 59 32 >[575,] 59 32 >[576,] 59 32 >[577,] 59 32 >[578,] 59 32 >[579,] 59 32 >[580,] 59 32 >[581,] 60 32 >[582,] 60 33 >[583,] 60 33 >[584,] 60 33 >[585,] 60 33 >[586,] 60 33 >[587,] 60 33 >[588,] 60 33 >[589,] 60 33 >[590,] 60 33 >[591,] 61 33 >[592,] 61 33 >[593,] 61 33 >[594,] 61 33 >[595,] 61 33 >[596,] 61 33 >[597,] 61 33 >[598,] 61 33 >[599,] 61 33 >[600,] 61 33 >[601,] 62 34 > >Is it possible somehow to count the similarities between the first and second >column and put them on a third column like this? > >59 32 3 >60 33 5 >62 34 1 > >where (3,5,1 are the frequencies for (59,32), (60,33) and (62,34) > >Best Regards >Alex > > > > > [[alternative HTML version deleted]] > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] count frequency
You are pretty good. Worked nicely :) Thanks! Alex From: Henrique Dallazuanna Cc: Rhelp Sent: Fri, September 17, 2010 4:58:50 PM Subject: Re: [R] count frequency So try this : aggregate(rep(1, nrow(x)), as.data.frame(x), sum) My bad :( >unfortunately does not work correct. > >This is some of the output of the table >.. >[494,] 50 27 >[495,] 50 27 >[496,] 50 27 >[497,] 50 28 >[498,] 50 28 >[499,] 50 28 >[500,] 50 28 >[501,] 51 28 >[502,] 51 28 >[503,] 51 28 >[504,] 51 28 >[505,] 51 28 >[506,] 51 28 >[507,] 51 28 >[508,] 51 28 >[509,] 51 28 >[510,] 5128 >[511,] 52 28 >[512,] 52 28 >[513,] 52 28 >[514,] 52 28 >[515,] 52 28 >[516,] 52 29 >[517,] 52 29 >[518,] 52 29 > > > > >I need the following results >50 27: 3 >50 28: 4 >51 28: 10 >52 28: 5 >52 29: 3 > > >So far nothing worked. >Best Regards >Alex > > > > > From: Henrique Dallazuanna > >Cc: Rhelp >Sent: Fri, September 17, 2010 3:09:36 PM > >Subject: Re: [R] count frequency > > >Try this: > >aggregate(rep(1, nrow(x)), x, sum) > > > >Hello everyone, >>please consider the following lines of a matrix >> >> >>[574,] 59 32 >>[575,] 59 32 >>[576,] 59 32 >>[577,] 59 32 >>[578,] 59 32 >>[579,] 59 32 >>[580,] 59 32 >>[581,] 60 32 >>[582,] 60 33 >>[583,] 60 33 >>[584,] 60 33 >>[585,] 60 33 >>[586,] 60 33 >>[587,] 60 33 >>[588,] 60 33 >>[589,] 60 33 >>[590,] 60 33 >>[591,] 61 33 >>[592,] 61 33 >>[593,] 61 33 >>[594,] 61 33 >>[595,] 61 33 >>[596,] 61 33 >>[597,] 61 33 >>[598,] 61 33 >>[599,] 61 33 >>[600,] 61 33 >>[601,] 62 34 >> >>Is it possible somehow to count the similarities between the first and second >>column and put them on a third column like this? >> >>59 32 3 >>60 33 5 >>62 34 1 >> >>where (3,5,1 are the frequencies for (59,32), (60,33) and (62,34) >> >>Best Regards >>Alex >> >> >> >> >> [[alternative HTML version deleted]] >> >>__ >>R-help@r-project.org mailing list >>https://stat.ethz.ch/mailman/listinfo/r-help >>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>and provide commented, minimal, self-contained, reproducible code. >> > > >-- >Henrique Dallazuanna >Curitiba-Paraná-Brasil >25° 25' 40" S 49° 16' 22" O > > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Interpolate? a line
I would like to thank you very much for making this clear. It seems that the solution you suggested is right one as the second attempt does find all the cells that are touched. Now I ll try to find out how much the line gets into one of this cell as every cell affects acts like a weight. The more you are in a cell the higher the weight will be. Best Regards Alex From: David Winsemius Cc: Rhelp list Sent: Fri, September 17, 2010 6:36:51 PM Subject: Re: [R] Interpolate? a line On Sep 17, 2010, at 7:22 AM, Alaios wrote: > I would like to thank you again for your help. > But it seems that the floor function (ceiling, round too) create more dots in >the matrix that line really "touches". You