Re: [R] How to test if there is a subvector in a longer vector
On 28-09-2012, at 07:41, Atte Tenkanen wrote: > Sorry. I should have mentioned that the order of the components is important. > > So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of > c(2,1,1,6,4,3). > > How to test this? See this discussion for a variety of solutions. http://r.789695.n4.nabble.com/matching-a-sequence-in-a-vector-td4389523.html#a4393453 Berend __ 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] Running different Regressions using for loops
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10 regressions in "summ.list
<- lapply(lm.list2, summary)"
And now once I enter" sum.list "it gives me the output for all
the 10 regressions...
I wanted to access a beta coefficient of one of the regressionssay
"Price2+Media1+Trend+Seasonality"...the result of which is stored in "
sum.list[2] "
I entered the below statement for accessing the Beta coefficient for
Price2...
> summ.list[2]$coefficients[2]
NULL
But this is giving me " NULL " as the output...
What I am looking for, is to access a beta value of a particular variable
from a particular regression output and use it for further analysis.
Can you please help me out with this. Greatly appreciate, you guys
efforts.
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 27 September 2012 21:55
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Inline.
Em 27-09-2012 13:52, Krunal Nanavati escreveu:
> Hi,
>
> Thanks for all your help. I am stuck again, but with a new problem, on
> similar lines.
>
> I have taken the problem to the next step now...i have now added 2 "for"
> loops... 1 for the Price variable...and another for the Media variable
>
> I have taken 5 price variables...and 2 media variables with the "trend
> and seasonality"(appearing in all of them)so in all there will be
> 10 regression to run now
>
> Price 1, Media 1
>
> Price 1, Media 2
>
> Price 2, Media 1'
>
> Price 2, Media 2
>
> ...and so on
>
> I have built up a code for it...
>
>
>
>
>> tryout=read.table("C:\\Users\\Krunal\\Desktop\\R
> tryout.csv",header=T,sep=",")
>> cnames <- names(tryout)
>> price <- cnames[grep("Price", cnames)] media <- cnames[grep("Media",
>> cnames)] resp <- cnames[1] regr <- cnames[7:8] lm.list <-
>> vector("list", 10) for(i in 1:5)
> + {
> + regress <- paste(price[i], paste(regr, collapse = "+"), sep = "+")
> + for(j in 1:2) {
> + regress1 <- paste(media[j],regress,sep="+") fmla <- paste(resp,
> + regress1, sep = "~") lm.list[[i]] <- lm(as.formula(fmla), data =
> + tryout) } }
>> summ.list <- lapply(lm.list, summary) summ.list
>
>
>
>
>
> But it is only running...5 regressions...only Media 1 along with the 5
> Price variables & Trend & Seasonality is regressed on Volume...giving
> only
> 5 outputs
>
> I feel there is something wrong with the" lm.list[[i]] <-
> lm(as.formula(fmla), data = tryout)" statement.
No, I don't think so. If it's giving you only 5 outputs the error is
probably in the fmla construction. Put print statements to see the results
of those paste() instructions.
Supposing your data.frame is now called tryout2,
price <- paste("Price", 1:5, sep = "")
media <- paste("Media", 1:2, sep = "")
pricemedia <- apply(expand.grid(price, media, stringsAsFactors = FALSE),
1, paste, collapse="+")
response <- "Volume"
trendseason <- "Trend+Seasonality" # do this only once
lm.list2 <- list()
for(i in seq_along(pricemedia)){
regr <- paste(pricemedia[i], trendseason, sep = "+")
fmla <- paste(response, regr, sep = "~")
lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
The trick is to use ?expand.grid
Hope this helps,
Rui Barradas
> I am not sure about its
> placement...whether it should be in loop 2 or in loop 1
>
> Can you please help me out??
>
>
>
>
>
>
>
>
>
>
> Thanks & Regards,
>
> Krunal Nanavati
> 9769-919198
>
> -Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: 27 September 2012 16:22
> To: David Winsemius
> Cc: Krunal Nanavati; r-help@r-project.org
> Subject: Re: [R] Running different Regressions using for loops
>
> Hello,
>
> Just to add that you can also
>
> lapply(lm.list, coef)
>
> with a different output.
>
> Rui Barradas
> Em 27-09-2012 09:24, David Winsemius escreveu:
>> On Sep 26, 2012, at 10:31 PM, Krunal Nanavati wrote:
>>
>>> Dear Rui,
>>>
>>> Thanks for your time.
>>>
>>> I have a question though, when I run the 5 regression, whose outputs
>>> are stored in "lm.list[i]", I only get the coefficients for the
>>> Intercept, Price, Trend & Seasonality as below
>>>
>>>
lm.list[1]
>>> [[1]]
>>>
>>> Call:
>>>
>>> lm(formula = as.formula(fmla), data = tryout)
>>>
>>> Coefficients:
>>>
>>> (Intercept) Price4Trend Seasonality
>>>
>>> 9923123 -260682664616 551392
>> summ.list <- lapply(lm.list, summary) coef.list <- lapply(summ.list,
>> coef) coef.list
>>
>>> I am also looking out for t stats and p value and R squared.
>> For the r.squared
>>
>> rsq.vec <- sapply(summ.list, "$", "r.squared") adj.rsq <-
>> sapply(summ.list, "$", "adj.r.squared")
>>
>>> Do you know,
>>> how can I get all these statistics. Also, why is " as.formula " used
>>> in the lm function. It should work without that as well, right?
>> No.
>>> Can you pleas
[R] blank plot----how do I make symbols appear
Hi,
I am trying to create a scatterplot, coding each point to one of 5
populations. I was successful when I did this for one set of data, yet
when I try plotting other data a blank plot appears (although the axes are
labelled and I can fit the regression lines from each population). I have
tried a variety of things to fix this but nothing seems to work.
I can plot the points if I do not specify that I want each population to
have a particular symbol. However, once I add the command [grip$Morph] to
my symbol parameter (e.g., pch=c(2,6,5,19,15) [grip$morph] ), I loose all
the points. As I mentioned above, I was able to create a plot
successfully using other data points from the same table (different
columns), so I know the data are fine.
Has anyone come across this before?
R-script used:
HAND<-AllMal[,c(2,4,5)]
na.omit(HAND)->HAND
write.csv(HAND, "grip.csv")
read.csv("grip.csv")->grip
grip
class(grip)
class(HAND)
grip$morph<-as.character(grip$Morph)
morph<- grip$morph
BML<-grip$BML
grip$MCF->MCF
reg1<-lm(BML~MCF,data=subset(grip,morph=="mel"));reg1
reg2<-lm(BML~MCF,data=subset(grip,morph=="tham"));reg2
reg3<-lm(BML~MCF,data=subset(grip,morph=="A"));reg3
reg4<-lm(BML~MCF,data=subset(grip,morph=="B"));reg4
reg5<-lm(BML~MCF,data=subset(grip,morph=="C"));reg5
plot(MCF,BML,pch=c(2,6,5,19,15)[grip$morph],xlab="Residual Metacarpal
Length",ylab="Residual Hand Strength (Broad Dowel)", main="Males")
abline(reg1,lty=1)
abline(reg2,lty=2)
abline(reg3,lty=3)
abline(reg4,lty=4)
abline(reg5,lty=6)
--
*Jessica da Silva*
PhD Candidate
Molecular Ecology & Evolution Program
Applied Biodiversity Research
Kirstenbosch Research Centre
South African National Biodiversity Institute
Postal address:
3 Sangster Road
Howick, KZN
3290
Home/Fax: +27 33 330 2230
Cell: +27 79 045 1781
Email: jessica.m.dasi...@gmail.com
j.dasi...@sanbi.org.za
Website: http://jmdasilva.doodlekit.com/home/home
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] blank plot----how do I make symbols appear
Jessica da Silva gmail.com> writes:
> I am trying to create a scatterplot, coding each point to
one of 5
> populations. I was successful when I did this for one
set of data, yet
> when I try plotting other data a blank plot appears
(although the axes are
> labelled and I can fit the regression lines from each
population). I
However, once I add the command [grip$Morph] to
> my symbol parameter (e.g., pch=c(2,6,5,19,15) [grip$morph] ),
I loose all
> the points. As I mentioned above, I was able
to create a plot
> successfully using other data points from the
same table (different
> columns), so I know the data are fine.
>
Try
grip$morph<-unclass(grip$Morph)
instead. Look at what
as.character(factor(letters[1:3]))
gives you.
> R-script used:
>
> HAND<-AllMal[,c(2,4,5)]
> na.omit(HAND)->HAND
>
> write.csv(HAND, "grip.csv")
>
> read.csv("grip.csv")->grip
> grip
> class(grip)
> class(HAND)
>
> grip$morph<-as.character(grip$Morph)
>
> morph<- grip$morph
> BML<-grip$BML
> grip$MCF->MCF
>
> reg1<-lm(BML~MCF,data=subset(grip,morph=="mel"));reg1
> reg2<-lm(BML~MCF,data=subset(grip,morph=="tham"));reg2
> reg3<-lm(BML~MCF,data=subset(grip,morph=="A"));reg3
> reg4<-lm(BML~MCF,data=subset(grip,morph=="B"));reg4
> reg5<-lm(BML~MCF,data=subset(grip,morph=="C"));reg5
>
> plot(MCF,BML,pch=c(2,6,5,19,15)[grip$morph],xlab="Residual Metacarpal
> Length",ylab="Residual Hand Strength (Broad Dowel)", main="Males")
> abline(reg1,lty=1)
> abline(reg2,lty=2)
> abline(reg3,lty=3)
> abline(reg4,lty=4)
> abline(reg5,lty=6)
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running different Regressions using for loops
Hello, Krunal, try summ.list[[2]]$coefficients[2] Note the double square brackets (as summ.list is a list)! Hth, Gerrit On Fri, 28 Sep 2012, Krunal Nanavati wrote: Hi Rui, Excellent!! This is what I was looking for. Thanks for the help. So, now I have stored the result of the 10 regressions in "summ.list <- lapply(lm.list2, summary)" And now once I enter" sum.list "it gives me the output for all the 10 regressions... I wanted to access a beta coefficient of one of the regressionssay "Price2+Media1+Trend+Seasonality"...the result of which is stored in " sum.list[2] " I entered the below statement for accessing the Beta coefficient for Price2... summ.list[2]$coefficients[2] NULL But this is giving me " NULL " as the output... What I am looking for, is to access a beta value of a particular variable from a particular regression output and use it for further analysis. <<>> __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running different Regressions using for loops
Hello,
To access list elements you need `[[`, like this:
summ.list[[2]]$coefficients
Or Use the extractor function,
coef(summ.list[[2]])
Rui Barradas
Em 28-09-2012 07:23, Krunal Nanavati escreveu:
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10 regressions in "summ.list
<- lapply(lm.list2, summary)"
And now once I enter" sum.list "it gives me the output for all
the 10 regressions...
I wanted to access a beta coefficient of one of the regressionssay
"Price2+Media1+Trend+Seasonality"...the result of which is stored in"
sum.list[2] "
I entered the below statement for accessing the Beta coefficient for
Price2...
summ.list[2]$coefficients[2]
NULL
But this is giving me " NULL " as the output...
What I am looking for, is to access a beta value of a particular variable
from a particular regression output and use it for further analysis.
Can you please help me out with this. Greatly appreciate, you guys
efforts.
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 27 September 2012 21:55
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Inline.
Em 27-09-2012 13:52, Krunal Nanavati escreveu:
Hi,
Thanks for all your help. I am stuck again, but with a new problem, on
similar lines.
I have taken the problem to the next step now...i have now added 2 "for"
loops... 1 for the Price variable...and another for the Media variable
I have taken 5 price variables...and 2 media variables with the "trend
and seasonality"(appearing in all of them)so in all there will be
10 regression to run now
Price 1, Media 1
Price 1, Media 2
Price 2, Media 1'
Price 2, Media 2
...and so on
I have built up a code for it...
tryout=read.table("C:\\Users\\Krunal\\Desktop\\R
tryout.csv",header=T,sep=",")
cnames <- names(tryout)
price <- cnames[grep("Price", cnames)] media <- cnames[grep("Media",
cnames)] resp <- cnames[1] regr <- cnames[7:8] lm.list <-
vector("list", 10) for(i in 1:5)
+ {
+ regress <- paste(price[i], paste(regr, collapse = "+"), sep = "+")
+ for(j in 1:2) {
+ regress1 <- paste(media[j],regress,sep="+") fmla <- paste(resp,
+ regress1, sep = "~") lm.list[[i]] <- lm(as.formula(fmla), data =
+ tryout) } }
summ.list <- lapply(lm.list, summary) summ.list
But it is only running...5 regressions...only Media 1 along with the 5
Price variables & Trend & Seasonality is regressed on Volume...giving
only
5 outputs
I feel there is something wrong with the" lm.list[[i]] <-
lm(as.formula(fmla), data = tryout)" statement.
No, I don't think so. If it's giving you only 5 outputs the error is
probably in the fmla construction. Put print statements to see the results
of those paste() instructions.
Supposing your data.frame is now called tryout2,
price <- paste("Price", 1:5, sep = "")
media <- paste("Media", 1:2, sep = "")
pricemedia <- apply(expand.grid(price, media, stringsAsFactors = FALSE),
1, paste, collapse="+")
response <- "Volume"
trendseason <- "Trend+Seasonality" # do this only once
lm.list2 <- list()
for(i in seq_along(pricemedia)){
regr <- paste(pricemedia[i], trendseason, sep = "+")
fmla <- paste(response, regr, sep = "~")
lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
The trick is to use ?expand.grid
Hope this helps,
Rui Barradas
I am not sure about its
placement...whether it should be in loop 2 or in loop 1
Can you please help me out??
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 27 September 2012 16:22
To: David Winsemius
Cc: Krunal Nanavati; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Just to add that you can also
lapply(lm.list, coef)
with a different output.
Rui Barradas
Em 27-09-2012 09:24, David Winsemius escreveu:
On Sep 26, 2012, at 10:31 PM, Krunal Nanavati wrote:
Dear Rui,
Thanks for your time.
I have a question though, when I run the 5 regression, whose outputs
are stored in "lm.list[i]", I only get the coefficients for the
Intercept, Price, Trend & Seasonality as below
lm.list[1]
[[1]]
Call:
lm(formula = as.formula(fmla), data = tryout)
Coefficients:
(Intercept) Price4Trend Seasonality
9923123 -260682664616 551392
summ.list <- lapply(lm.list, summary) coef.list <- lapply(summ.list,
coef) coef.list
I am also looking out for t stats and p value and R squared.
For the r.squared
rsq.vec <- sapply(summ.list, "$", "r.squared") adj.rsq <-
sapply(summ.list, "$", "adj.r.squared")
Do you know,
how can I get all these statistics. Also, why is " as.formula " used
in the lm function. It should work without that as well, right?
No.
Can you please tell me, why the code that I had written, does
Re: [R] Drawing asymmetric error bars
On 09/27/2012 08:59 PM, Alexandra Howe wrote: Hello, I have data which I have arcsin transformed to analyse. I want to plot my data with error bars however as my data is back-transformed my standard errors are uneven. Is there a simple way to draw these asymmetric error bars in R? Hi Alexandra, Have a look at the "dispersion" function in the plotrix package. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Crosstable-like analysis (ks test) of dataframe
Hi, I have a dataframe with multiple (appr. 20) columns containing vectors of different values (different distributions). Now I'd like to create a crosstable where I compare the distribution of each vector (df-column) with each other. For the comparison I want to use the ks.test(). The result should contain as row and column names the column names of the input dataframe and the cells should be populated with the p-value of the ks.test for each pairwise analysis. My data.frame looks like: df <- data.frame(X=rnorm(1000,2),Y=rnorm(1000,1),Z=rnorm(1000,2)) And the test for one single case is: ks <- ks.test(df$X,df$Z) where the p value is: ks[2] How can I create an automatized way of this pairwise analysis? Any suggestions? I guess that is a quite common analysis (probably with other tests). cheers, Johannes __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Crosstable-like analysis (ks test) of dataframe
Hello,
Try the following.
f <- function(x, y, ...,
alternative = c("two.sided", "less", "greater"), exact = NULL){
#w <- getOption("warn")
#options(warn = -1) # ignore warnings
p <- ks.test(x, y, ..., alternative = alternative, exact =
exact)$p.value
#options(warn = w)
p
}
n <- 1e1
dat <- data.frame(X=rnorm(n), Y=runif(n), Z=rchisq(n, df=3))
apply(dat, 2, function(x) apply(dat, 2, function(y) f(x, y)))
Hope this helps,
Rui Barradas
Em 28-09-2012 11:10, Johannes Radinger escreveu:
Hi,
I have a dataframe with multiple (appr. 20) columns containing
vectors of different values (different distributions).
Now I'd like to create a crosstable
where I compare the distribution of each vector (df-column) with
each other. For the comparison I want to use the ks.test().
The result should contain as row and column names the column names
of the input dataframe and the cells should be populated with
the p-value of the ks.test for each pairwise analysis.
My data.frame looks like:
df <- data.frame(X=rnorm(1000,2),Y=rnorm(1000,1),Z=rnorm(1000,2))
And the test for one single case is:
ks <- ks.test(df$X,df$Z)
where the p value is:
ks[2]
How can I create an automatized way of this pairwise analysis?
Any suggestions? I guess that is a quite common analysis (probably with
other tests).
cheers,
Johannes
__
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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.
__
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] Running different Regressions using for loops
Hello,
Try
names(lm.list2[[2]]$coefficient[2] )
Rui Barradas
Em 28-09-2012 11:29, Krunal Nanavati escreveu:
Ok...this solves a part of my problem
When I type " lm.list2[2] " ...I get the following output
[[1]]
Call:
lm(formula = as.formula(fmla), data = tryout2)
Coefficients:
(Intercept) Price2 Media1 Distri1Trend
Seasonality
13491232 -5759030-15203437048628
445351
When I enter " lm.list2[[2]]$coefficient[2] " it gives me the below
output
Price2
-5759030
And when I enter " lm.list2[[2]]$coefficient[[2]] " ...I get the
number...which is -5759030
I am looking out for a way to get just the " Price2 "is there a
statement for that??
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 28 September 2012 15:18
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
To access list elements you need `[[`, like this:
summ.list[[2]]$coefficients
Or Use the extractor function,
coef(summ.list[[2]])
Rui Barradas
Em 28-09-2012 07:23, Krunal Nanavati escreveu:
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10 regressions in
"summ.list
<- lapply(lm.list2, summary)"
And now once I enter" sum.list "it gives me the output for
all
the 10 regressions...