said "cells" not "dots". Are you trying to change the problem now? My concern is rather that it can still miss cells. > > unique( floor( cbind( seq(2,62, by=0.1), linefn(seq(2,62, by=0.1)) ) ) ) > > You can see that in the picture below > > http://yfrog.com/5blineswj > > So, how to select the only the cells that the line "touches"? If you had taken my suggestion of overlaying a grid rather than plotting dots that fail to represent a "cell" (which I was taking to be a square of dimension 1 x 1) you would see that my solution was correct (at least to the point of not missing any cells so defined that were touched up to a tolerance of 0.01 cell units. If you want to define "cells" differently, then it's your turn to step up and get mathematically precise. Calculus still works if you define neighborhoods as hyperspheres s rather than epsilon by delta hyper-rectangles. # Here is the the illustrated sequence of getting to what I am calling my "final answer", # even though it could still miss an occasional cell. interp <- approx(c(2, 62), c(3, 34), method="linear", xout=2:62) m <- matrix(c(interp$x, round(interp$y)), ncol=2) tie <- m[,2] == c(-Inf, m[-nrow(m),2]) m <- m[ !tie, ] plot(m) # plots points lines(c(2,62), c(3, 34)) # overlay line for comparison #you can add a grid with abline(v=2:62, h=3:34) ## First attempt at integer values of x linefn <- function(x) 3+((34-3)/(62-2)) *(x-2) findInterval(linefn(2:62), 3:34) # Second attempt at 0.1 intervals # cellidxs <- unique( floor( cbind( seq(2,62, by=0.1), # There will be many duplicates after rounding down linefn(seq(2,62, by=0.1)) ) ) ) # the same function that just gets a y value rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1, col="red") #redraw line : lines(2:62, 3+(34-3)/(62-2)*(0:60)) # That is the first plot with coarse tolerances #Third attempt: # Now calculate a set of cell ids with tolerances that at ten-fold more numerous cellid2 <-unique( floor(cbind(seq(2,62, by=0.01), linefn(seq(2,62, by=0.01) )) ) ) NROW(cellid2) # 91 cells rect(cellid2[,1], cellid2[,2], cellid2[,1]+1, cellid2[,2]+1, col="blue") rect(cellidxs[,1], cellidxs[,2], cellidxs[,1]+1, cellidxs[,2]+1, col="red") lines(2:62, 3+(34-3)/(62-2)*(0:60)) --Best David. > > I would like to thank you in advance for your help > best Regards > Alex > > From: David Winsemius > Cc: Rhelp list > Sent: Wed, September 15, 2010 1:55:10 PM > Subject: Re: [R] Interpolate? a line > > > On Sep 15, 2010, at 7:24 AM, David Winsemius wrote: > > > Replacing context: > > > >>> Hello everyone. > >>> I have created a 100*100 matrix in R. > >>> Let's now say that I have a line that starts from (2,3) point and ends to >the > >>> (62,34) point. In other words this line starts at cell (2,3) and ends at >cell > >>> (62,34). > >>> > >>> Is it possible to get by some R function all the matrix's cells that this >line > >>> transverses? > >>> > >>> I would like to thank you for your feedback. > >>> > >>> Best Regards > >>> Alex > > > > On Sep 15, 2010, at 6:52 AM, Michael Bedward wrote: > > > >> Hello Alex, > >> > >> Here is one way to do it. It works but it's not pretty :) > > > > If you want an alternative, consider that produces the Y cell indices > > (since >the x cell indices are already 2:62): > > > > > linefn <- function(x) 3+((34-3)/(62-2)) *(x-2) > > > findInterval(linefn(2:62), 3:34) > > [1] 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 >13 13 14 > > [28] 14 15 15 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 > > 26 >27 27 28 > > [55] 28 29 29 30 30 31 32 > > # that seems "off" by two > > > linefn(62) >
[R] Indexing sublists inside lists.
I would like to thank you very much for your reply. Actually I would like to ask you if there is a small list called fred: fred <- list(happy = 1:10, name = "squash") and a big list called bigfred that included fred list 5 times bigfred <- rep(fred,5) Is it possible somehow to index all these sublists(fred) inside bigfred with a more direct way like bigfred[1] shows the first sublist fred bigfred[2][2] shows the second sublist fred, the second element of the fred list So far I found some way to do this by refering to the sublists by the following: bigfred[1+index*length(fred)] where index shows the beginning of a sublist. I would like to thank you in advance for your help Best Regards Alex From: David Winsemius Cc: Dennis Murphy ; Rhelp Sent: Tue, September 14, 2010 3:55:39 PM Subject: Re: [R] Object oriented programming in R. On Sep 14, 2010, at 9:29 AM, Alaios wrote: > I would like to thank you very much all that you helped me so far. > So I tried to check how the following works > > fred <- list(happy = 1:10, name = "squash") > rep(fred, 5) > > This returns the following : > >> fred[1] > $happy > [1] 1 2 3 4 5 6 7 8 9 10 > > >> fred[2] > $name > [1] "squash" Not on my machine: > fred <- list(happy = 1:10, name = "squash") > rep(fred, 5) $happy [1] 1 2 3 4 5 6 7 8 9 10 $name [1] "squash" $happy [1] 1 2 3 4 5 6 7 8 9 10 $name [1] "squash" $happy [1] 1 2 3 4 5 6 7 8 9 10 $name [1] "squash" $happy [1] 1 2 3 4 5 6 7 8 9 10 $name [1] "squash" $happy [1] 1 2 3 4 5 6 7 8 9 10 $name [1] "squash" > > What I am trying to do is to address the number 5 of the fred[1] $happy value. > I tried something like fred[1][5] fred[1,5] > but it didn't work Almost: > fred[[1]][5] [1] 5 > > I would like to thank you in advance for your help > > Best Regards > Alex > > > From: Dennis Murphy > > Cc: Rhelp > Sent: Tue, September 14, 2010 3:13:37 PM > Subject: Re: [R] Object oriented programming in R. > > Hi: > > You could create a list of lists, where the outer list would be between agents > and the inner list within agents. The inner list could have the 'matrices and > one list' as separate components for each agent. Of course, you would have to >be > able to keep all of this straight :) > > HTH, > Dennis > > > Here are some more information: >> I would like to create some agents that span over a specific area map.Every >> agent needs to have its own data structures like one or two matrices and one >> list. >> >> I think that the best way to do this is to create objects and every instance >of >> an object will be used for a single agent. >> >> The number of agents is not predetermined and it varies for any execution. >> So I read this value from the command line interface and then I would like to >> initiate so many objects as the agents. I think that the best way to do that >is >> to create using a for loop a list containing as many objects as the agents >are. >> >> >> I would like to thank you in advance for your help >> >> Best Regards >> Alex >> >> >> From: jim holtman >> >> Cc: Tal Galili ; Rhelp >> Sent: Tue, September 14, 2010 1:40:37 PM >> >> Subject: Re: [R] Object oriented programming in R. >> >> >> It depends on what you mean by objects. If you are just looking at >> creating many named variables that are going to hold values (e.g., >> reading in data from several files that you want to correlate >> separately), then consider the use of 'lists'. Can you provide a >> little more detail on exactly the problem that you are trying to >> solve, and then maybe we can propose a solution. >> >> >> >>> Thank you very much. I checked the tutorials that on that list but still I do >>> not know how to create many objects of the same type. Can you please help me >>> with that? >>> >>> Best Regards >>> Alex >>> >>> >>> >>> >>> >>> From: Tal Galili >>> >>> Cc: Rhelp >>> Sent: Tue, September 14, 2010 10:11:36 AM >>> Subject: Re: [R] Object oriented programming in R. >>> >>> >>> Hello Alaios, >>> I see a bunch of good materials here: >>>http://www.google.