I wanted to access a beta coefficient of one of the regressionssay
"Price2+Media1+Trend+Seasonality"...the result of which is stored in"
sum.list[2] "
I entered the below statement for accessing the Beta coefficient for
Price2...
summ.list[2]$coefficients[2]
NULL
But this is giving me " NULL " as the output...
What I am looking for, is to access a beta value of a particular
variable from a particular regression output and use it for further
analysis.
Can you please help me out with this. Greatly appreciate, you guys
efforts.
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 27 September 2012 21:55
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Inline.
Em 27-09-2012 13:52, Krunal Nanavati escreveu:
Hi,
Thanks for all your help. I am stuck again, but with a new problem,
on similar lines.
I have taken the problem to the next step now...i have now added 2
"for"
loops... 1 for the Price variable...and another for the Media
variable
I have taken 5 price variables...and 2 media variables with the
"trend and seasonality"(appearing in all of them)so in all there
will be
10 regression to run now
Price 1, Media 1
Price 1, Media 2
Price 2, Media 1'
Price 2, Media 2
...and so on
I have built up a code for it...
tryout=read.table("C:\\Users\\Krunal\\Desktop\\R
tryout.csv",header=T,sep=",")
cnames <- names(tryout)
price <- cnames[grep("Price", cnames)] media <- cnames[grep("Media",
cnames)] resp <- cnames[1] regr <- cnames[7:8] lm.list <-
vector("list", 10) for(i in 1:5)
+ {
+ regress <- paste(price[i], paste(regr, collapse = "+"), sep = "+")
+ for(j in 1:2) {
+ regress1 <- paste(media[j],regress,sep="+") fmla <- paste(resp,
+ regress1, sep = "~") lm.list[[i]] <- lm(as.formula(fmla), data =
+ tryout) } }
summ.list <- lapply(lm.list, summary) summ.list
But it is only running...5 regressions...only Media 1 along with the
5 Price variables & Trend & Seasonality is regressed on
Volume...giving only
5 outputs
I feel there is something wrong with the" lm.list[[i]] <-
lm(as.formula(fmla), data = tryout)" statement.
No, I don't think so. If it's giving you only 5 outputs the error is
probably in the fmla construction. Put print statements to see the
results of those paste() instructions.
Supposing your data.frame is now called tryout2,
price <- paste("Price", 1:5, sep = "") media <- paste("Media", 1:2,
sep = "") pricemedia <- apply(expand.grid(price, media,
stringsAsFactors = FALSE), 1, paste, collapse="+")
response <- "Volume"
trendseason <- "Trend+Seasonality" # do this only once
lm.list2 <- list()
for(i in seq_along(pricemedia)){
regr <- paste(pricemedia[i], trendseason, sep = "+")
fmla <- paste(response, regr, sep = "~")
lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
The trick is to use ?expand.grid
Hope this helps,
Rui Barradas
I am not sure about its
placement...whether it should be in loop 2 or in loop 1
Can you please help me out??
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 27 September 2012 16:22
To: David Winsemius
Cc: Krunal Nanavati; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Just to add that you can also
lapply(lm.list, coef)
with a different output.
Re: [R] changing outlier shapes of boxplots using lattice
I would guess that if you find the bit that says pch="|" and change it to
pch=1 it will solve your question, and that reading ?par will tell you why.
Sarah
On Thursday, September 27, 2012, Elaine Kuo wrote:
> Hello
>
> This is Elaine.
>
> I am using package lattice to generate boxplots.
> Using Richard's code, the display was almost perfect except the outlier
> shape.
> Based on the following code, the outliers are vertical lines.
> However, I want the outliers to be empty circles.
> Please kindly help how to modify the code to change the outlier shapes.
> Thank you.
>
> code
> package (lattice)
>
> dataN <- data.frame(GE_distance=rnorm(260),
>
> Diet_B=factor(rep(1:13, each=20)))
>
> Diet.colors <- c("forestgreen", "darkgreen","chocolate1","darkorange2",
>
> "sienna2","red2","firebrick3","saddlebrown","coral4",
>
> "chocolate4","darkblue","navy","grey38")
>
> levels(dataN$Diet_B) <- Diet.colors
>
> bwplot(GE_distance ~ Diet_B, data=dataN,
>
>xlab=list("Diet of Breeding Ground", cex = 1.4),
>
>ylab = list(
>
> "Distance between Centers of B and NB Range (1000 km)",
>
> cex = 1.4),
>
>panel=panel.bwplot.intermediate.hh,
>
>col=Diet.colors,
>
>pch=rep("|",13),
>
>scales=list(x=list(rot=90)),
>
>par.settings=list(box.umbrella=list(lty=1)))
>
> [[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.
>
--
Sarah Goslee
http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org
[[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] Running different Regressions using for loops
Ok, if I'm understanding it well, you want the mean value of Price1, ,
Price5? I don't know if it makes any sense, the coefficients already are
mean values, but see if this is it.
price.coef <- sapply(lm.list, function(x) coef(x)[2])
mean(price.coef)
Rui Barradas
Em 28-09-2012 12:07, Krunal Nanavati escreveu:
Hi,
Yes the thing that you provided...works finebut probably I should have
asked for some other thing.
Here is what I am trying to do
I am trying to get the mean of Price variableso I am entering the
below function:
mean(names(lm.list2[[2]]$coefficient[2] ))
but this gives me an error
[1] NA
Warning message:
In mean.default(names(lm.list2[[2]]$coefficient[2])) :
argument is not numeric or logical: returning NA
I thought by getting the text from the list variable...will help me
generate the mean for that text...which is a variable in the data...say
Price 1, Media 2and so on
Is this a proper approach...if it is...then something more needs to be
done with the function that you provided.
If not, is there a better way...to generate the mean of a particular
variable inside the " for loop " used earlier...given below:
lm.list2 <- list()
for(i in seq_along(pricemedia)){
regr <- paste(pricemedia[i], trendseason, sep = "+")
fmla <- paste(response, regr, sep = "~")
lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 28 September 2012 16:02
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Try
names(lm.list2[[2]]$coefficient[2] )
Rui Barradas
Em 28-09-2012 11:29, Krunal Nanavati escreveu:
Ok...this solves a part of my problem
When I type " lm.list2[2] " ...I get the following output
[[1]]
Call:
lm(formula = as.formula(fmla), data = tryout2)
Coefficients:
(Intercept) Price2 Media1 Distri1Trend
Seasonality
13491232 -5759030-15203437048628
445351
When I enter " lm.list2[[2]]$coefficient[2] " it gives me the below
output
Price2
-5759030
And when I enter " lm.list2[[2]]$coefficient[[2]] " ...I get the
number...which is -5759030
I am looking out for a way to get just the " Price2 "is there a
statement for that??
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 28 September 2012 15:18
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
To access list elements you need `[[`, like this:
summ.list[[2]]$coefficients
Or Use the extractor function,
coef(summ.list[[2]])
Rui Barradas
Em 28-09-2012 07:23, Krunal Nanavati escreveu:
Hi Rui,
Excellent!! This is what I was looking for. Thanks for the help.
So, now I have stored the result of the 10 regressions in
"summ.list
<- lapply(lm.list2, summary)"
And now once I enter" sum.list "it gives me the output for
all
the 10 regressions...
I wanted to access a beta coefficient of one of the
regressionssay "Price2+Media1+Trend+Seasonality"...the result of
which is stored in"
sum.list[2] "
I entered the below statement for accessing the Beta coefficient for
Price2...
summ.list[2]$coefficients[2]
NULL
But this is giving me " NULL " as the output...
What I am looking for, is to access a beta value of a particular
variable from a particular regression output and use it for further
analysis.
Can you please help me out with this. Greatly appreciate, you guys
efforts.
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 27 September 2012 21:55
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Inline.
Em 27-09-2012 13:52, Krunal Nanavati escreveu:
Hi,
Thanks for all your help. I am stuck again, but with a new problem,
on similar lines.
I have taken the problem to the next step now...i have now added 2
"for"
loops... 1 for the Price variable...and another for the Media
variable
I have taken 5 price variables...and 2 media variables with the
"trend and seasonality"(appearing in all of them)so in all there
will be
10 regression to run now
Price 1, Media 1
Price 1, Media 2
Price 2, Media 1'
Price 2, Media 2
...and so on
I have built up a code for it...
tryout=read.table("C:\\Users\\Krunal\\Desktop\\R
tryout.csv",header=T,sep=",")
cnames <- names(tryout)
price <- cnames[grep("Price", cnames)] media <-
cnames[grep("Media", cnames)] resp <- cnames[1] regr <- cnames[7:8]
lm.list <- vector("list", 10) for(i in 1:5)
+ {
+ regress <- paste(price[i], paste(regr, collapse = "+"), sep = "+")
+ for(j in 1:2) {
+ regre
Re: [R] What to use for ti in back-transforming summary statistics from F-T double square-root transformation in 'metafor'
Dear Chunyan, One possibility would be to use the harmonic mean of the person-time at risk values. You will have to do this manually though at the moment. Here is an example: ### let's just use the treatment group data from dat.warfarin data(dat.warfarin) dat <- escalc(xi=x1i, ti=t1i, measure="IRFT", data=dat.warfarin, append=TRUE) dat ### check if back-transformation of individual IRFT values works transf.iirft(dat$yi, ti=dat$t1i) escalc(xi=x1i, ti=t1i, measure="IR", data=dat.warfarin)$yi ### random-effects models res <- rma(yi, vi, data=dat) res ### harmonic mean of the ti's ti.hm <- 1/(mean(1/dat$t1i)) ### back-transformation using the harmonic mean transf.iirft(res$b, ti=ti.hm) transf.iirft(res$ci.lb, ti=ti.hm) transf.iirft(res$ci.ub, ti=ti.hm) Best, Wolfgang -- Wolfgang Viechtbauer, Ph.D., Statistician Department of Psychiatry and Psychology School for Mental Health and Neuroscience Faculty of Health, Medicine, and Life Sciences Maastricht University, P.O. Box 616 (VIJV1) 6200 MD Maastricht, The Netherlands +31 (43) 388-4170 | http://www.wvbauer.com From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of Liu, Chunyan [chunyan@cchmc.org] Sent: Thursday, September 27, 2012 10:48 PM To: r-help@R-project.org Subject: [R] What to use for ti in back-transforming summary statistics from F-T double square-root transformation in 'metafor' Hi Dr. Viechtbauer, I'm doing meta-analysis using your package 'metafor'. I used the 'IRFT' to transform the incident rate. But when I tried to back-transform the summary estimates from function rma, I don't know what's the appropriate ti to feed in function transf.iirft. I searched and found your post about using harmonic mean for ni to back-transform the double arcsine transformation. I'm hoping I can get your help on ti too. Thanks. Chunyan Liu 513-636-9763 Biostatistician II Department of Biostatistics and Epidemiology Cincinnati Children's Hospital Medical Center __ 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] Anova and tukey-grouping
Hello, I am really new to R and it's still a challenge to me. Currently I'm working on my Master's Thesis. My supervisor works with SAS and is not familiar with R at all. I want to run an Anova, a tukey-test and as a result I want to have the tukey-grouping ( something like A - AB - B) I came across the HSD.test in the agricolae-package, but... unfortunately I do not get an output (like here in the answer http://stats.stackexchange.com/questions/31547/how-to-obtain-the-results-of-a-tukey-hsd-post-hoc-test-in-a-table-showing-groupe ) I did it like this: ## ANOVA anova.typabunmit<-aov(ds.typabunmit$abun ~ ds.typabunmit$typ) summary(anova.typabunmit) summary.lm(anova.typabunmit) ## post HOC tukey.typabunmit<-TukeyHSD(anova.typabunmit) tukey.typabunmit ## HSD HSD.test(anova.typabunmit, "abun", group=TRUE) and the ONLY output is this: Name: abun ds.typabunmit$typ I would be very pleased about some ides..:! -- View this message in context: http://r.789695.n4.nabble.com/Anova-and-tukey-grouping-tp4644485.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Is it possible to enter in a function wich is within a library ?
Hello, I'd like to know if it is Ipossible to enter in a function wich is included in a library ? I know how to debug function wich is in a R file (but not in a library). But it is not the case when the function is included in a library. I want to go step by step in this function in order to test objects 'values. I tried debug(the_function) but the program does not stop at the_function (it only shows the body of the function). Thanks for your help. -- View this message in context: http://r.789695.n4.nabble.com/Is-it-possible-to-enter-in-a-function-wich-is-within-a-library-tp4644488.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running different Regressions using for loops
Ok...this solves a part of my problem
When I type " lm.list2[2] " ...I get the following output
[[1]]
Call:
lm(formula = as.formula(fmla), data = tryout2)
Coefficients:
(Intercept) Price2 Media1 Distri1Trend
Seasonality
13491232 -5759030-15203437048628
445351
When I enter " lm.list2[[2]]$coefficient[2] " it gives me the below
output
Price2
-5759030
And when I enter " lm.list2[[2]]$coefficient[[2]] " ...I get the
number...which is -5759030
I am looking out for a way to get just the " Price2 "is there a
statement for that??
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 28 September 2012 15:18
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
To access list elements you need `[[`, like this:
summ.list[[2]]$coefficients
Or Use the extractor function,
coef(summ.list[[2]])
Rui Barradas
Em 28-09-2012 07:23, Krunal Nanavati escreveu:
> Hi Rui,
>
> Excellent!! This is what I was looking for. Thanks for the help.
>
> So, now I have stored the result of the 10 regressions in
"summ.list
> <- lapply(lm.list2, summary)"
>
> And now once I enter" sum.list "it gives me the output for
all
> the 10 regressions...
>
> I wanted to access a beta coefficient of one of the regressionssay
> "Price2+Media1+Trend+Seasonality"...the result of which is stored in"
> sum.list[2] "
>
> I entered the below statement for accessing the Beta coefficient for
> Price2...
>
>> summ.list[2]$coefficients[2]
> NULL
>
> But this is giving me " NULL " as the output...
>
> What I am looking for, is to access a beta value of a particular
> variable from a particular regression output and use it for further
analysis.
>
> Can you please help me out with this. Greatly appreciate, you guys
> efforts.
>
>
>
>
> Thanks & Regards,
>
> Krunal Nanavati
> 9769-919198
>
> -Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: 27 September 2012 21:55
> To: Krunal Nanavati
> Cc: David Winsemius; r-help@r-project.org
> Subject: Re: [R] Running different Regressions using for loops
>
> Hello,
>
> Inline.
> Em 27-09-2012 13:52, Krunal Nanavati escreveu:
>> Hi,
>>
>> Thanks for all your help. I am stuck again, but with a new problem,
>> on similar lines.
>>
>> I have taken the problem to the next step now...i have now added 2
"for"
>> loops... 1 for the Price variable...and another for the Media
>> variable
>>
>> I have taken 5 price variables...and 2 media variables with the
>> "trend and seasonality"(appearing in all of them)so in all there
>> will be
>> 10 regression to run now
>>
>> Price 1, Media 1
>>
>> Price 1, Media 2
>>
>> Price 2, Media 1'
>>
>> Price 2, Media 2
>>
>> ...and so on
>>
>> I have built up a code for it...
>>
>>
>>
>>
>>> tryout=read.table("C:\\Users\\Krunal\\Desktop\\R
>> tryout.csv",header=T,sep=",")
>>> cnames <- names(tryout)
>>> price <- cnames[grep("Price", cnames)] media <- cnames[grep("Media",
>>> cnames)] resp <- cnames[1] regr <- cnames[7:8] lm.list <-
>>> vector("list", 10) for(i in 1:5)
>> + {
>> + regress <- paste(price[i], paste(regr, collapse = "+"), sep = "+")
>> + for(j in 1:2) {
>> + regress1 <- paste(media[j],regress,sep="+") fmla <- paste(resp,
>> + regress1, sep = "~") lm.list[[i]] <- lm(as.formula(fmla), data =
>> + tryout) } }
>>> summ.list <- lapply(lm.list, summary) summ.list
>>
>>
>>
>>
>> But it is only running...5 regressions...only Media 1 along with the
>> 5 Price variables & Trend & Seasonality is regressed on
>> Volume...giving only
>> 5 outputs
>>
>> I feel there is something wrong with the" lm.list[[i]] <-
>> lm(as.formula(fmla), data = tryout)" statement.
> No, I don't think so. If it's giving you only 5 outputs the error is
> probably in the fmla construction. Put print statements to see the
> results of those paste() instructions.
>
> Supposing your data.frame is now called tryout2,
>
>
> price <- paste("Price", 1:5, sep = "") media <- paste("Media", 1:2,
> sep = "") pricemedia <- apply(expand.grid(price, media,
> stringsAsFactors = FALSE), 1, paste, collapse="+")
>
> response <- "Volume"
> trendseason <- "Trend+Seasonality" # do this only once
>
> lm.list2 <- list()
> for(i in seq_along(pricemedia)){
> regr <- paste(pricemedia[i], trendseason, sep = "+")
> fmla <- paste(response, regr, sep = "~")
> lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
>
> The trick is to use ?expand.grid
>
> Hope this helps,
>
> Rui Barradas
>
>>I am not sure about its
>> placement...whether it should be in loop 2 or in loop 1
>>
>> Can you please help me out??
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> Thanks & Regards,
>>
>> Krunal Nanavati
>> 9769-919198
>>
>> -Original Message-
>> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
>> Sent: 27 September 2012 16:22
>> To: David Winsemius
>> C
Re: [R] Running different Regressions using for loops
Hi,
Yes the thing that you provided...works finebut probably I should have
asked for some other thing.
Here is what I am trying to do
I am trying to get the mean of Price variableso I am entering the
below function:
mean(names(lm.list2[[2]]$coefficient[2] ))
but this gives me an error
[1] NA
Warning message:
In mean.default(names(lm.list2[[2]]$coefficient[2])) :
argument is not numeric or logical: returning NA
I thought by getting the text from the list variable...will help me
generate the mean for that text...which is a variable in the data...say
Price 1, Media 2and so on
Is this a proper approach...if it is...then something more needs to be
done with the function that you provided.
If not, is there a better way...to generate the mean of a particular
variable inside the " for loop " used earlier...given below:
> lm.list2 <- list()
> for(i in seq_along(pricemedia)){
> regr <- paste(pricemedia[i], trendseason, sep = "+")
> fmla <- paste(response, regr, sep = "~")
> lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 28 September 2012 16:02
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Hello,
Try
names(lm.list2[[2]]$coefficient[2] )
Rui Barradas
Em 28-09-2012 11:29, Krunal Nanavati escreveu:
> Ok...this solves a part of my problem
>
> When I type " lm.list2[2] " ...I get the following output
>
> [[1]]
>
> Call:
> lm(formula = as.formula(fmla), data = tryout2)
>
> Coefficients:
> (Intercept) Price2 Media1 Distri1Trend
> Seasonality
> 13491232 -5759030-15203437048628
> 445351
>
>
>
>
> When I enter " lm.list2[[2]]$coefficient[2] " it gives me the below
> output
>
> Price2
> -5759030
>
> And when I enter " lm.list2[[2]]$coefficient[[2]] " ...I get the
> number...which is -5759030
>
>
> I am looking out for a way to get just the " Price2 "is there a
> statement for that??
>
>
>
> Thanks & Regards,
>
> Krunal Nanavati
> 9769-919198
>
>
> -Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: 28 September 2012 15:18
> To: Krunal Nanavati
> Cc: David Winsemius; r-help@r-project.org
> Subject: Re: [R] Running different Regressions using for loops
>
> Hello,
>
> To access list elements you need `[[`, like this:
>
> summ.list[[2]]$coefficients
>
> Or Use the extractor function,
>
> coef(summ.list[[2]])
>
> Rui Barradas
> Em 28-09-2012 07:23, Krunal Nanavati escreveu:
>> Hi Rui,
>>
>> Excellent!! This is what I was looking for. Thanks for the help.
>>
>> So, now I have stored the result of the 10 regressions in
> "summ.list
>> <- lapply(lm.list2, summary)"
>>
>> And now once I enter" sum.list "it gives me the output for
> all
>> the 10 regressions...
>>
>> I wanted to access a beta coefficient of one of the
>> regressionssay "Price2+Media1+Trend+Seasonality"...the result of
which is stored in"
>> sum.list[2] "
>>
>> I entered the below statement for accessing the Beta coefficient for
>> Price2...
>>
>>> summ.list[2]$coefficients[2]
>> NULL
>>
>> But this is giving me " NULL " as the output...
>>
>> What I am looking for, is to access a beta value of a particular
>> variable from a particular regression output and use it for further
> analysis.
>> Can you please help me out with this. Greatly appreciate, you guys
>> efforts.
>>
>>
>>
>>
>> Thanks & Regards,
>>
>> Krunal Nanavati
>> 9769-919198
>>
>> -Original Message-
>> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
>> Sent: 27 September 2012 21:55
>> To: Krunal Nanavati
>> Cc: David Winsemius; r-help@r-project.org
>> Subject: Re: [R] Running different Regressions using for loops
>>
>> Hello,
>>
>> Inline.
>> Em 27-09-2012 13:52, Krunal Nanavati escreveu:
>>> Hi,
>>>
>>> Thanks for all your help. I am stuck again, but with a new problem,
>>> on similar lines.
>>>
>>> I have taken the problem to the next step now...i have now added 2
> "for"
>>> loops... 1 for the Price variable...and another for the Media
>>> variable
>>>
>>> I have taken 5 price variables...and 2 media variables with the
>>> "trend and seasonality"(appearing in all of them)so in all there
>>> will be
>>> 10 regression to run now
>>>
>>> Price 1, Media 1
>>>
>>> Price 1, Media 2
>>>
>>> Price 2, Media 1'
>>>
>>> Price 2, Media 2
>>>
>>> ...and so on
>>>
>>> I have built up a code for it...
>>>
>>>
>>>
>>>
tryout=read.table("C:\\Users\\Krunal\\Desktop\\R
>>> tryout.csv",header=T,sep=",")
cnames <- names(tryout)
price <- cnames[grep("Price", cnames)] media <-
cnames[grep("Media", cnames)] resp <- cnames[1] regr <- cnames[7:8]
lm.list <- vector("list", 10) for(i in 1:5)
>>> + {
>>> + regress <- paste(price[i], paste(regr, collapse = "+"), sep = "
Re: [R] Running different Regressions using for loops
Ok...I am sorry for the misunderstanding
what I am trying to do is
>> lm.list2 <- list()
>> for(i in seq_along(pricemedia)){
>>regr <- paste(pricemedia[i], trendseason, sep = "+")
>>fmla <- paste(response, regr, sep = "~")
>>lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
When I run...this set of statementsthe 1st regression to be run, will
have Price 1, Media 1...as X variablesand in the second loop it will
have Price 1 & Media 2
So, what I was thinking is...if I can generate inside the for loopthe
mean for Price 1 and Media 1 during the 1st loopand then mean for
Price 1 and Media 2 during the second loop...and so on...for all the 10
regressions
Is the method that I was trying appropriate...or is there a better method
there...I am sorry for the earlier explanation, I hope this one makes it
more understandable
Thanks for your time...and all the quick replies
Thanks & Regards,
Krunal Nanavati
9769-919198
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 28 September 2012 16:49
To: Krunal Nanavati
Cc: David Winsemius; r-help@r-project.org
Subject: Re: [R] Running different Regressions using for loops
Ok, if I'm understanding it well, you want the mean value of Price1, ,
Price5? I don't know if it makes any sense, the coefficients already are
mean values, but see if this is it.
price.coef <- sapply(lm.list, function(x) coef(x)[2])
mean(price.coef)
Rui Barradas
Em 28-09-2012 12:07, Krunal Nanavati escreveu:
> Hi,
>
> Yes the thing that you provided...works finebut probably I should
> have asked for some other thing.
>
> Here is what I am trying to do
>
> I am trying to get the mean of Price variableso I am entering the
> below function:
>
> mean(names(lm.list2[[2]]$coefficient[2] ))
>
> but this gives me an error
>
> [1] NA
> Warning message:
> In mean.default(names(lm.list2[[2]]$coefficient[2])) :
> argument is not numeric or logical: returning NA
>
> I thought by getting the text from the list variable...will help me
> generate the mean for that text...which is a variable in the
> data...say Price 1, Media 2and so on
>
> Is this a proper approach...if it is...then something more needs to be
> done with the function that you provided.
>
> If not, is there a better way...to generate the mean of a particular
> variable inside the " for loop " used earlier...given below:
>
>> lm.list2 <- list()
>> for(i in seq_along(pricemedia)){
>>regr <- paste(pricemedia[i], trendseason, sep = "+")
>>fmla <- paste(response, regr, sep = "~")
>>lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
>
>
>
> Thanks & Regards,
>
> Krunal Nanavati
> 9769-919198
>
>
> -Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: 28 September 2012 16:02
> To: Krunal Nanavati
> Cc: David Winsemius; r-help@r-project.org
> Subject: Re: [R] Running different Regressions using for loops
>
> Hello,
>
> Try
>
> names(lm.list2[[2]]$coefficient[2] )
>
> Rui Barradas
> Em 28-09-2012 11:29, Krunal Nanavati escreveu:
>> Ok...this solves a part of my problem
>>
>> When I type " lm.list2[2] " ...I get the following output
>>
>> [[1]]
>>
>> Call:
>> lm(formula = as.formula(fmla), data = tryout2)
>>
>> Coefficients:
>> (Intercept) Price2 Media1 Distri1Trend
>> Seasonality
>> 13491232 -5759030-15203437048628
>> 445351
>>
>>
>>
>>
>> When I enter " lm.list2[[2]]$coefficient[2] " it gives me the below
>> output
>>
>> Price2
>> -5759030
>>
>> And when I enter " lm.list2[[2]]$coefficient[[2]] " ...I get the
>> number...which is -5759030
>>
>>
>> I am looking out for a way to get just the " Price2 "is there a
>> statement for that??
>>
>>
>>
>> Thanks & Regards,
>>
>> Krunal Nanavati
>> 9769-919198
>>
>>
>> -Original Message-
>> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
>> Sent: 28 September 2012 15:18
>> To: Krunal Nanavati
>> Cc: David Winsemius; r-help@r-project.org
>> Subject: Re: [R] Running different Regressions using for loops
>>
>> Hello,
>>
>> To access list elements you need `[[`, like this:
>>
>> summ.list[[2]]$coefficients
>>
>> Or Use the extractor function,
>>
>> coef(summ.list[[2]])
>>
>> Rui Barradas
>> Em 28-09-2012 07:23, Krunal Nanavati escreveu:
>>> Hi Rui,
>>>
>>> Excellent!! This is what I was looking for. Thanks for the help.
>>>
>>> So, now I have stored the result of the 10 regressions in
>> "summ.list
>>> <- lapply(lm.list2, summary)"
>>>
>>> And now once I enter" sum.list "it gives me the output for
>> all
>>> the 10 regressions...
>>>
>>> I wanted to access a beta coefficient of one of the
>>> regressionssay "Price2+Media1+Trend+Seasonality"...the result of
> which is stored in"
>>> sum.list[2] "
>>>
>>> I entered the below statement for accessing the Beta coefficient for
>>> Price2...
>>>
summ.list[2]$coeffi
[R] RES: Generating an autocorrelated binary variable
I think the package BinarySimCLF can help.
See http://cran.r-project.org/web/packages/binarySimCLF/binarySimCLF.pdf.
André Gabriel.
-Mensagem original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Em
nome de Rolf Turner
Enviada em: sexta-feira, 28 de setembro de 2012 00:02
Para: Simon Zehnder
Cc: r help
Assunto: Re: [R] Generating an autocorrelated binary variable
I have no idea what your code is doing, nor why you want correlated binary
variables. Correlation makes little or no sense in the context of binary
random variables --- or more generally in the context of discrete random
variables.
Be that as it may, it is an easy calculation to show that if X and Y are
binary random variables both with success probability of 0.5 then cor(X,Y) =
0.2 if and only if Pr(X=1 | Y = 1) = 0.6. So just generate X and Y using
that
fact:
set.seed(42)
X <- numeric(1000)
Y <- numeric(1000)
for(i in 1:1000) {
Y[i] <- rbinom(1,1,0.5)
X[i] <- if(Y[i]==1) rbinom(1,1,0.6) else rbinom(1,1,0.4) }
# Check:
cor(X,Y) # Get 0.2012336
Looks about right. Note that the sample proportions are 0.484 and
0.485 for X and Y respectively. These values do not differ significantly
from 0.5.
cheers,
Rolf Turner
On 28/09/12 08:26, Simon Zehnder wrote:
> Hi R-fellows,
>
> I am trying to simulate a multivariate correlated sample via the Gaussian
copula method. One variable is a binary variable, that should be
autocorrelated. The autocorrelation should be rho = 0.2. Furthermore, the
overall probability to get either outcome of the binary variable should be
0.5.
> Below you can see the R code (I use for simplicity a diagonal matrix in
rmvnorm even if it produces no correlated sample):
>
> "sampleCop" <- function(n = 1000, rho = 0.2) {
>
> require(splus2R)
> mvrs <- rmvnorm(n + 1, mean = rep(0, 3), cov = diag(3))
> pmvrs <- pnorm(mvrs, 0, 1)
> var1 <- matrix(0, nrow = n + 1, ncol = 1)
> var1[1] <- qbinom(pmvrs[1, 1], 1, 0.5)
> if(var1[1] == 0) var1[nrow(mvrs)] <- -1
> for(i in 1:(nrow(pmvrs) - 1)) {
> if(pmvrs[i + 1, 1] <= rho) var1[i + 1] <- var1[i]
> else var1[i + 1] <- var1[i] * (-1)
> }
> sample <- matrix(0, nrow = n, ncol = 4)
> sample[, 1] <- var1[1:nrow(var1) - 1]
> sample[, 2] <- var1[2:nrow(var1)]
> sample[, 3] <- qnorm(pmvrs[1:nrow(var1) - 1, 2], 0, 1, 1, 0)
> sample[, 4] <- qnorm(pmvrs[1:nrow(var1) - 1, 3], 0, 1, 1, 0)
>
> sample
>
> }
>
> Now, the code is fine, everything compiles. But when I compute the
autocorrelation of the binary variable, it is not 0.2, but 0.6. Does anyone
know why this happens?
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to write R package
On 27/09/2012 5:15 PM, Dr. Alireza Zolfaghari wrote: Hi List, Would you please send me a good link to talk me through on how to write a R package? See the ?package.skeleton help page. After you have run it, follow the instructions in the "Read-and-delete-me" file that it will create. For full details, see the Writing R Extensions manual. For modifying the package after you've finished the "Read-and-delete-me" instructions, just manually add *.R files where the rest of them are, and use the prompt() function to produce skeleton documentation. That's about it, but you can read more if you like in a tutorial I gave a few years ago at a UseR meeting in Dortmund: http://www.statistik.uni-dortmund.de/useR-2008/slides/Murdoch.pdf Duncan Murdoch __ 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] changing outlier shapes of boxplots using lattice
Elaine,
For panel.bwplot you see that the central dot and the outlier dots are
controlled by
the same pch argument. I initially set the pch="|" to match your first
example with the horizontal
indicator for the median. I would be inclined to use the default circle
for the outliers and
therefore also for the median.
Rich
On Fri, Sep 28, 2012 at 7:13 AM, Sarah Goslee wrote:
> I would guess that if you find the bit that says pch="|" and change it to
> pch=1 it will solve your question, and that reading ?par will tell you why.
>
> Sarah
>
> On Thursday, September 27, 2012, Elaine Kuo wrote:
>
> > Hello
> >
> > This is Elaine.
> >
> > I am using package lattice to generate boxplots.
> > Using Richard's code, the display was almost perfect except the outlier
> > shape.
> > Based on the following code, the outliers are vertical lines.
> > However, I want the outliers to be empty circles.
> > Please kindly help how to modify the code to change the outlier shapes.
> > Thank you.
> >
> > code
> > package (lattice)
> >
> > dataN <- data.frame(GE_distance=rnorm(260),
> >
> > Diet_B=factor(rep(1:13, each=20)))
> >
> > Diet.colors <- c("forestgreen", "darkgreen","chocolate1","darkorange2",
> >
> > "sienna2","red2","firebrick3","saddlebrown","coral4",
> >
> > "chocolate4","darkblue","navy","grey38")
> >
> > levels(dataN$Diet_B) <- Diet.colors
> >
> > bwplot(GE_distance ~ Diet_B, data=dataN,
> >
> >xlab=list("Diet of Breeding Ground", cex = 1.4),
> >
> >ylab = list(
> >
> > "Distance between Centers of B and NB Range (1000 km)",
> >
> > cex = 1.4),
> >
> >panel=panel.bwplot.intermediate.hh,
> >
> >col=Diet.colors,
> >
> >pch=rep("|",13),
> >
> >scales=list(x=list(rot=90)),
> >
> >par.settings=list(box.umbrella=list(lty=1)))
> >
> > [[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.
> >
>
>
> --
> Sarah Goslee
> http://www.stringpage.com
> http://www.sarahgoslee.com
> http://www.functionaldiversity.org
>
> [[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.
Re: [R] Anova and tukey-grouping
HI, I guess there is a mistake in your code. You should have used "typ" instead of "abun" as "abun" is the dependent variable. summary(fm1 <- aov(breaks ~ wool + tension, data = warpbreaks)) myresults <- TukeyHSD(fm1, "tension", ordered = TRUE) library(agricolae) HSD.test(fm1,"wool",group=TRUE) #Study: #HSD Test for breaks #Mean Square Error: 134.9578 #wool, means # breaks std.err replication #A 31.03704 3.050609 27 #B 25.25926 1.789963 27 #alpha: 0.05 ; Df Error: 50 #Critical Value of Studentized Range: 2.840532 #Honestly Significant Difference: 6.350628 #Means with the same letter are not significantly different. #Groups, Treatments and means #a A 31.037037037037 #a B 25.2592592592593 A.K. - Original Message - From: Landi To: r-help@r-project.org Cc: Sent: Friday, September 28, 2012 5:41 AM Subject: [R] Anova and tukey-grouping Hello, I am really new to R and it's still a challenge to me. Currently I'm working on my Master's Thesis. My supervisor works with SAS and is not familiar with R at all. I want to run an Anova, a tukey-test and as a result I want to have the tukey-grouping ( something like A - AB - B) I came across the HSD.test in the agricolae-package, but... unfortunately I do not get an output (like here in the answer http://stats.stackexchange.com/questions/31547/how-to-obtain-the-results-of-a-tukey-hsd-post-hoc-test-in-a-table-showing-groupe ) I did it like this: ## ANOVA anova.typabunmit<-aov(ds.typabunmit$abun ~ ds.typabunmit$typ) summary(anova.typabunmit) summary.lm(anova.typabunmit) ## post HOC tukey.typabunmit<-TukeyHSD(anova.typabunmit) tukey.typabunmit ## HSD HSD.test(anova.typabunmit, "abun", group=TRUE) and the ONLY output is this: Name: abun ds.typabunmit$typ I would be very pleased about some ides..:! -- View this message in context: http://r.789695.n4.nabble.com/Anova-and-tukey-grouping-tp4644485.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Crosstable-like analysis (ks test) of dataframe
Thank you Rui!
that works as I want it... :)
/Johannes
On Fri, Sep 28, 2012 at 12:30 PM, Rui Barradas wrote:
> Hello,
>
> Try the following.
>
>
> f <- function(x, y, ...,
> alternative = c("two.sided", "less", "greater"), exact = NULL){
> #w <- getOption("warn")
> #options(warn = -1) # ignore warnings
> p <- ks.test(x, y, ..., alternative = alternative, exact =
> exact)$p.value
> #options(warn = w)
> p
> }
>
> n <- 1e1
> dat <- data.frame(X=rnorm(n), Y=runif(n), Z=rchisq(n, df=3))
>
> apply(dat, 2, function(x) apply(dat, 2, function(y) f(x, y)))
>
> Hope this helps,
>
> Rui Barradas
> Em 28-09-2012 11:10, Johannes Radinger escreveu:
>>
>> Hi,
>>
>> I have a dataframe with multiple (appr. 20) columns containing
>> vectors of different values (different distributions).
>> Now I'd like to create a crosstable
>> where I compare the distribution of each vector (df-column) with
>> each other. For the comparison I want to use the ks.test().
>> The result should contain as row and column names the column names
>> of the input dataframe and the cells should be populated with
>> the p-value of the ks.test for each pairwise analysis.
>>
>> My data.frame looks like:
>> df <- data.frame(X=rnorm(1000,2),Y=rnorm(1000,1),Z=rnorm(1000,2))
>>
>> And the test for one single case is:
>> ks <- ks.test(df$X,df$Z)
>>
>> where the p value is:
>> ks[2]
>>
>> How can I create an automatized way of this pairwise analysis?
>> Any suggestions? I guess that is a quite common analysis (probably with
>> other tests).
>>
>> cheers,
>> Johannes
>>
>> __
>> 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.
>
>
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[R] Lattice bwplot(): Conditioning on one factor
I'm not able to create the proper syntax to specify a lattice bwplot() for
only one of two conditioning factors.