[R] efficient list indexing
Hello everyone, I need some help with lists inside lists (a good way to emulate a struct)\ Assume that there is a small list called fred: fred <- list(happy = 1:10, name = "squash") and a big list called bigfred that includes fred list 5 times bigfred <- rep(fred,5) Is it possible somehow to index all these sublists(fred) inside bigfred with a more direct way like this: bigfred[1] shows the first sublist fred bigfred[2][2] shows the second sublist fred, the second element of the fred list So far I found some way to do this by refering to the sublists by the following: bigfred[1+index*length(fred)] where index shows the beginning of a sublist. I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient list indexing
Great one! Thanks this will simplify a lot addresing. Best REgards From: Keith Jewell To: r-h...@stat.math.ethz.ch Sent: Wed, September 22, 2010 1:25:27 PM Subject: Re: [R] efficient list indexing Also, it may be that you want bigfred to be a list of 5 lists each of 2 elements (happy and name) rather than a list of 10 elements. Thus (also using double bracketing) fred<- list(happy = 1:10, name = "squash") bigfred <- replicate(5, fred, FALSE) bigfred[[2]][[2]] hth Keith J "Patrick Burns" wrote in message news:4c99e118.2010...@pburns.seanet.com... > I'm confused by what you are looking for. > There's some slight possibility that you > are looking for double bracket subscripting > with a vector: > > > list(a=1:5, b=letters)[[c(2,4)]] > [1] "d" > > > On 22/09/2010 10:58, Alaios wrote: >> Hello everyone, >> >> I need some help with lists inside lists (a good way to emulate a >> struct)\ >> >> Assume that there is a small list called fred: >> fred<- list(happy = 1:10, name = "squash") >> >> >> and a big list called bigfred that includes fred list 5 times >> bigfred<- rep(fred,5) >> >> Is it possible somehow to index all these sublists(fred) inside bigfred >> with a >> more direct way like this: >> bigfred[1] shows the first sublist fred >> bigfred[2][2] shows the second sublist fred, the second element of the >> fred list >> >> So far I found some way to do this by refering to the sublists by the >> following: >> bigfred[1+index*length(fred)] >> where index shows the beginning of a sublist. >> >> I would like to thank you in advance for your help >> Best Regards >> Alex >> >> >> >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > -- > Patrick Burns > pbu...@pburns.seanet.com > http://www.burns-stat.com > (home of 'Some hints for the R beginner' > and 'The R Inferno') > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient list indexing
For some reason I did not receive your email. Sorry for the inconvenience caused From: Dennis Murphy Cc: Rhelp Sent: Wed, September 22, 2010 1:46:28 PM Subject: Re: [R] efficient list indexing Hi: I believe we had this discussion yesterday, http://r.789695.n4.nabble.com/Object-oriented-programming-in-R-td2538541.html#a2538916 but since you chose to repeat that message, it clearly wasn't enough, so start with http://cran.r-project.org/doc/manuals/R-intro.html#Lists http://stackoverflow.com/questions/2050790/how-to-correctly-use-lists-in-r With the following example as a reference: Empl <- list(employee = "Anna", spouse = "Fred", children = 3, child.ages = c(4, 7, 9)) we quote Venables and Ripley (2002, p. 45): "It is important to appreciate the difference between [ and [[. The [ form extracts sub-vectors, so Empl[2] is a list of length one, whereas Empl[[2]] is a component (a character vector of length one)." In that context, consider the following: > Empl$employee [1] "Anna" > Empl$child.ages[2] [1] 7 > x <- "spouse"; Empl[[x]] [1] "Fred" > Empl[[3]] [1] 3 > Empl[[4]] [1] 4 7 9 > Empl[[4]][2] [1] 7 > Empl[4] $child.ages [1] 4 7 9 > Empl[4][2] # Why? $ NULL > Empl[2] $spouse [1] "Fred" > Empl[[2]] # Why the difference? [1] "Fred" A somewhat more expansive discussion is found on pp. 12-13 of Venables and Ripley (2000), S Programming. Dennis Hello everyone, > >I need some help with lists inside lists (a good way to emulate a struct)\ > >Assume that there is a small list called fred: >fred <- list(happy = 1:10, name = "squash") > > >and a big list called bigfred that includes fred list 5 times >bigfred <- rep(fred,5) > >Is it possible somehow to index all these sublists(fred) inside bigfred with a >more direct way like this: >bigfred[1] shows the first sublist fred >bigfred[2][2] shows the second sublist fred, the second element of the fred list > >So far I found some way to do this by refering to the sublists by the following: >bigfred[1+index*length(fred)] >where index shows the beginning of a sublist. > >I would like to thank you in advance for your help >Best Regards >Alex > > > > > [[alternative HTML version deleted]] > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] print-show-display a matrix
Hello. I want to print the value a matrix has. The matrix's size is not too big (100*100 cells). I tried print(matrixname) but as it does not fit very well on my screen R splits it into several small matrixes that do overflow my screen. IS it possible somehow to display this matrix as one (even if this needs to go fullscreen) . There might be some function for that like plot cells or display matrixes. I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Make a loop more efficient
Hello everyone
I need some advice if the following might be easier implemented.