The syntax that produces a box plot of each of the two conditioning
factors is:
bwplot(quant ~ param | era, data=mg.d, main='Dissolved Magnesium',
ylab='Concentration (mg/L)')
What I've tried unsuccessfully are:
bwplot(quant ~ param | factor(era=='Pre-mining'), data=mg.d,
main='Magnesium', ylab='Concentration (mg/L))
bwplot(quant ~ param | era, data=mg.d, main='Magnesium', ylab='Concentration
(mg/L)', subset=era('Pre-mining'))
plus slight variations of the above. None work.
Please point me to what I've missed in specifying only one of two
conditioning factors for the plot.
Rich
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Re: [R] Simple Question
Many thanks Dr. Winsemius , Kimmo and Pascal
All of them are working and really beautiful...
Best Regards,
Bhupendrasinh Thakre
*Disclaimer :*
The information contained in this communication is confidential and may be
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are hereby (a) notified that any disclosure, copying, distribution or
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strictly prohibited and may be unlawful, and (b) kindly requested to inform
the sender immediately and destroy any copies.
On Fri, Sep 28, 2012 at 1:36 AM, David Winsemius wrote:
>
> On Sep 27, 2012, at 11:13 PM, Bhupendrasinh Thakre wrote:
>
> >
> > Hi Everyone,
> >
> > I am trying a very simple task to append the Timestamp with a variable
> name so something like
> > a_2012_09_27_00_12_30 <- rnorm(1,2,1).
>
> If you want to assign a value to a character-name you need to use ...
> `assign`. You cannot just stick a numeric value which is what you get with
> sys.Time() on the LHS of a "<-" and expect R to intuit what you intend.
>
> ?assign
> assign( "a_2012_09_27_00_12_30" , rnorm(1,2,1) )
> assign( as.character(unclass(Sys.time())) , rnorm(1,2,1) )
>
> (I would have thought you wanted to format that sys.Time result:)
>
> > format(Sys.time(), "%Y_%m_%d_%H_%M_%S")
> [1] "2012_09_27_23_32_40"
>
> > assign(format(Sys.time(), "%Y_%m_%d_%H_%M_%S"), rnorm(1,2,1) )
> > grep("^2012", ls(), value=TRUE)
> [1] "2012_09_27_23_33_45"
>
>
> >
> > Tried some commands but it doesn't work out well. Hope someone has some
> answer on it.
> >
> > Session Info
> >
> > R version 2.15.1 (2012-06-22)
> > Platform: i386-apple-darwin9.8.0/i386 (32-bit)
> >
> > locale:
> > [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
> >
> > attached base packages:
> > [1] stats graphics grDevices utils datasets methods base
> >
> > other attached packages:
> > [1] chron_2.3-42twitteR_0.99.19 rjson_0.2.9 RCurl_1.91-1
> bitops_1.0-4.1 tm_0.5-7.1 RMySQL_0.9-3DBI_0.2-5
> >
> > loaded via a namespace (and not attached):
> > [1] slam_0.1-24 tools_2.15.1
> >
> > Statement I tried :
> >
> > b <- unclass(Sys.time())
> > b = 1348812597
> > c_b <- rnorm(1,2,1)
> >
> > Works perfect but doesn't show me c_1348812597.
> >
> > Best Regards,
> >
> >
> > Bhupendrasinh Thakre
> > [[alternative HTML version deleted]]
>
> BT; Please learn to post in plain text. It's really very simple with gmail.
>
> --
>
> David Winsemius, MD
> Alameda, CA, USA
>
>
[[alternative HTML version deleted]]
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Re: [R] [R-sig-hpc] Quickest way to make a large "empty" file on disk?
Jonathan,
ff has a utility function file.resize() which allows to give a new filesize
in bytes using doubles.
See ?file.resize
Regards
Jens Oehlschlägel
Gesendet: Donnerstag, 27. September 2012 um 21:17 Uhr
Von: "Jonathan Greenberg"
An: r-help , r-sig-...@r-project.org
Betreff: Re: [R-sig-hpc] Quickest way to make a large "empty" file on disk?
Folks:
Asked this question some time ago, and found what appeared (at first) to be
the best solution, but I'm now finding a new problem. First off, it seemed
like ff as Jens suggested worked:
# outdata_ncells = the number of rows * number of columns * number of bands
in an image:
out<-ff(vmode="double",length=outdata_ncells,filename=filename)
finalizer(out) <- close
close(out)
This was working fine until I attempted to set length to a VERY large
number: outdata_ncells = 17711913600. This would create a file that is
131.964GB. Big, but not obscenely so (and certainly not larger than the
filesystem can handle). However, length appears to be restricted
by .Machine$integer.max (I'm on a 64-bit windows box):
> .Machine$integer.max
[1] 2147483647
Any suggestions on how to solve this problem for much larger file sizes?
--j
OnThu, May 3, 2012 at 10:44 AM, Jonathan Greenberg
wrote:
> Thanks, all! I'll try these out. I'm trying to work up something that is
> platform independent (if possible) for use with mmap. I'll do some tests
> on these suggestions and see which works best. I'll try to report back in
a
> few days. Cheers!
>
> --j
>
>
>
> 2012/5/3 "Jens Oehlschlägel"
>
>> Jonathan,
>>
>> On some filesystems (e.g. NTFS, see below) it is possible to create
>> 'sparse' memory-mapped files, i.e. reserving the space without the cost
of
>> actually writing initial values.
>> Package 'ff' does this automatically and also allows to access the file
>> in parallel. Check the example below and see how big file creation is
>> immediate.
>>
>> Jens Oehlschlägel
>>
>>
>> > library(ff)
>> > library(snowfall)
>> > ncpus <- 2
>> > n <- 1e8
>> > system.time(
>> + x <- ff(vmode="double", length=n, filename="c:/Temp/x.ff")
>> + )
>> User System verstrichen
>> 0.01 0.00 0.02
>> > # check finalizer, with an explicit filename we should have a 'close'
>> finalizer
>> > finalizer(x)
>> [1] "close"
>> > # if not, set it to 'close' inorder to not let slaves delete x on slave
>> shutdown
>> > finalizer(x) <- "close"
>> > sfInit(parallel=TRUE, cpus=ncpus, type="SOCK")
>> R Version: R version 2.15.0 (2012-03-30)
>>
>> snowfall 1.84 initialized (using snow 0.3-9): parallel execution on 2
>> CPUs.
>>
>> > sfLibrary(ff)
>> Library ff loaded.
>> Library ff loaded in cluster.
>>
>> Warnmeldung:
>> In library(package = "ff", character.only = TRUE, pos = 2, warn.conflicts
>> = TRUE, :
>> 'keep.source' is deprecated and will be ignored
>> > sfExport("x") # note: do not export the same ff multiple times
>> > # explicitely opening avoids a gc problem
>> > sfClusterEval(open(x, caching="mmeachflush")) # opening with
>> 'mmeachflush' inststead of 'mmnoflush' is a bit slower but prevents OS
>> write storms when the file is larger than RAM
>> [[1]]
>> [1] TRUE
>>
>> [[2]]
>> [1] TRUE
>>
>> > system.time(
>> + sfLapply( chunk(x, length=ncpus), function(i){
>> + x[i] <- runif(sum(i))
>> + invisible()
>> + })
>> + )
>> User System verstrichen
>> 0.00 0.00 30.78
>> > system.time(
>> + s <- sfLapply( chunk(x, length=ncpus), function(i) quantile(x[i],
>> c(0.05, 0.95)) )
>> + )
>> User System verstrichen
>> 0.00 0.00 4.38
>> > # for completeness
>> > sfClusterEval(close(x))
>> [[1]]
>> [1] TRUE
>>
>> [[2]]
>> [1] TRUE
>>
>> > csummary(s)
>> 5% 95%
>> Min. 0.04998 0.95
>> 1st Qu. 0.04999 0.95
>> Median 0.05001 0.95
>> Mean 0.05001 0.95
>> 3rd Qu. 0.05002 0.95
>> Max. 0.05003 0.95
>> > # stop slaves
>> > sfStop()
>>
>> Stopping cluster
>>
>> > # with the close finalizer we are responsible for deleting the file
>> explicitely (unless we want to keep it)
>> > delete(x)
>> [1] TRUE
>> > # remove r-side metadata
>> > rm(x)
>> > # truly free memory
>> > gc()
>>
>>
>>
>> *Gesendet:* Donnerstag, 03. Mai 2012 um 00:23 Uhr
>> *Von:* "Jonathan Greenberg"
>> *An:* r-help , r-sig-...@r-project.org
>> *Betreff:* [R-sig-hpc] Quickest way to make a large "empty" file on
>> disk?
>> R-helpers:
>>
>> What would be the absolute fastest way to make a large "empty" file (e.g.
>> filled with all zeroes) on disk, given a byte size and a given number
>> number of empty values. I know I can use writeBin, but the "object" in
>> this case may be far too large to store in main
Re: [R] Anova and tukey-grouping
Hello ! Thanks for your advice. I tried it, but the output is the same: > HSD.test(anova.typabunmit, "typ", group=TRUE) Name: typ ds.typabunmit$typ I don't get the values...!?!? -- View this message in context: http://r.789695.n4.nabble.com/Anova-and-tukey-grouping-tp4644485p4644513.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] List of Variables in Original Order
AK: Thanks, that was very helpful. It led me to think of the function
names(base) which provided the vector of names in the correct order. I
then used the same matrix formatting and everything worked out exactly
as planned.
Dick
On 9/28/2012 1:09 AM, arun kirshna [via R] wrote:
>
>
> HI,
> May be this helps you:
> set.seed(1)
> mat1<-matrix(rnorm(60,5),nrow=5,ncol=12)
> colnames(mat1)<-paste0("Var",1:12)
> vec2<-format(c(1,cor(mat1[,1],mat1[,2:12])),digits=4)
> vec3<-colnames(mat1)
> arr2<-array(rbind(vec3,vec2),dim=c(2,3,4))
> res<-data.frame(do.call(rbind,lapply(1:dim(arr2)[3],function(i)
> arr2[,,i])))
> res
> #X1 X2 X3
> #1 Var1 Var2 Var3
> #2 1.0 0.27890 -0.61497
> #3 Var4 Var5 Var6
> #4 0.24916 -0.76155 0.30853
> #5 Var7 Var8 Var9
> #6 -0.46413 0.79287 0.05191
> #7Var10Var11Var12
> #8 -0.06940 -0.53251 0.06766
>
> A.K.
>
>
> - Original Message -
> From: rkulp <[hidden email]
> >
> To: [hidden email]
> Cc:
> Sent: Thursday, September 27, 2012 6:26 PM
> Subject: [R] List of Variables in Original Order
>
> I am trying to Sweave the output of calculating correlations between one
> variable and several others. I wanted to print a table where the
> odd-numbered rows contain the variable names and the even-numbered rows
> contain the correlations. So if VarA is correlated with all the
> variables in
> mydata.df, then it would look like
>
> var1var2 var3
> corr1 corr2 corr3
> var4 var5var6
> corr4 corr5 corr6
> .
> .
> etc.
> I tried using a matrix for the correlations and another one for the
> variable
> names. I built the correlation matrix using
> x = matrix(format(cor(mydata.df[,1],mydata.df[,c(2:79)]),digits=4),nc=3)
> and the variable names matrix using
> y = matrix(ls(mydata.df[c(2:79)]),nc=3).
> The problem is the function ls returns the names in alphabetical order,
> columnar order.
> How do I get the names in columnar order? Is there a better way to
> display
> the correlation of a single variable with a large number of other
> variables?
> If there is, how do I do it? I appreciate any help I can get. This is my
> first project in R so I don't know much about it yet.
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/List-of-Variables-in-Original-Order-tp4644436.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> [hidden email]
> 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.
>
>
> __
> [hidden email]
> 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.
>
>
>
> If you reply to this email, your message will be added to the
> discussion below:
> http://r.789695.n4.nabble.com/List-of-Variables-in-Original-Order-tp4644436p4644469.html
>
>
> To unsubscribe from List of Variables in Original Order, click here
> .
> NAML
>
>
>
rkulp.vcf (418 bytes)
--
View this message in context:
http://r.789695.n4.nabble.com/List-of-Variables-in-Original-Order-tp4644436p4644516.html
Sent from the R help mailing list archive at Nabble.com.
[[alternative HTML version deleted]]
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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] Running different Regressions using for loops
On Sep 28, 2012, at 4:35 AM, Krunal Nanavati wrote:
> Ok...I am sorry for the misunderstanding
>
> what I am trying to do is
Perhaps (and that is a really large 'perhaps'):
>>> lm.list2 <- list()
lm.means <- list()
>>> for(i in seq_along(pricemedia)){
>>> regr <- paste(pricemedia[i], trendseason, sep = "+")
>>> fmla <- paste(response, regr, sep = "~")
>>> lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
lm.means[[i]] <- mean(lm.list2[[i]]$coefficients[c("Price1",
"Media1")]
}
>
> When I run...this set of statementsthe 1st regression to be run, will
> have Price 1, Media 1...as X variablesand in the second loop it will
> have Price 1 & Media 2
>
> So, what I was thinking is...if I can generate inside the for loopthe
> mean for Price 1 and Media 1 during the 1st loopand then mean for
> Price 1 and Media 2 during the second loop...and so on...for all the 10
> regressions
>
>
> Is the method that I was trying appropriate...or is there a better method
> there...I am sorry for the earlier explanation, I hope this one makes it
> more understandable
One generally want ones methods to be determinate while allowing the results to
be approximate.
Had you followed the posting guide a offered a reproducible example it would
have been much more "understandable".
>
>
> Thanks for your time...and all the quick replies
>
>
> -Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: 28 September 2012 16:49
> To: Krunal Nanavati
> Cc: David Winsemius; r-help@r-project.org
> Subject: Re: [R] Running different Regressions using for loops
>
> Ok, if I'm understanding it well, you want the mean value of Price1, ,
> Price5? I don't know if it makes any sense, the coefficients already are
> mean values, but see if this is it.
>
> price.coef <- sapply(lm.list, function(x) coef(x)[2])
> mean(price.coef)
>
> Rui Barradas
> Em 28-09-2012 12:07, Krunal Nanavati escreveu:
>> Hi,
>>
>> Yes the thing that you provided...works finebut probably I should
>> have asked for some other thing.
>>
>> Here is what I am trying to do
>>
>> I am trying to get the mean of Price variableso I am entering the
>> below function:
>>
>> mean(names(lm.list2[[2]]$coefficient[2] ))
>>
>> but this gives me an error
>>
>> [1] NA
>> Warning message:
>> In mean.default(names(lm.list2[[2]]$coefficient[2])) :
>> argument is not numeric or logical: returning NA
>>
>> I thought by getting the text from the list variable...will help me
>> generate the mean for that text...which is a variable in the
>> data...say Price 1, Media 2and so on
>>
>> Is this a proper approach...if it is...then something more needs to be
>> done with the function that you provided.
>>
>> If not, is there a better way...to generate the mean of a particular
>> variable inside the " for loop " used earlier...given below:
>>
>>> lm.list2 <- list()
>>> for(i in seq_along(pricemedia)){
>>> regr <- paste(pricemedia[i], trendseason, sep = "+")
>>> fmla <- paste(response, regr, sep = "~")
>>> lm.list2[[i]] <- lm(as.formula(fmla), data = tryout2) }
>>
>>
>>
>> Thanks & Regards,
>>
>> Krunal Nanavati
>> 9769-919198
>>
>>
>> -Original Message-
>> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
>> Sent: 28 September 2012 16:02
>> To: Krunal Nanavati
>> Cc: David Winsemius; r-help@r-project.org
>> Subject: Re: [R] Running different Regressions using for loops
>>
>> Hello,
>>
>> Try
>>
>> names(lm.list2[[2]]$coefficient[2] )
>>
>> Rui Barradas
>> Em 28-09-2012 11:29, Krunal Nanavati escreveu:
>>> Ok...this solves a part of my problem
>>>
>>> When I type " lm.list2[2] " ...I get the following output
>>>
>>> [[1]]
>>>
>>> Call:
>>> lm(formula = as.formula(fmla), data = tryout2)
>>>
>>> Coefficients:
>>> (Intercept) Price2 Media1 Distri1Trend
>>> Seasonality
>>> 13491232 -5759030-15203437048628
>>> 445351
>>>
>>>
>>>
>>>
>>> When I enter " lm.list2[[2]]$coefficient[2] " it gives me the below
>>> output
>>>
>>> Price2
>>> -5759030
>>>
>>> And when I enter " lm.list2[[2]]$coefficient[[2]] " ...I get the
>>> number...which is -5759030
>>>
>>>
>>> I am looking out for a way to get just the " Price2 "is there a
>>> statement for that??
>>>
>>>
>>>
>>> Thanks & Regards,
>>>
>>> Krunal Nanavati
>>> 9769-919198
>>>
>>>
>>> -Original Message-
>>> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
>>> Sent: 28 September 2012 15:18
>>> To: Krunal Nanavati
>>> Cc: David Winsemius; r-help@r-project.org
>>> Subject: Re: [R] Running different Regressions using for loops
>>>
>>> Hello,
>>>
>>> To access list elements you need `[[`, like this:
>>>
>>> summ.list[[2]]$coefficients
>>>
>>> Or Use the extractor function,
>>>
>>> coef(summ.list[[2]])
>>>
>>> Rui Barradas
>>> Em 28-09-2012 07:23, Krunal Nanava
[R] max & summary contradict each other
why does summary report max 27600 and not 27603? > x <- c(27603, 1) > max(x) [1] 27603 > summary(x) Min. 1st Qu. MedianMean 3rd Qu.Max. 16902 13800 13800 20700 27600 -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://memri.org http://pmw.org.il http://dhimmi.com http://iris.org.il http://mideasttruth.com Vegetarians eat Vegetables, Humanitarians are scary. __ 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] max & summary contradict each other
On 28/09/2012 12:14 PM, Sam Steingold wrote: why does summary report max 27600 and not 27603? > x <- c(27603, 1) > max(x) [1] 27603 > summary(x) Min. 1st Qu. MedianMean 3rd Qu.Max. 16902 13800 13800 20700 27600 Because you asked for 3 digit accuracy. See ?summary. Duncan Murdoch __ 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] Lattice bwplot(): Conditioning on one factor
On Sep 28, 2012, at 7:49 AM, Rich Shepard wrote:
> I'm not able to create the proper syntax to specify a lattice bwplot() for
> only one of two conditioning factors.
Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts a
subset argument) or using the 'subset' function to pass the desired rows to the
data argument if it doesn't?
>
> The syntax that produces a box plot of each of the two conditioning
> factors is:
>
> bwplot(quant ~ param | era, data=mg.d, main='Dissolved Magnesium',
> ylab='Concentration (mg/L)')
>
> What I've tried unsuccessfully are:
>
> bwplot(quant ~ param | factor(era=='Pre-mining'), data=mg.d,
> main='Magnesium', ylab='Concentration (mg/L))
>
> bwplot(quant ~ param | era, data=mg.d, main='Magnesium', ylab='Concentration
> (mg/L)', subset=era('Pre-mining'))
>
> plus slight variations of the above. None work.
>
> Please point me to what I've missed in specifying only one of two
> conditioning factors for the plot.
>
>
--
David Winsemius, MD
Alameda, CA, USA
__
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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] Simple Question
Hi Everyone,
Sorry for coming back again with a new problem.
Editing question, session info and data so you don't have to scroll till
the end of page.
*Situation :*
I have a data frame and it's name is df. Now I want to add Time Stamp to
the end of *"name" of "data Frame" i.e. "df_system_time"*. Previously it
was running great and thanks to Dr. Winsemius , Kimmo and Pascal and I
believe as the function which i used was scalar.
*Data :*
dput(df)structure(list(x = 1:10, y = 1:10), .Names = c("x", "y"),
row.names = c(NA,
-10L), class = "data.frame")
*Session Info :*
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices datasets utils methods
[7] base
other attached packages:
[1] rcom_2.2-5 rscproxy_2.0-5
loaded via a namespace (and not attached):
[1] colorspace_1.1-1 dichromat_1.2-4digest_0.5.2
[4] ggplot2_0.9.2.1grid_2.15.1gtable_0.1.1
[7] labeling_0.1 MASS_7.3-18memoise_0.1
[10] munsell_0.3plyr_1.7.1 proto_0.3-9.2
[13] RColorBrewer_1.0-5 reshape2_1.2.1 scales_0.2.2
[16] stringr_0.6.1 tools_2.15.1
It's kind of very easy in SQL but I love doing all the work in R so don't
want to leave for just changing the name.
Best Regards,
Bhupendrasinh Thakre
Best Regards,
Bhupendrasinh Thakre
*Disclaimer :*
The information contained in this communication is confidential and may be
legally privileged. It is intended solely for the use of the individual or
entity to whom it is adressed. If you are not the intended recipient you
are hereby (a) notified that any disclosure, copying, distribution or
taking any action with respect to the content of this information is
strictly prohibited and may be unlawful, and (b) kindly requested to inform
the sender immediately and destroy any copies.
On Fri, Sep 28, 2012 at 10:13 AM, Bhupendrasinh Thakre <
vickytha...@gmail.com> wrote:
> Many thanks Dr. Winsemius , Kimmo and Pascal
> All of them are working and really beautiful...
>
> Best Regards,
>
>
> Bhupendrasinh Thakre
>
> *Disclaimer :*
>
> The information contained in this communication is confidential and may be
> legally privileged. It is intended solely for the use of the individual or
> entity to whom it is adressed. If you are not the intended recipient you
> are hereby (a) notified that any disclosure, copying, distribution or
> taking any action with respect to the content of this information is
> strictly prohibited and may be unlawful, and (b) kindly requested to inform
> the sender immediately and destroy any copies.
>
>
>
> On Fri, Sep 28, 2012 at 1:36 AM, David Winsemius
> wrote:
>
>>
>> On Sep 27, 2012, at 11:13 PM, Bhupendrasinh Thakre wrote:
>>
>> >
>> > Hi Everyone,
>> >
>> > I am trying a very simple task to append the Timestamp with a variable
>> name so something like
>> > a_2012_09_27_00_12_30 <- rnorm(1,2,1).
>>
>> If you want to assign a value to a character-name you need to use ...
>> `assign`. You cannot just stick a numeric value which is what you get with
>> sys.Time() on the LHS of a "<-" and expect R to intuit what you intend.
>>
>> ?assign
>> assign( "a_2012_09_27_00_12_30" , rnorm(1,2,1) )
>> assign( as.character(unclass(Sys.time())) , rnorm(1,2,1) )
>>
>> (I would have thought you wanted to format that sys.Time result:)
>>
>> > format(Sys.time(), "%Y_%m_%d_%H_%M_%S")
>> [1] "2012_09_27_23_32_40"
>>
>> > assign(format(Sys.time(), "%Y_%m_%d_%H_%M_%S"), rnorm(1,2,1) )
>> > grep("^2012", ls(), value=TRUE)
>> [1] "2012_09_27_23_33_45"
>>
>>
>> >
>> > Tried some commands but it doesn't work out well. Hope someone has some
>> answer on it.
>> >
>> > Session Info
>> >
>> > R version 2.15.1 (2012-06-22)
>> > Platform: i386-apple-darwin9.8.0/i386 (32-bit)
>> >
>> > locale:
>> > [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
>> >
>> > attached base packages:
>> > [1] stats graphics grDevices utils datasets methods base
>> >
>> > other attached packages:
>> > [1] chron_2.3-42twitteR_0.99.19 rjson_0.2.9 RCurl_1.91-1
>> bitops_1.0-4.1 tm_0.5-7.1 RMySQL_0.9-3DBI_0.2-5
>> >
>> > loaded via a namespace (and not attached):
>> > [1] slam_0.1-24 tools_2.15.1
>> >
>> > Statement I tried :
>> >
>> > b <- unclass(Sys.time())
>> > b = 1348812597
>> > c_b <- rnorm(1,2,1)
>> >
>> > Works perfect but doesn't show me c_1348812597.
>> >
>> > Best Regards,
>> >
>> >
>> > Bhupendrasinh Thakre
>> > [[alternative HTML version deleted]]
>>
>> BT; Please learn to post in plain text. It's really very simple with
>> gmail.
>>
>> --
>>
>> David Winsemius, MD
>> Alameda, CA, USA
>>
>>
>
[[alternative HTML version deleted]]
__
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https
Re: [R] Lattice bwplot(): Conditioning on one factor
A small reproducible example, as requested bythe posting guide, would
have been very helpful here (if you provide one, use ?dput to provide
the data). You have also not told us what you mean by "unsuccessful,"
so we are left to guess what sort of problems you experienced. "None
work" is completely useless to help diagnose the problem. This means
we waste time going back and forth trying to elucidate what you mean.
Please consider these things if/when you post in future.
In any case, my guess is that param is numeric and it should be a
factor, so, e.g.
bwplot(quant ~ factor(param) | era, data=mg.d, main='Dissolved
Magnesium', ylab='Concentration (mg/L)')
might be what you want. But of course, it may be completely wrong.
Cheers,
Bert
On Fri, Sep 28, 2012 at 9:25 AM, David Winsemius wrote:
>
> On Sep 28, 2012, at 7:49 AM, Rich Shepard wrote:
>
>> I'm not able to create the proper syntax to specify a lattice bwplot() for
>> only one of two conditioning factors.
>
> Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts a
> subset argument) or using the 'subset' function to pass the desired rows to
> the data argument if it doesn't?
>
>>
>> The syntax that produces a box plot of each of the two conditioning
>> factors is:
>>
>> bwplot(quant ~ param | era, data=mg.d, main='Dissolved Magnesium',
>> ylab='Concentration (mg/L)')
>>
>> What I've tried unsuccessfully are:
>>
>> bwplot(quant ~ param | factor(era=='Pre-mining'), data=mg.d,
>> main='Magnesium', ylab='Concentration (mg/L))
>>
>> bwplot(quant ~ param | era, data=mg.d, main='Magnesium', ylab='Concentration
>> (mg/L)', subset=era('Pre-mining'))
>>
>> plus slight variations of the above. None work.
>>
>> Please point me to what I've missed in specifying only one of two
>> conditioning factors for the plot.
>>
>>
> --
>
> David Winsemius, MD
> Alameda, CA, USA
>
> __
> 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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
__
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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] Simple Question
Hi Everyone,
Sorry for coming back again with a new problem.
Editing question, session info and data so you don't have to scroll till
the end of page.
*Situation :*
I have a data frame and it's name is df. Now I want to add Time Stamp to
the end of *"name" of "data Frame" i.e. "df_system_time"*. Previously it
was running great and thanks to Dr. Winsemius , Kimmo and Pascal and I
believe as the function which i used was scalar.
*Data :*
dput(df)structure(list(x = 1:10, y = 1:10), .Names = c("x", "y"),
row.names = c(NA,
-10L), class = "data.frame")
*Session Info :*
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices datasets utils methods
[7] base
other attached packages:
[1] rcom_2.2-5 rscproxy_2.0-5
loaded via a namespace (and not attached):
[1] colorspace_1.1-1 dichromat_1.2-4digest_0.5.2
[4] ggplot2_0.9.2.1grid_2.15.1gtable_0.1.1
[7] labeling_0.1 MASS_7.3-18memoise_0.1
[10] munsell_0.3plyr_1.7.1 proto_0.3-9.2
[13] RColorBrewer_1.0-5 reshape2_1.2.1 scales_0.2.2
[16] stringr_0.6.1 tools_2.15.1
It's kind of very easy in SQL but I love doing all the work in R so don't
want to leave for just changing the name.
Best Regards,
Bhupendrasinh Thakre
Best Regards,
Bhupendrasinh Thakre
*Disclaimer :*
The information contained in this communication is confidential and may be
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strictly prohibited and may be unlawful, and (b) kindly requested to inform
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On Fri, Sep 28, 2012 at 10:13 AM, Bhupendrasinh Thakre <
vickytha...@gmail.com> wrote:
> Many thanks Dr. Winsemius , Kimmo and Pascal
> All of them are working and really beautiful...
>
> Best Regards,
>
>
> Bhupendrasinh Thakre
>
> *Disclaimer :*
>
> The information contained in this communication is confidential and may be
> legally privileged. It is intended solely for the use of the individual or
> entity to whom it is adressed. If you are not the intended recipient you
> are hereby (a) notified that any disclosure, copying, distribution or
> taking any action with respect to the content of this information is
> strictly prohibited and may be unlawful, and (b) kindly requested to inform
> the sender immediately and destroy any copies.
>
>
>
> On Fri, Sep 28, 2012 at 1:36 AM, David Winsemius
> wrote:
>
>>
>> On Sep 27, 2012, at 11:13 PM, Bhupendrasinh Thakre wrote:
>>
>> >
>> > Hi Everyone,
>> >
>> > I am trying a very simple task to append the Timestamp with a variable
>> name so something like
>> > a_2012_09_27_00_12_30 <- rnorm(1,2,1).
>>
>> If you want to assign a value to a character-name you need to use ...
>> `assign`. You cannot just stick a numeric value which is what you get with
>> sys.Time() on the LHS of a "<-" and expect R to intuit what you intend.
>>
>> ?assign
>> assign( "a_2012_09_27_00_12_30" , rnorm(1,2,1) )
>> assign( as.character(unclass(Sys.time())) , rnorm(1,2,1) )
>>
>> (I would have thought you wanted to format that sys.Time result:)
>>
>> > format(Sys.time(), "%Y_%m_%d_%H_%M_%S")
>> [1] "2012_09_27_23_32_40"
>>
>> > assign(format(Sys.time(), "%Y_%m_%d_%H_%M_%S"), rnorm(1,2,1) )
>> > grep("^2012", ls(), value=TRUE)
>> [1] "2012_09_27_23_33_45"
>>
>>
>> >
>> > Tried some commands but it doesn't work out well. Hope someone has some
>> answer on it.
>> >
>> > Session Info
>> >
>> > R version 2.15.1 (2012-06-22)
>> > Platform: i386-apple-darwin9.8.0/i386 (32-bit)
>> >
>> > locale:
>> > [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
>> >
>> > attached base packages:
>> > [1] stats graphics grDevices utils datasets methods base
>> >
>> > other attached packages:
>> > [1] chron_2.3-42twitteR_0.99.19 rjson_0.2.9 RCurl_1.91-1
>> bitops_1.0-4.1 tm_0.5-7.1 RMySQL_0.9-3DBI_0.2-5
>> >
>> > loaded via a namespace (and not attached):
>> > [1] slam_0.1-24 tools_2.15.1
>> >
>> > Statement I tried :
>> >
>> > b <- unclass(Sys.time())
>> > b = 1348812597
>> > c_b <- rnorm(1,2,1)
>> >
>> > Works perfect but doesn't show me c_1348812597.
>> >
>> > Best Regards,
>> >
>> >
>> > Bhupendrasinh Thakre
>> > [[alternative HTML version deleted]]
>>
>> BT; Please learn to post in plain text. It's really very simple with
>> gmail.
>>
>> --
>>
>> David Winsemius, MD
>> Alameda, CA, USA
>>
>>
>
[[alternative HTML version deleted]]
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https
Re: [R] max & summary contradict each other
Hi, Try this: summary(x,digits=max(5)) # Min. 1st Qu. Median Mean 3rd Qu. Max. # 1.0 6901.5 13802.0 13802.0 20702.0 27603.0 A.K. - Original Message - From: Sam Steingold To: r-help@r-project.org Cc: Sent: Friday, September 28, 2012 12:14 PM Subject: [R] max & summary contradict each other why does summary report max 27600 and not 27603? > x <- c(27603, 1) > max(x) [1] 27603 > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 1 6902 13800 13800 20700 27600 -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://memri.org http://pmw.org.il http://dhimmi.com http://iris.org.il http://mideasttruth.com Vegetarians eat Vegetables, Humanitarians are scary. __ 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@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] Lattice bwplot(): Conditioning on one factor
On Fri, 28 Sep 2012, David Winsemius wrote:
Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts
a subset argument) or using the 'subset' function to pass the desired rows
to the data argument if it doesn't?
David,
That's what I tried:
bwplot(quant ~ param | era, data=mg.d, main='Magnesium', ylab='Concentration
(mg/L)', subset=era('Pre-mining'))
Perhaps I didn't write it correctly.
Thanks,
Rich
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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] Simple Question
On 28-09-2012, at 18:40, Bhupendrasinh Thakre wrote:
> Hi Everyone,
>
> Sorry for coming back again with a new problem.
> Editing question, session info and data so you don't have to scroll till
> the end of page.
>
> *Situation :*
>
> I have a data frame and it's name is df. Now I want to add Time Stamp to
> the end of *"name" of "data Frame" i.e. "df_system_time"*. Previously it
> was running great and thanks to Dr. Winsemius , Kimmo and Pascal and I
> believe as the function which i used was scalar.
>
> *Data :*
>
> dput(df)structure(list(x = 1:10, y = 1:10), .Names = c("x", "y"),
> row.names = c(NA,
> -10L), class = "data.frame")
>
You have been given the answer.
It only needs a minor variation:
newname.df <- paste0("df_", format(Sys.time(), "%Y_%m_%d_%H_%M_%S") )
assign(newname.df,df)
and if you wish
rm(list=c('df','newname.df'))
Or install package memisc (found by doing findFn("rename") from package sos)
and use function rename(0; I have not tried this.
Berend
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] [R-sig-hpc] Quickest way to make a large "empty" file on disk?
Rui:
Quick follow-up -- it looks like seek does do what I want (I see Simon
suggested it some time ago) -- what do mean by "trash your disk"? What I'm
trying to accomplish is getting parallel, asynchronous writes to a large
binary image (just a binary file) working. Each node writes to a different
sector of the file via mmap, "filling in" the values as the process runs,
but the file needs to be pre-created before I can mmap it. Running a
writeBin with a bunch of 0s would mean I'd basically have to write the file
twice, but the seek/ff trick seems to be much faster.
Do I risk doing some damage to my filesystem if I use seek? I see there is
a strongly worded warning in the help for ?seek:
"Use of seek on Windows is discouraged. We have found so many errors in the
Windows implementation of file positioning that users are advised to use it
only at their own risk, and asked not to waste the *R* developers' time
with bug reports on Windows' deficiencies." --> there's no detail here on
which errors people have experienced, so I'm not sure if doing something as
simple as just "creating" a file using seek falls under the "discouraging"
category.
As a note, we are trying to work this up on both Windows and *nix systems,
hence our wanting to have a single approach that works on both OSs.
--j
On Thu, Sep 27, 2012 at 3:49 PM, Rui Barradas wrote:
> Hello,
>
> If you really need to trash your disk, why not use seek()?
>
> > fl <- file("Test.txt", open = "wb")
> > seek(fl, where = 1024, origin = "start", rw = "write")
> [1] 0
> > writeChar(character(1), fl, nchars = 1, useBytes = TRUE)
> Warning message:
> In writeChar(character(1), fl, nchars = 1, useBytes = TRUE) :
> writeChar: more characters requested than are in the string - will
> zero-pad
> > close(fl)
>
>
> File "Test.txt" is now 1Kb in size.
>
> Hope this helps,
>
> Rui Barradas
> Em 27-09-2012 20:17, Jonathan Greenberg escreveu:
>
> Folks:
>
> Asked this question some time ago, and found what appeared (at first) to be
> the best solution, but I'm now finding a new problem. First off, it seemed
> like ff as Jens suggested worked:
>
> # outdata_ncells = the number of rows * number of columns * number of bands
> in an image:
> out<-ff(vmode="double",length=outdata_ncells,filename=filename)
> finalizer(out) <- close
> close(out)
>
> This was working fine until I attempted to set length to a VERY large
> number: outdata_ncells = 17711913600. This would create a file that is
> 131.964GB. Big, but not obscenely so (and certainly not larger than the
> filesystem can handle). However, length appears to be restricted
> by .Machine$integer.max (I'm on a 64-bit windows box):
>
> .Machine$integer.max
>
> [1] 2147483647
>
> Any suggestions on how to solve this problem for much larger file sizes?
>
> --j
>
>
> On Thu, May 3, 2012 at 10:44 AM, Jonathan Greenberg
> wrote:
>
>
> Thanks, all! I'll try these out. I'm trying to work up something that is
> platform independent (if possible) for use with mmap. I'll do some tests
> on these suggestions and see which works best. I'll try to report back in a
> few days. Cheers!
>
> --j
>
>
>
> 2012/5/3 "Jens Oehlschlägel"
>
>
> Jonathan,
>
> On some filesystems (e.g. NTFS, see below) it is possible to create
> 'sparse' memory-mapped files, i.e. reserving the space without the cost of
> actually writing initial values.
> Package 'ff' does this automatically and also allows to access the file
> in parallel. Check the example below and see how big file creation is
> immediate.