There are n matrixes
each matrix needs to calculate one value for the rest n-1 matrixes (but not for
itself). I implemented this one by
two nested loops
for (i in c(1:length(CRX))) {
for (j in c(1:length(CRX)))
if (i!=j) # where i and j are the two indexes. If i==j then is the same
matrix.
calculations();
}
Do you think there might be a more efficient R-wise way to implement this?
I would like to thank you in advance for your help
Best Regards
Alex
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and provide commented, minimal, self-contained, reproducible code.
[R] Linear Integration
Hello I would like to calculate a weighted line integral. The integral is calculated by the cells that this lines trasverses (the small cells belong to matrix (m*n) that represent the value that a specific area has. I need to calculate the weights by finding out how much the line touches or impinges inside a cell. The weight is less if a line just touches one of the four edges of a square and much more if it traverses inside the cell. I think there might be some function to calculate this in any image processing packages. So far I could not find anything appropriate. Could you please help me with that? I would like to thank you in advance for your help Best Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] print-show-display a matrix
Just to confirm that it works. Thanks for help. -- View this message in context: http://r.789695.n4.nabble.com/print-show-display-a-matrix-tp2954168p2964344.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Make a loop more efficient
Thank you very much both of you. I will check and post back later if needed. Best Regards Alex -- View this message in context: http://r.789695.n4.nabble.com/Make-a-loop-more-efficient-tp2955912p2964346.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Linear Integration
I would like to thank you for your reply. Yes I had this conversation of how to find the cells that are touched. I did that with these two lines: temp<-(floor(cbind(seq(x[1],xr[1],by=0.01),lineeq(x,xr #. cellid2 <-unique( floor(cbind(seq(x[1],xr[1], by=0.01), lineeq(x,xr)) ) ) # cell ids that are touched You can find in the picture below how the cells look like for a line that spans from (2,11) to (3,8) http://img824.imageshack.us/img824/9914/lineu.jpg In the picture attached you can see that there are some points missing denoted by * which I do not know how to find out. I need them to find the proportion a line gets into each cell. You mentioned something like "densified vertices of the line" but I am not sure what does it mean. In the picture are depicted the areas that I need to calculate somehow. Of course the easiest way would be to use some already implemented function which does not seem to exist. I would like to thank everyone that contributed to this so far. Best Regards Alex -- View this message in context: http://r.789695.n4.nabble.com/Linear-Integration-tp2956145p2964694.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Looking for a book/tutorial with the following context:
Hello everyone. It is time to start writing more and more function and I want to read in a good reference -book ( I can buy one, especially if it is second handed :P) -online tutorial -any other guide -How functions really work in R -How to write bigger R programs -If there are local function variables. -Global ones and how to treat them -How to include function files in order to keep your all increasing code into separate files -How to debug an R program step by step. -How to print debugging messages. I would like to thank you in advance for your help Best Regards Alex -- View this message in context: http://r.789695.n4.nabble.com/Looking-for-a-book-tutorial-with-the-following-context-tp2965206p2965206.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Linear Integration
No, because I thought something that might be easier. If you see the image again you might notice that the proportions I am looking for might also be found by using the hypotenuse which is the same (as all squares are triangles) by finding the adjacent. The adjacent are easier to be found by tracking when the x value changes. This happens when the x value increments by one and when y also increments by one. As I know the line equation y=a*x+b it is very easy all these points. Then it is straightforward to find the proportion of these vectors? Best Regards Alex -- View this message in context: http://r.789695.n4.nabble.com/Linear-Integration-tp2956145p2966318.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