>
> Jens Oehlschlägel
>
>
>
> library(ff)
> library(snowfall)
> ncpus <- 2
> n <- 1e8
> system.time(
>
> + x <- ff(vmode="double", length=n, filename="c:/Temp/x.ff")
> + )
>User System verstrichen
>0.010.000.02
>
> # check finalizer, with an explicit filename we should have a 'close'
>
> finalizer
>
> finalizer(x)
>
> [1] "close"
>
> # if not, set it to 'close' inorder to not let slaves delete x on slave
>
> shutdown
>
> finalizer(x) <- "close"
> sfInit(parallel=TRUE, cpus=ncpus, type="SOCK")
>
> R Version: R version 2.15.0 (2012-03-30)
>
> snowfall 1.84 initialized (using snow 0.3-9): parallel execution on 2
> CPUs.
>
>
> sfLibrary(ff)
>
> Library ff loaded.
> Library ff loaded in cluster.
>
> Warnmeldung:
> In library(package = "ff", character.only = TRUE, pos = 2, warn.conflicts
> = TRUE, :
> 'keep.source' is deprecated and will be ignored
>
> sfExport("x") # note: do not export the same ff multiple times
> # explicitely opening avoids a gc problem
> sfClusterEval(open(x, caching="mmeachflush")) # opening with
>
> 'mmeachflush' inststead of 'mmnoflush' is a bit slower but prevents OS
> write storms when the file is larger than RAM
> [[1]]
> [1] TRUE
>
> [[2]]
> [1] TRUE
>
>
> system.time(
>
> + sfLapply( chunk(x, length=ncpus), function(i){
> + x[i] <- runif(sum(i))
> + invisible()
> + })
> + )
>User System verstrichen
>0.000.00 30.78
>
>
Re: [R] Lattice bwplot(): Conditioning on one factor
On Sep 28, 2012, at 9:56 AM, Rich Shepard wrote:
> On Fri, 28 Sep 2012, David Winsemius wrote:
>
>> Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts
>> a subset argument) or using the 'subset' function to pass the desired rows
>> to the data argument if it doesn't?
>
> David,
>
> That's what I tried:
>
>>> bwplot(quant ~ param | era, data=mg.d, main='Magnesium', ylab='Concentration
>>> (mg/L)', subset=era('Pre-mining'))
Sigh. If I were testing that strategy (which I did not try because you were too
busy to have included a working example) I would have written it:
bwplot(quant ~ param , data=mg.d, main='Magnesium', ylab='Concentration
(mg/L)', subset= era=='Pre-mining' )
That passes a logical vector which will "work" only if bwplot created an local
environment where column names of the 'data' argument have been added to the
local namespce. I do not know if that is true. I just looked at the bwplot help
page and do not see a subset argument documented there.
The other suggestion which it seems you were also to busy too have tried was:
bwplot(quant ~ param , main='Magnesium', ylab='Concentration
(mg/L)', data = subset( mg.dsubset, era=='Pre-mining' ) )
Wrapping a column name around a factor level with parentheses (which R takes to
mean there is a function named 'era' to be applied) and expecting R to
understand the you want a subset seems doomed to failure.
It makes no sense to me to condition on a factor that you know for certainty
has only one level in the data being offered.
--
David Winsemius, MD
Alameda, CA, USA
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Re: [R] Lattice bwplot(): Conditioning on one factor
On Fri, 28 Sep 2012, David Winsemius wrote: bwplot(quant ~ param , data=mg.d, main='Magnesium', ylab='Concentration (mg/L)', subset= era=='Pre-mining' ) David, Don: Thank you. I tried subset= and era== separately, not together. Now I know. Much appreciated, Rich __ 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] Lattice bwplot(): Conditioning on one factor
Yes. Now I understand what was wanted.
1. the subset argument is certainly documented on the Help page:
subset
An expression that evaluates to a logical or integer indexing vector.
Like groups, it is evaluated in data. Only the resulting rows of data
are used for the plot. If subscripts is TRUE, the subscripts provided
to the panel function will be indices referring to the rows of data
prior to the subsetting. Whether levels of factors in the data frame
that are unused after the subsetting will be dropped depends on the
drop.unused.levels argument.
Had the OP read this carefully, he would have presumably recognized
the errors in his specification.
2. Here is a small reproducible example to show how it should be done
(probably unnecessary now):
> df <-expand.grid(a = letters[1:3],b=LETTERS[1:2])
> df <- df[rep(1:6,10),]
> df$y <- runif(60)
> bwplot(y~a|b, dat=df,subset = (b=="A"))
## The logical condition is parenthesized only for clarity
Cheers,
Bert
On Fri, Sep 28, 2012 at 10:10 AM, David Winsemius
wrote:
>
> On Sep 28, 2012, at 9:56 AM, Rich Shepard wrote:
>
>> On Fri, 28 Sep 2012, David Winsemius wrote:
>>
>>> Wouldn't that involve specifying the 'subset' parameter (if bwplot accepts
>>> a subset argument) or using the 'subset' function to pass the desired rows
>>> to the data argument if it doesn't?
>>
>> David,
>>
>> That's what I tried:
>>
bwplot(quant ~ param | era, data=mg.d, main='Magnesium',
ylab='Concentration
(mg/L)', subset=era('Pre-mining'))
>
> Sigh. If I were testing that strategy (which I did not try because you were
> too busy to have included a working example) I would have written it:
>
> bwplot(quant ~ param , data=mg.d, main='Magnesium', ylab='Concentration
> (mg/L)', subset= era=='Pre-mining' )
>
> That passes a logical vector which will "work" only if bwplot created an
> local environment where column names of the 'data' argument have been added
> to the local namespce. I do not know if that is true. I just looked at the
> bwplot help page and do not see a subset argument documented there.
>
> The other suggestion which it seems you were also to busy too have tried was:
>
> bwplot(quant ~ param , main='Magnesium', ylab='Concentration
> (mg/L)', data = subset( mg.dsubset, era=='Pre-mining' ) )
>
> Wrapping a column name around a factor level with parentheses (which R takes
> to mean there is a function named 'era' to be applied) and expecting R to
> understand the you want a subset seems doomed to failure.
>
> It makes no sense to me to condition on a factor that you know for certainty
> has only one level in the data being offered.
> --
>
> David Winsemius, MD
> Alameda, CA, USA
>
> __
> 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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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Re: [R] install.packages on windows
On 28.09.2012 00:32, Duncan Murdoch wrote: On 12-09-27 2:53 PM, Anju R wrote: Sometimes when I try to install certain packages I get a warning message. For example, I tried to install the package "Imtest" on windows R version 2.15.1 and got the following message: Warning message: package ‘Imtest’ is not available (for R version 2.15.1) How can I install the above package? Why do I get the above Warning message? It probably means exactly what it says, except that the information is about the mirror you are using. I would try another mirror. If that doesn't solve it, then it probably means that the package is really not available for 2.15.1. You can look on the cran.r-project.org website for information about it, and probably download the source from there, but you will probably need to fix whatever is wrong with it before it will work. Or in other words: There is no such package "Imtest" on CRAN, perhaps you are looking for "lmtest"? Uwe Ligges Duncan Murdoch __ 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@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] Anova and tukey-grouping
Hi, As I mentioned earlier, these are just guess work until you provide a subset of your data with dput(). Also, please check the structure of the data with str(). A.K. - Original Message - From: Landi To: r-help@r-project.org Cc: Sent: Friday, September 28, 2012 10:35 AM Subject: Re: [R] Anova and tukey-grouping Hello ! Thanks for your advice. I tried it, but the output is the same: > HSD.test(anova.typabunmit, "typ", group=TRUE) Name: typ ds.typabunmit$typ I don't get the values...!?!? -- View this message in context: http://r.789695.n4.nabble.com/Anova-and-tukey-grouping-tp4644485p4644513.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-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] How to write R package
On 28.09.2012 14:22, Duncan Murdoch wrote: On 27/09/2012 5:15 PM, Dr. Alireza Zolfaghari wrote: Hi List, Would you please send me a good link to talk me through on how to write a R package? See the ?package.skeleton help page. After you have run it, follow the instructions in the "Read-and-delete-me" file that it will create. For full details, see the Writing R Extensions manual. For modifying the package after you've finished the "Read-and-delete-me" instructions, just manually add *.R files where the rest of them are, and use the prompt() function to produce skeleton documentation. That's about it, but you can read more if you like in a tutorial I gave a few years ago at a UseR meeting in Dortmund: http://www.statistik.uni-dortmund.de/useR-2008/slides/Murdoch.pdf ... and there are others who gave talks or tutorials about it (inlcuding myself). Nevertheless, I'd recommend to look into the manual "Writing R Extensions" which is updated with R and with the changes in the package related mechanisms --- while all our talks and tutorials won't get updated. Probably Duncan's is still correct, but I want to make this remark for the list's archives. Best, Uwe Ligges Duncan Murdoch __ 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@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] Simple Question
Thanks a ton Berend. That worked like a charm..
R comes with thousands of Sweet Surprises everyday
Bhupendrasinh Thakre
On Sep 28, 2012, at 12:00 PM, Berend Hasselman wrote:
>
> On 28-09-2012, at 18:40, Bhupendrasinh Thakre wrote:
>
>> Hi Everyone,
>>
>> Sorry for coming back again with a new problem.
>> Editing question, session info and data so you don't have to scroll till
>> the end of page.
>>
>> *Situation :*
>>
>> I have a data frame and it's name is df. Now I want to add Time Stamp to
>> the end of *"name" of "data Frame" i.e. "df_system_time"*. Previously it
>> was running great and thanks to Dr. Winsemius , Kimmo and Pascal and I
>> believe as the function which i used was scalar.
>>
>> *Data :*
>>
>> dput(df)structure(list(x = 1:10, y = 1:10), .Names = c("x", "y"),
>> row.names = c(NA,
>> -10L), class = "data.frame")
>>
>
> You have been given the answer.
> It only needs a minor variation:
>
> newname.df <- paste0("df_", format(Sys.time(), "%Y_%m_%d_%H_%M_%S") )
> assign(newname.df,df)
>
> and if you wish
>
> rm(list=c('df','newname.df'))
>
> Or install package memisc (found by doing findFn("rename") from package sos)
> and use function rename(0; I have not tried this.
>
> Berend
[[alternative HTML version deleted]]
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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] Simple Question
On Fri, Sep 28, 2012 at 11:15 AM, Bhupendrasinh Thakre
wrote:
> Thanks a ton Berend. That worked like a charm..
> R comes with thousands of Sweet Surprises everyday
-- Not for those who read the docs. :-o
-- Bert
>
>
> Bhupendrasinh Thakre
>
>
>
>
> On Sep 28, 2012, at 12:00 PM, Berend Hasselman wrote:
>
>>
>> On 28-09-2012, at 18:40, Bhupendrasinh Thakre wrote:
>>
>>> Hi Everyone,
>>>
>>> Sorry for coming back again with a new problem.
>>> Editing question, session info and data so you don't have to scroll till
>>> the end of page.
>>>
>>> *Situation :*
>>>
>>> I have a data frame and it's name is df. Now I want to add Time Stamp to
>>> the end of *"name" of "data Frame" i.e. "df_system_time"*. Previously it
>>> was running great and thanks to Dr. Winsemius , Kimmo and Pascal and I
>>> believe as the function which i used was scalar.
>>>
>>> *Data :*
>>>
>>> dput(df)structure(list(x = 1:10, y = 1:10), .Names = c("x", "y"),
>>> row.names = c(NA,
>>> -10L), class = "data.frame")
>>>
>>
>> You have been given the answer.
>> It only needs a minor variation:
>>
>> newname.df <- paste0("df_", format(Sys.time(), "%Y_%m_%d_%H_%M_%S") )
>> assign(newname.df,df)
>>
>> and if you wish
>>
>> rm(list=c('df','newname.df'))
>>
>> Or install package memisc (found by doing findFn("rename") from package sos)
>> and use function rename(0; I have not tried this.
>>
>> Berend
>
>
> [[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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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Re: [R] How to test if there is a subvector in a longer vector
Thank you! ___ Lähettäjä: Berend Hasselman [b...@xs4all.nl] Lähetetty: 28. syyskuuta 2012 10:47 Vastaanottaja: Atte Tenkanen Cc: R help Aihe: Re: [R] How to test if there is a subvector in a longer vector On 28-09-2012, at 07:41, Atte Tenkanen wrote: > Sorry. I should have mentioned that the order of the components is important. > > So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of > c(2,1,1,6,4,3). > > How to test this? See this discussion for a variety of solutions. http://r.789695.n4.nabble.com/matching-a-sequence-in-a-vector-td4389523.html#a4393453 Berend __ 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] Arules - predict function issues - subscript out of bounds
Hi Ankur, I am running into the exact same issue you have described above. Were you able to find out why it didn't work on your data set and resolve it? If yes, could you share? Much thanks & regards, Alice -- View this message in context: http://r.789695.n4.nabble.com/Arules-predict-function-issues-subscript-out-of-bounds-tp4634422p4644546.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] [R-sig-hpc] Quickest way to make a large "empty" file on disk?
Hello,
I've written a function to try to answer to your op request, but I've
run into a problem. See in the end.
In the mean time, inline.
Em 28-09-2012 17:44, Jonathan Greenberg escreveu:
Rui:
Quick follow-up -- it looks like seek does do what I want (I see Simon
suggested it some time ago) -- what do mean by "trash your disk"?
Nothing special, just that sometimes there are good ways of doing so.
mmap seems to be safe.
What I'm
trying to accomplish is getting parallel, asynchronous writes to a large
binary image (just a binary file) working. Each node writes to a different
sector of the file via mmap, "filling in" the values as the process runs,
but the file needs to be pre-created before I can mmap it. Running a
writeBin with a bunch of 0s would mean I'd basically have to write the file
twice, but the seek/ff trick seems to be much faster.
Do I risk doing some damage to my filesystem if I use seek? I see there is
a strongly worded warning in the help for ?seek:
"Use of seek on Windows is discouraged. We have found so many errors in the
Windows implementation of file positioning that users are advised to use it
only at their own risk, and asked not to waste the *R* developers' time
with bug reports on Windows' deficiencies." --> there's no detail here on
which errors people have experienced, so I'm not sure if doing something as
simple as just "creating" a file using seek falls under the "discouraging"
category.
I'm not a great system programmer but in 20+ years of using seek on
Windows has shown nothing of the sort. In fact, I've just found a
problem with ubuntu 12.04, where seek gives the expected result on
Windows, it goes up to a certain point on ubuntu and then "stops
seeking", or whatever is happening. I installed ubuntu very recently so
I really don't know why the behavior that you can see in the example run
below. But I do that Windows 7 is causing no problem, as expected.
As a note, we are trying to work this up on both Windows and *nix systems,
hence our wanting to have a single approach that works on both OSs.
--j
#
# Function: creates a file of ascii nulls using seek/writeBin. File size
can be big.
#
createBig <- function(filename, size){
if(size == 0) return(0)
chunk <- .Machine$integer.max
nchunks <- as.integer(size / chunk)
rest <- size - as.double(nchunks)*as.double(chunk)
fl <- file(filename, open = "wb")
for(i in seq_len(nchunks)){
seek(fl, where = chunk - 1, origin = "current", rw = "write")
writeBin(raw(1), fl)
# -- debug --
print(seek(fl, where = NA))
}
if(rest > 0){
seek(fl, where = rest - 1, origin = "current", rw = "write")
writeBin(raw(1), fl)
}
close(fl)
}
As you can see from the debug prints, on Windows 7, everything works as
planned while on ubuntu 12.04 when it reaches 17Gb seek stops seeking.
The increments in file size become 1 byte at a time, explained by the
writeBin instruction. (The different, slightly larger, size is
irrelevant, the code was ran several times all with the same result: at
17179869176 bytes it no longer works.)
#
#
# System: Windows 7 / R 2.15.1
size <- 10*.Machine$integer.max + sample(.Machine$integer.max, 1)
size
[1] 22195364413
createBig("Test.txt", size)
[1] 2147483647
[1] 4294967294
[1] 6442450941
[1] 8589934588
[1] 10737418235
[1] 12884901882
[1] 15032385529
[1] 17179869176
[1] 19327352823
[1] 21474836470
file.info("Test.txt")$size
[1] 22195364413
file.info("Test.txt")$size %/% .Machine$integer.max
[1] 10
file.info("Test.txt")$size %% .Machine$integer.max
[1] 720527943
sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=Portuguese_Portugal.1252 LC_CTYPE=Portuguese_Portugal.1252
[3] LC_MONETARY=Portuguese_Portugal.1252 LC_NUMERIC=C
[5] LC_TIME=Portuguese_Portugal.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] fortunes_1.5-0
#
#
# System: ubuntu 12.04 precise pangolim / R 2.15.1
size <- 10*.Machine$integer.max + sample(.Machine$integer.max, 1)
size
[1] 23091487381
createBig("Test.txt", size)
[1] 2147483647
[1] 4294967294
[1] 6442450941
[1] 8589934588
[1] 10737418235
[1] 12884901882
[1] 15032385529
[1] 17179869176
[1] 17179869177
[1] 17179869178
file.info("Test.txt")$size
[1] 17179869179
file.info("Test.txt")$size %/% .Machine$integer.max
[1] 8
file.info("Test.txt")$size %% .Machine$integer.max
[1] 3
sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=pt_PT.UTF-8 LC_NUMERIC=C
[3] LC_TIME=pt_PT.UTF-8LC_COLLATE=pt_PT.UTF-8
[5] LC_MONETARY=pt_PT.UTF-8LC_MESSAGES=pt_PT.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C
Re: [R] changing outlier shapes of boxplots using lattice
On Fri, Sep 28, 2012 at 6:57 AM, Richard M. Heiberger wrote:
> Elaine,
>
> For panel.bwplot you see that the central dot and the outlier dots are
> controlled by
> the same pch argument.
??? I don't think so...
bwplot(rgamma(20,.1,1)~gl(2,10), pch=rep(17,2),
panel = lattice::panel.bwplot)
I think you mean panel.bwplot.intermidiate.hh ?
BTW thank you for the useful HH package but in this case OP is using it
with no "at" argument, so why not
Diet.colors <- c("forestgreen", "darkgreen","chocolate1","darkorange2",
"sienna2","red2","firebrick3","saddlebrown","coral4","chocolate4","darkblue","navy","grey38")
bwplot(rgamma(20*13,1,.1)~gl(13,20),
fill = Diet.colors, pch = "|",
par.settings = list(box.umbrella=list(lty=1)))
cheers
I initially set the pch="|" to match your first
> example with the horizontal
> indicator for the median. I would be inclined to use the default circle
> for the outliers and
> therefore also for the median.
>
> Rich
>
> On Fri, Sep 28, 2012 at 7:13 AM, Sarah Goslee >wrote:
>
> > I would guess that if you find the bit that says pch="|" and change it to
> > pch=1 it will solve your question, and that reading ?par will tell you
> why.
> >
> > Sarah
> >
> > On Thursday, September 27, 2012, Elaine Kuo wrote:
> >
> > > Hello
> > >
> > > This is Elaine.
> > >
> > > I am using package lattice to generate boxplots.
> > > Using Richard's code, the display was almost perfect except the outlier
> > > shape.
> > > Based on the following code, the outliers are vertical lines.
> > > However, I want the outliers to be empty circles.
> > > Please kindly help how to modify the code to change the outlier shapes.
> > > Thank you.
> > >
> > > code
> > > package (lattice)
> > >
> > > dataN <- data.frame(GE_distance=rnorm(260),
> > >
> > > Diet_B=factor(rep(1:13, each=20)))
> > >
> > > Diet.colors <- c("forestgreen", "darkgreen","chocolate1","darkorange2",
> > >
> > > "sienna2","red2","firebrick3","saddlebrown","coral4",
> > >
> > > "chocolate4","darkblue","navy","grey38")
> > >
> > > levels(dataN$Diet_B) <- Diet.colors
> > >
> > > bwplot(GE_distance ~ Diet_B, data=dataN,
> > >
> > >xlab=list("Diet of Breeding Ground", cex = 1.4),
> > >
> > >ylab = list(
> > >
> > > "Distance between Centers of B and NB Range (1000 km)",
> > >
> > > cex = 1.4),
> > >
> > >panel=panel.bwplot.intermediate.hh,
> > >
> > >col=Diet.colors,
> > >
> > >pch=rep("|",13),
> > >
> > >scales=list(x=list(rot=90)),
> > >
> > >par.settings=list(box.umbrella=list(lty=1)))
> > >
> > > [[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.
> > >
> >
> >
> > --
> > Sarah Goslee
> > http://www.stringpage.com
> > http://www.sarahgoslee.com
> > http://www.functionaldiversity.org
> >
> > [[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.
>
[[alternative HTML version deleted]]
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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.
[R] Better way of Grouping?
Hello R users,
This is more of a convenience question that I hope others might find useful
if there is a better answer. I work with large datasets that requires
multiple parsing stages for different analysis. For example, compare group
3 vs. group 4. A more complicated comparison would be time B in group 3 of
group L with B in group 4 of group L. I normally subset each group with
the following type of code.
data=read(...)
#L v D
L=data[LvD %in% c("L"),]
D=data[LvD %in% c("D"),]
#Groups 3 and 4 within L and D
group3L=L[group %in% c("3"),]
group4L=L[group %in% c("3"),]
group3D=D[group %in% c("3"),]
group4D=D[group %in% c("3"),]
#Times B, S45, FR2, FR8
you get the idea
Is there a more efficient way to subset groups? Thanks for any insight.
Regards,
Charles
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] [R-sig-hpc] Quickest way to make a large "empty" file on disk?
On Sep 28, 2012, at 12:44 PM, Jonathan Greenberg wrote:
> Rui:
>
> Quick follow-up -- it looks like seek does do what I want (I see Simon
> suggested it some time ago) -- what do mean by "trash your disk"?
I can't speak for Rui, but the difference between seeking and explicit write is
that the FS can optimize the former by not actually writing anything to disk
(which is why it's so fast on some OS/FS combos). However, what this means that
the layout on the disk may not be sequential depending on the write patterns of
the actual data blocks, because the FS may keep a mask of unused blocks and
don't write them. But that is just a FS issue and thus varies vasty by OS and
FS. For your use this probably doesn't matter as you probably don't need to
stream the resulting file at the end.
> What I'm
> trying to accomplish is getting parallel, asynchronous writes to a large
> binary image (just a binary file) working. Each node writes to a different
> sector of the file via mmap, "filling in" the values as the process runs,
> but the file needs to be pre-created before I can mmap it. Running a
> writeBin with a bunch of 0s would mean I'd basically have to write the file
> twice, but the seek/ff trick seems to be much faster.
>
> Do I risk doing some damage to my filesystem if I use seek? I see there is
> a strongly worded warning in the help for ?seek:
>
> "Use of seek on Windows is discouraged. We have found so many errors in the
> Windows implementation of file positioning that users are advised to use it
> only at their own risk, and asked not to waste the *R* developers' time
> with bug reports on Windows' deficiencies." --> there's no detail here on
> which errors people have experienced, so I'm not sure if doing something as
> simple as just "creating" a file using seek falls under the "discouraging"
> category.
>
Quick search in my mail shows issues that were related to what Windows reports
as the seek location on text files when querying. AFAICS it did not affect the
side-effect of seek which is what you're interested in.
Cheers,
Simon
> As a note, we are trying to work this up on both Windows and *nix systems,
> hence our wanting to have a single approach that works on both OSs.
>
> --j
>
>
> On Thu, Sep 27, 2012 at 3:49 PM, Rui Barradas wrote:
>
>> Hello,
>>
>> If you really need to trash your disk, why not use seek()?
>>
>>> fl <- file("Test.txt", open = "wb")
>>> seek(fl, where = 1024, origin = "start", rw = "write")
>> [1] 0
>>> writeChar(character(1), fl, nchars = 1, useBytes = TRUE)
>> Warning message:
>> In writeChar(character(1), fl, nchars = 1, useBytes = TRUE) :
>> writeChar: more characters requested than are in the string - will
>> zero-pad
>>> close(fl)
>>
>>
>> File "Test.txt" is now 1Kb in size.
>>
>> Hope this helps,
>>
>> Rui Barradas
>> Em 27-09-2012 20:17, Jonathan Greenberg escreveu:
>>
>> Folks:
>>
>> Asked this question some time ago, and found what appeared (at first) to be
>> the best solution, but I'm now finding a new problem. First off, it seemed
>> like ff as Jens suggested worked:
>>
>> # outdata_ncells = the number of rows * number of columns * number of bands
>> in an image:
>> out<-ff(vmode="double",length=outdata_ncells,filename=filename)
>> finalizer(out) <- close
>> close(out)
>>
>> This was working fine until I attempted to set length to a VERY large
>> number: outdata_ncells = 17711913600. This would create a file that is
>> 131.964GB. Big, but not obscenely so (and certainly not larger than the
>> filesystem can handle). However, length appears to be restricted
>> by .Machine$integer.max (I'm on a 64-bit windows box):
>>
>> .Machine$integer.max
>>
>> [1] 2147483647
>>
>> Any suggestions on how to solve this problem for much larger file sizes?
>>
>> --j
>>
>>
>> On Thu, May 3, 2012 at 10:44 AM, Jonathan Greenberg
>> wrote:
>>
>>
>> Thanks, all! I'll try these out. I'm trying to work up something that is
>> platform independent (if possible) for use with mmap. I'll do some tests
>> on these suggestions and see which works best. I'll try to report back in a
>> few days. Cheers!
>>
>> --j
>>
>>
>>
>> 2012/5/3 "Jens Oehlschlägel"
>>
>>
>> Jonathan,
>>
>> On some filesystems (e.g. NTFS, see below) it is possible to create
>> 'sparse' memory-mapped files, i.e. reserving the space without the cost of
>> actually writing initial values.
>> Package 'ff' does this automatically and also allows to access the file
>> in parallel. Check the example below and see how big file creation is
>> immediate.
>>
>> Jens Oehlschlägel
>>
>>
>>
>> library(ff)
>> library(snowfall)
>> ncpus <- 2
>> n <- 1e8
>> system.time(
>>
>> + x <- ff(vmode="double", length=n, filename="c:/Temp/x.ff")
>> + )
>> User System verstrichen
>> 0.010.000.02
>>
>> # check finalizer, with an explicit filename we should have a 'close'
>>
>> finalizer
>>
>> finalizer(x)
>>
>> [1] "close"
>>
>> # if not, set it
[R] Select Original and Duplicates
I would like to select a all the duplicate rows of a data frame including
the original. Any help would be much appreciated. This is where I'm at so
far. Thanks.
#Sample data frame:
df <- read.table(header=T, con <- textConnection('
label value
A 4
B 3
C 6
B 3
B 1
A 2
A 4
A 4
'))
close(con)
# Duplicate entries
df[duplicated(df),]
# label value
# B 3
# A 4
# A 4
#I want to select all the rows that are duplicated including the original
#This is the output I want
# label value
# B 3
# B 3
# A 4
# A 4
# A 4
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Select Original and Duplicates
Hello,
Try the following.
idx <- duplicated(df) | duplicated(df, fromLast = TRUE)
df[idx, ]
Note that they are returned in their original order in the df.
Hope this helps,
Rui Barradas
Em 28-09-2012 21:11, Adam Gabbert escreveu:
I would like to select a all the duplicate rows of a data frame including
the original. Any help would be much appreciated. This is where I'm at so
far. Thanks.
#Sample data frame:
df <- read.table(header=T, con <- textConnection('
label value
A 4
B 3
C 6
B 3
B 1
A 2
A 4
A 4
'))
close(con)
# Duplicate entries
df[duplicated(df),]
# label value
# B 3
# A 4
# A 4
#I want to select all the rows that are duplicated including the original
#This is the output I want
# label value
# B 3
# B 3
# A 4
# A 4
# A 4
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
__
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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] Better way of Grouping?
You have not specified the objective function you are trying to optimize with
your term "efficient", or what you do with all of these subsets once you have
them.
For notational simplification and completeness of coverage (not necessarily
computational speedup) you might want to look at "tapply" or ddply/dlply from
the plyr package. If you build lists of subsets you can index into them
according to grouping value. You can use expand.grid to build all permutations
of grouping values to use as indexes into those lists of subsets.
To reiterate, you have not indicated what you want to do with these subsets, so
there could be special-purpose functions that do what you want. As always,
reproducible code leads to reproducible answers. :)
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Charles Determan Jr wrote:
>Hello R users,
>
>This is more of a convenience question that I hope others might find
>useful
>if there is a better answer. I work with large datasets that requires
>multiple parsing stages for different analysis. For example, compare
>group
>3 vs. group 4. A more complicated comparison would be time B in group
>3 of
>group L with B in group 4 of group L. I normally subset each group
>with
>the following type of code.
>
>data=read(...)
>
>#L v D
>L=data[LvD %in% c("L"),]
>D=data[LvD %in% c("D"),]
>
>#Groups 3 and 4 within L and D
>group3L=L[group %in% c("3"),]
>group4L=L[group %in% c("3"),]
>
>group3D=D[group %in% c("3"),]
>group4D=D[group %in% c("3"),]
>
>#Times B, S45, FR2, FR8
>you get the idea
>
>
>Is there a more efficient way to subset groups? Thanks for any
>insight.
>
>Regards,
>Charles
>
> [[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@r-project.org mailing list
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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.
[R] Merging multiple columns into one column
Good Evening- I have a dataframe that has 10 columns that has a header and 7306 rows in each column, I want to combine these columns into one. I utilized the stack function but it only returned 3/4 of the data...my code is: where nfcuy_bw is the dataframe with 7305 obs. and 10 variables Once I apply this code I only receive a data frame with 58440 obs. of 2 variables, of which there should be 73,050 obs. of 2 variables, just wondering what is happening here? View(nfcuy_bw) attach(nfcuy_bw) cuyahoga_nf<-data.frame(s5,s10,s25,s27,s33,s41,s51,his_c) cuy_nf<-stack(cuyahoga_nf) Thanks Meredith -- Doctoral Candidate Department of Civil and Environmental Engineering Michigan Technological University [[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] Better way of Grouping?
On Sep 28, 2012, at 11:59 AM, Charles Determan Jr wrote:
> Hello R users,
>
> This is more of a convenience question that I hope others might find useful
> if there is a better answer. I work with large datasets that requires
> multiple parsing stages for different analysis. For example, compare group
> 3 vs. group 4. A more complicated comparison would be time B in group 3 of
> group L with B in group 4 of group L. I normally subset each group with
> the following type of code.
>
> data=read(...)
>
> #L v D
> L=data[LvD %in% c("L"),]
> D=data[LvD %in% c("D"),]
>
> #Groups 3 and 4 within L and D
> group3L=L[group %in% c("3"),]
> group4L=L[group %in% c("3"),]
Assume you meant to have a "4" there
>
> group3D=D[group %in% c("3"),]
> group4D=D[group %in% c("3"),]
Ditto. Only makes sense with a "4".
The usual way is to use:
lapply( split(data, interaction(data$LvD, data$group)) ,
fun( subdf) {} )
That way you do not end up littering you workspace with subsidiary subsets of
you main data object.
>
> #Times B, S45, FR2, FR8
> you get the idea
>
>
> Is there a more efficient way to subset groups? Thanks for any insight.
>
--
David Winsemius, MD
Alameda, CA, USA
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Re: [R] Merging multiple columns into one column
On Sep 28, 2012, at 2:51 PM, Meredith Ballard LaBeau wrote: > Good Evening- > I have a dataframe that has 10 columns that has a header and 7306 rows in > each column, I want to combine these columns into one. I utilized the stack > function but it only returned 3/4 of the data...my code is: > where nfcuy_bw is the dataframe with 7305 obs. and 10 variables > Once I apply this code I only receive a data frame with 58440 obs. of 2 > variables, of which there should be 73,050 obs. of 2 variables, just > wondering what is happening here? > > View(nfcuy_bw) > > attach(nfcuy_bw) Using 'attach' is a great way to produce confusing errors. > > cuyahoga_nf<-data.frame(s5,s10,s25,s27,s33,s41,s51,his_c) > > cuy_nf<-stack(cuyahoga_nf) Unable to do much else in the absence of a dataset, much less a summary of these objects, whose creation is your responsibility, not ours. -- David Winsemius, MD Alameda, CA, USA __ 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] Heatmap Colors
Hello R-Users! I'm using a heatmap to visualize a matrix of values between -1 and 3. How can I set the colors so that white is zero, below zero is blue of increasing intensity towards -1 and above zero is red of increasing intensity towards red? I tried like this (using the marray and gplots packages from bioconductor): mcol <- maPalette(low="blue", mid="white", high="red",k=100) heatmap.2(my_matrix, col=mcol) But white does not correspond to zero, because the value distribution is not symmetrical, so that zero is not in the middle. Is it somehow possible to create a color palette with white centered at zero? Nick Fankhauser __ 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] Merging multiple columns into one column
?unlist (A data frame is a list, as ?data.frame explains. Also the Intro to R tutorial, which should be read by everyone beginning with R). -- Bert On Fri, Sep 28, 2012 at 2:51 PM, Meredith Ballard LaBeau wrote: > Good Evening- > I have a dataframe that has 10 columns that has a header and 7306 rows in > each column, I want to combine these columns into one. I utilized the stack > function but it only returned 3/4 of the data...my code is: > where nfcuy_bw is the dataframe with 7305 obs. and 10 variables > Once I apply this code I only receive a data frame with 58440 obs. of 2 > variables, of which there should be 73,050 obs. of 2 variables, just > wondering what is happening here? > > View(nfcuy_bw) > > attach(nfcuy_bw) > > cuyahoga_nf<-data.frame(s5,s10,s25,s27,s33,s41,s51,his_c) > > cuy_nf<-stack(cuyahoga_nf) > > Thanks > Meredith > > -- > Doctoral Candidate > Department of Civil and Environmental Engineering > Michigan Technological University > > [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] changing outlier shapes of boxplots using lattice
Hello Ilai,
Thank you for the response.
It did help a lot.
However, a beginner to lattice has three questions.
Q1
Please kindly explain why "in this case OP is using it with no "at"
argument,""
so it is possible to display the median and the outliers with different pch?
Q2.
what is the relationship between package "HH" and graphic-drawing?
I checked ??HH and found little explanation on its function of
graphic-drawing.
Q3
Please kindly advise how to make outliers empty circle (pch=2) in this case
as the code below.
Thank you.
code
Diet.colors <-
c("forestgreen","darkgreen","chocolate1","darkorange2","sienna2",
"red2","firebrick3","saddlebrown","coral4","chocolate4","darkblue","navy","grey38")
levels(dataN$Diet_B) <- diet.code
bwplot(MS_midpoint_lat~Diet_B, data=dataN,
xlab=list("Diet of Breeding Ground", cex = 1.4),
ylab = list("Latitudinal Midpoint Breeding Ground ",cex = 1.4),
lwd=1.5,
cex.lab=1.4, cex.axis=1.2,
font.axis=2,
cex=1.5,
las=1,
panel=panel.bwplot.intermediate.hh,
bty="l",
col=Diet.colors,
pch=rep("l",13),
scales=list(x=list(rot=90)),
par.settings=list(plot.symbol = list(pch = 2, cex =
2),box.umbrella=list(lty=1)))
Elaine
On Sat, Sep 29, 2012 at 2:44 AM, ilai wrote:
> On Fri, Sep 28, 2012 at 6:57 AM, Richard M. Heiberger wrote:
>
>> Elaine,
>>
>> For panel.bwplot you see that the central dot and the outlier dots are
>> controlled by
>> the same pch argument.
>
>
> ??? I don't think so...
>
> bwplot(rgamma(20,.1,1)~gl(2,10), pch=rep(17,2),
> panel = lattice::panel.bwplot)
>
> I think you mean panel.bwplot.intermidiate.hh ?
>
> BTW thank you for the useful HH package but in this case OP is using it
> with no "at" argument, so why not
>
> Diet.colors <- c("forestgreen", "darkgreen","chocolate1","darkorange2",
> "sienna2","red2","firebrick3","saddlebrown","coral4","chocolate4","darkblue","navy","grey38")
> bwplot(rgamma(20*13,1,.1)~gl(13,20),
> fill = Diet.colors, pch = "|",
> par.settings = list(box.umbrella=list(lty=1)))
>
> cheers
>
>
>
> I initially set the pch="|" to match your first
>> example with the horizontal
>> indicator for the median. I would be inclined to use the default circle
>> for the outliers and
>> therefore also for the median.
>>
>> Rich
>>
>> On Fri, Sep 28, 2012 at 7:13 AM, Sarah Goslee > >wrote:
>>
>> > I would guess that if you find the bit that says pch="|" and change it
>> to
>> > pch=1 it will solve your question, and that reading ?par will tell you
>> why.
>> >
>> > Sarah
>> >
>> > On Thursday, September 27, 2012, Elaine Kuo wrote:
>> >
>> > > Hello
>> > >
>> > > This is Elaine.
>> > >
>> > > I am using package lattice to generate boxplots.
>> > > Using Richard's code, the display was almost perfect except the
>> outlier
>> > > shape.
>> > > Based on the following code, the outliers are vertical lines.
>> > > However, I want the outliers to be empty circles.
>> > > Please kindly help how to modify the code to change the outlier
>> shapes.
>> > > Thank you.
>> > >
>> > > code
>> > > package (lattice)
>> > >
>> > > dataN <- data.frame(GE_distance=rnorm(260),
>> > >
>> > > Diet_B=factor(rep(1:13, each=20)))
>> > >
>> > > Diet.colors <- c("forestgreen",
>> "darkgreen","chocolate1","darkorange2",
>> > >
>> > > "sienna2","red2","firebrick3","saddlebrown","coral4",
>> > >
>> > > "chocolate4","darkblue","navy","grey38")
>> > >
>> > > levels(dataN$Diet_B) <- Diet.colors
>> > >
>> > > bwplot(GE_distance ~ Diet_B, data=dataN,
>> > >
>> > >xlab=list("Diet of Breeding Ground", cex = 1.4),
>> > >
>> > >ylab = list(
>> > >
>> > > "Distance between Centers of B and NB Range (1000 km)",
>> > >
>> > > cex = 1.4),
>> > >
>> > >panel=panel.bwplot.intermediate.hh,
>> > >
>> > >col=Diet.colors,
>> > >
>> > >pch=rep("|",13),
>> > >
>> > >scales=list(x=list(rot=90)),
>> > >
>> > >par.settings=list(box.umbrella=list(lty=1)))
>> > >
>> > > [[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.
>> > >
>> >
>> >
>> > --
>> > Sarah Goslee
>> > http://www.stringpage.com
>> > http://www.sarahgoslee.com
>> > http://www.functionaldiversity.org
>> >
>> > [[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]]
>>
>> __
Re: [R] Heatmap Colors
On Sep 28, 2012, at 3:16 PM, Nick Fankhauser wrote:
> Hello R-Users!
>
> I'm using a heatmap to visualize a matrix of values between -1 and 3.
> How can I set the colors so that white is zero, below zero is blue of
> increasing intensity towards -1 and above zero is red of increasing intensity
> towards red?
>
> I tried like this (using the marray and gplots packages from bioconductor):
> mcol <- maPalette(low="blue", mid="white", high="red",k=100)
> heatmap.2(my_matrix, col=mcol)
>
> But white does not correspond to zero, because the value distribution is not
> symmetrical, so that zero is not in the middle.
> Is it somehow possible to create a color palette with white centered at zero?
The way you stated it at the beginning, I thought you should want the palette
centered at 1 rather than 0:
test <- seq(-1,3, len=20)
shift.BR <- colorRamp(c("blue","white", "red"), bias=2)((1:16)/16)
tpal <- rgb(shift.BR, maxColorValue=255)
barplot(test,col = tpal)
Perhaps I was being led astray by a somewhat similar question on StackOverflow.
--
David Winsemius, MD
Alameda, CA, USA
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Heatmap Colors
On Sep 28, 2012, at 4:52 PM, David Winsemius wrote:
>
> On Sep 28, 2012, at 3:16 PM, Nick Fankhauser wrote:
>
>> Hello R-Users!
>>
>> I'm using a heatmap to visualize a matrix of values between -1 and 3.
>> How can I set the colors so that white is zero, below zero is blue of
>> increasing intensity towards -1 and above zero is red of increasing
>> intensity towards red?
>>
>> I tried like this (using the marray and gplots packages from bioconductor):
>> mcol <- maPalette(low="blue", mid="white", high="red",k=100)
>> heatmap.2(my_matrix, col=mcol)
>>
>> But white does not correspond to zero, because the value distribution is not
>> symmetrical, so that zero is not in the middle.
>> Is it somehow possible to create a color palette with white centered at zero?
>
> The way you stated it at the beginning, I thought you should want the palette
> centered at 1 rather than 0:
Oopps ... should have the number of breaks match the number of colors:
test <- seq(-1,3, len=20)
shift.BR <- colorRamp(c("blue","white", "red"), bias=2)((1:20)/20)
tpal <- rgb(shift.BR, maxColorValue=255)
barplot(test,col = tpal)
>
--
David Winsemius, MD
Alameda, CA, USA
__
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[R] Errors in if statement
Hi guys, I have many rows (>1000) and columns (>30) of "geno" matrix. I use the
following loop and condition statement (adapted from someone else code). I
always have an error below. I was wondering if anyone knows what's the problem
& how to fix it.
Thanks,Zhengyu ### geno matrix P1 P2 P3 P4
1 2 2 3 2
2 2 2 1 1
1 2 1 2 NANA 2 3 4 5 ###
for(i in 1:4) {
cat(i,"")
if(sum(geno[i,]!=2)>3 && sum(geno[i,]==1)>=1 && sum(geno[i,]==3)>=1){
tmp = 1
}
} ### 1 2 Error in if (sum(geno[i, ] != 2) > 3 && sum(geno[i, ] == 1)
>= 1 && sum(geno[i, :
missing value where TRUE/FALSE needed
[[alternative HTML version deleted]]
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Re: [R] Better way of Grouping?
Hi,
You can also use grep() to subset:
LD<-paste0(rep(rep(c(3,4),each=4),2),c(rep("L",8),rep("D",8)))
set.seed(1)
dat1<-data.frame(LD=LD,value=sample(1:15,16,replace=TRUE))
dat2<-within(dat1,{LD<-as.character(LD)})
dat2[grepl(".*L",dat2$LD),] # subset all L values
dat2[grepl(".*D",dat2$LD),] # subset all D values
dat2[grepl("3D",dat2$LD),]
dat2[grepl("4D",dat2$LD),]
A.K.
- Original Message -
From: Charles Determan Jr
To: r-help@r-project.org
Cc:
Sent: Friday, September 28, 2012 2:59 PM
Subject: [R] Better way of Grouping?
Hello R users,
This is more of a convenience question that I hope others might find useful
if there is a better answer. I work with large datasets that requires
multiple parsing stages for different analysis. For example, compare group
3 vs. group 4. A more complicated comparison would be time B in group 3 of
group L with B in group 4 of group L. I normally subset each group with
the following type of code.
data=read(...)
#L v D
L=data[LvD %in% c("L"),]
D=data[LvD %in% c("D"),]
#Groups 3 and 4 within L and D
group3L=L[group %in% c("3"),]
group4L=L[group %in% c("3"),]
group3D=D[group %in% c("3"),]
group4D=D[group %in% c("3"),]
#Times B, S45, FR2, FR8
you get the idea
Is there a more efficient way to subset groups? Thanks for any insight.
Regards,
Charles
[[alternative HTML version deleted]]
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[R] Text mining? Text manipulation? Both? Predicting KRAS test results in cancer patients
Happy Friday Everyone, Hope Friday afternoon doesn't turn out to be a terrible time to post a question. I've been doing a little data mining of patient text medical records as of late. I started out trying to predict whether or not cancer patients had received KRAS mutation testing and did quite well with that. Now I'm trying to predict the results of KRAS testing (mutated vs. wild type). This is proving to be a little more difficult. With the first classification task, I created counts of terms (e.g., ""kras", "mutated") in the text medical records using the tm package and then used those counts to predict whether or not patients had had KRAS mutation testing. I tried a few different analyses here, but found that random forests worked the best. Predicting the results of testing is harder though because of the way physicians and other healthcare professionals write about testing. For example, I'm finding phrases like "KRAS mutation returned wild-type". In this example, if we're counting, we get 1 instance of "kras", 1 instance of "mutated", and one instance of "wild". So you can see how it might be difficult to accurately predict the results of testing based on counts alone. My question is how best to deal with this. Are there any R text mining packages or related software that would be particularly suited to my problem? I took a look at the CRAN Task View: Natural Language Processing and there were so many options I didn't really know where to start (and it's not even clear that an R-based solution will work best for my problem). Alternatively, is there any real chance one could simply write code that would be able to identify true references to the results of KRAS testing and then create counts only of what are likely to be true references? I'd greatly appreciate it if someone could point me in the right direction. Thanks, Paul __ 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 Original and Duplicates
That works. Thank you!
On Fri, Sep 28, 2012 at 4:22 PM, Rui Barradas wrote:
> Hello,
>
> Try the following.
>
>
> idx <- duplicated(df) | duplicated(df, fromLast = TRUE)
> df[idx, ]
>
> Note that they are returned in their original order in the df.
>
> Hope this helps,
>
> Rui Barradas
>
> Em 28-09-2012 21:11, Adam Gabbert escreveu:
>
>> I would like to select a all the duplicate rows of a data frame including
>> the original. Any help would be much appreciated. This is where I'm at
>> so
>> far. Thanks.
>>
>> #Sample data frame:
>> df <- read.table(header=T, con <- textConnection('
>> label value
>> A 4
>> B 3
>> C 6
>> B 3
>> B 1
>> A 2
>> A 4
>> A 4
>> '))
>> close(con)
>>
>> # Duplicate entries
>> df[duplicated(df),]
>>
>> # label value
>> # B 3
>> # A 4
>> # A 4
>>
>> #I want to select all the rows that are duplicated including the original
>> #This is the output I want
>> # label value
>> # B 3
>> # B 3
>> # A 4
>> # A 4
>> # A 4
>>
>> [[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] Select Original and Duplicates
HI,
You can also try:
idx<-data.frame(t(sapply(df,function(x) !is.na(match(x,x[duplicated(x)])
df1<-df[sapply(idx,function(x) all(x==TRUE)),]
df1
# label value
#1 A 4
#2 B 3
#4 B 3
#7 A 4
#8 A 4
A.K.
- Original Message -
From: Rui Barradas
To: Adam Gabbert
Cc: r-help@r-project.org
Sent: Friday, September 28, 2012 4:22 PM
Subject: Re: [R] Select Original and Duplicates
Hello,
Try the following.
idx <- duplicated(df) | duplicated(df, fromLast = TRUE)
df[idx, ]
Note that they are returned in their original order in the df.
Hope this helps,
Rui Barradas
Em 28-09-2012 21:11, Adam Gabbert escreveu:
> I would like to select a all the duplicate rows of a data frame including
> the original. Any help would be much appreciated. This is where I'm at so
> far. Thanks.
>
> #Sample data frame:
> df <- read.table(header=T, con <- textConnection('
> label value
> A 4
> B 3
> C 6
> B 3
> B 1
> A 2
> A 4
> A 4
> '))
> close(con)
>
> # Duplicate entries
> df[duplicated(df),]
>
> # label value
> # B 3
> # A 4
> # A 4
>
> #I want to select all the rows that are duplicated including the original
> #This is the output I want
> # label value
> # B 3
> # B 3
> # A 4
> # A 4
> # A 4
>
> [[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.
__
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__
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and provide commented, minimal, self-contained, reproducible code.
[R] Converting array to matrix
Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact") I basically want to plot this probability with filled. contour. Many thanks. method p n coverage 1 exact 0.01 5 0.9990199 2 exact 0.01 10 0.9957338 3 exact 0.01 15 0.9903702 4 exact 0.01 20 0.9831407 5 exact 0.01 25 0.9980493 6 exact 0.01 30 0.9966823 7 exact 0.01 35 0.9948463 8 exact 0.01 40 0.9925026 9 exact 0.01 45 0.9896219 10 exact 0.01 50 0.9861827 11 exact 0.01 55 0.9821712 12 exact 0.01 60 0.9775798 13 exact 0.01 65 0.9958308 14 exact 0.01 70 0.9945711 15 exact 0.01 75 0.9930800 16 exact 0.01 80 0.9913408 17 exact 0.01 85 0.9893386 18 exact 0.01 90 0.9870598 19 exact 0.01 95 0.9844924 20 exact 0.01 100 0.9816260 21 exact 0.02 5 0.9961576 22 exact 0.02 10 0.9838224 23 exact 0.02 15 0.9969606 24 exact 0.02 20 0.9929313 25 exact 0.02 25 0.9867566 26 exact 0.02 30 0.9782822 27 exact 0.02 35 0.9948918 28 exact 0.02 40 0.9917591 29 exact 0.02 45 0.9875780 30 exact 0.02 50 0.9822419 31 exact 0.02 55 0.9756698 32 exact 0.02 60 0.9929754 33 exact 0.02 65 0.9902072 34 exact 0.02 70 0.9867702 35 exact 0.02 75 0.9826010 36 exact 0.02 80 0.9776446 37 exact 0.02 85 0.9927058 38 exact 0.02 90 0.9904482 39 exact 0.02 95 0.9877327 40 exact 0.02 100 0.9845164 41 exact 0.03 5 0.9915279 42 exact 0.03 10 0.9972351 43 exact 0.03 15 0.9906286 44 exact 0.03 20 0.9789916 45 exact 0.03 25 0.9938142 46 exact 0.03 30 0.9880954 47 exact 0.03 35 0.9797802 48 exact 0.03 40 0.9933299 49 exact 0.03 45 0.9890462 50 exact 0.03 50 0.9831894 51 exact 0.03 55 0.9755598 52 exact 0.03 60 0.9908560 53 exact 0.03 65 0.9866943 54 exact 0.03 70 0.9813629 55 exact 0.03 75 0.9926775 56 exact 0.03 80 0.9896911 57 exact 0.03 85 0.9859049 58 exact 0.03 90 0.9812172 59 exact 0.03 95 0.9755343 60 exact 0.03 100 0.9893762 61 exact 0.04 5 0.9852420 62 exact 0.04 10 0.9937863 63 exact 0.04 15 0.9797082 64 exact 0.04 20 0.9925871 65 exact 0.04 25 0.9834784 66 exact 0.04 30 0.9936800 67 exact 0.04 35 0.9877867 68 exact 0.04 40 0.9789777 69 exact 0.04 45 0.9912599 70 exact 0.04 50 0.9855896 71 exact 0.04 55 0.9777638 72 exact 0.04 60 0.9901122 73 exact 0.04 65 0.9849824 74 exact 0.04 70 0.9781965 75 exact 0.04 75 0.9897956 76 exact 0.04 80 0.9852643 77 exact 0.04 85 0.9794261 78 exact 0.04 90 0.9899813 79 exact 0.04 95 0.9653302 80 exact 0.04 100 0.9641378 81 exact 0.05 5 0.9774075 82 exact 0.05 10 0.9884964 83 exact 0.05 15 0.9945327 84 exact 0.05 20 0.9840985 85 exact 0.05 25 0.9928351 86 exact 0.05 30 0.9843645 87 exact 0.05 35 0.9927483 88 exact 0.05 40 0.9861231 89 exact 0.05 45 0.9761385 90 exact 0.05 50 0.9882136 91 exact 0.05 55 0.9806825 92 exact 0.05 60 0.9902109 93 exact 0.05 65 0.9844774 94 exact 0.05 70 0.9766393 95 exact 0.05 75 0.9662306 96 exact 0.05 80 0.9650815 97 exact 0.05 85 0.9772934 98 exact 0.05 90 0.9755923 99 exact 0.05 95 0.9718140 100 exact 0.05 100 0.9826071 101 exact 0.06 5 0.9980297 102 exact 0.06 10 0.9811622 103 exact 0.06 15 0.9896401 104 exact 0.06 20 0.9943659 105 exact 0.06 25 0.9849507 106 exact 0.06 30 0.9920548 107 exact 0.06 35 0.9831689 108 exact 0.06 40 0.9909419 109 exact 0.06 45 0.9829932 110 exact 0.06 50 0.9906217 111 exact 0.06 55 0.9836566 112 exact 0.06 60 0.9663670 113 exact 0.06 65 0.9668145 114 exact 0.06 70 0.9630279 115 exact 0.06 75 0.9763348 116 exact 0.06 80 0.9716289 117 exact 0.06 85 0.9820840 118 exact 0.06 90 0.9772655 119 exact 0.06 95 0.9687703 120 exact 0.06 100 0.9680765 121 exact 0.07 5 0.9969201 122 exact 0.07 10 0.9964239 123 exact 0.07 15 0.9824673 124 exact 0.07 20 0.9892932 125 exact 0.07 25 0.9934691 126 exact 0.07 30 0.9837683 127 exact 0.07 35 0.9902956 128 exact 0.07 40 0.9801496 129 exact 0.07 45 0.9879752 130 exact 0.07 50 0.9779901 131 exact 0.07 55 0.9679391 132 exact 0.07 60 0.9640110 133 exact 0.07 65 0.9765091 134 exact 0.07 70 0.9702320 135 exact 0.07 75 0.9806132 136 exact 0.07 80 0.9553953 137 exact 0.07 85 0.9692733 138 exact 0.07 90 0.9656231 139 exact 0.07 95 0.9765780 140 exact 0.07 100 0.9715796 141 exact 0.08 5 0.9954747 142 exact 0.08 10 0.9941987 143 exact 0.08 15 0.9950303 144 exact 0.08 20 0.9816556 145 exact 0.08 25 0.9877073 146 exact
Re: [R] Converting array to matrix
On Sep 28, 2012, at 3:59 PM, farnoosh sheikhi wrote: > Hi, > > I have a 3d array as below, I want to make this array to a matrix of > p=50(rows) and n=20(columns) with the coverage values . > The code before the array is: ?matrix mat <- matrix(datfrm$coverage, 50, 20) filled.contour(mat) # untested -- David > library(binom) > Loading required package: lattice > pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) > no.seq<-seq(from = 5, to = 100, by = 5) > cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = > "exact") > > > I basically want to plot this probability with filled. contour. > Many thanks. > > methodp n coverage > 1 exact 0.01 5 0.9990199 > 2 exact 0.01 10 0.9957338 > 3 exact 0.01 15 0.9903702 > 4 exact 0.01 20 0.9831407 > 5 exact 0.01 25 0.9980493 > 6 exact 0.01 30 0.9966823 > 7 exact 0.01 35 0.9948463 > 8 exact 0.01 40 0.9925026 > 9 exact 0.01 45 0.9896219 > 10exact 0.01 50 0.9861827 > 11exact 0.01 55 0.9821712 > 12exact 0.01 60 0.9775798 > 13exact 0.01 65 0.9958308 > 14exact 0.01 70 0.9945711 > 15exact 0.01 75 0.9930800 > 16exact 0.01 80 0.9913408 > 17exact 0.01 85 0.9893386 > 18exact 0.01 90 0.9870598 > 19exact 0.01 95 0.9844924 > 20exact 0.01 100 0.9816260 > 21exact 0.02 5 0.9961576 > 22exact 0.02 10 0.9838224 > 23exact 0.02 15 0.9969606 > 24exact 0.02 20 0.9929313 > 25exact 0.02 25 0.9867566 > 26exact 0.02 30 0.9782822 > 27exact 0.02 35 0.9948918 > 28exact 0.02 40 0.9917591 > 29exact 0.02 45 0.9875780 > 30exact 0.02 50 0.9822419 > 31exact 0.02 55 0.9756698 > 32exact 0.02 60 0.9929754 > 33exact 0.02 65 0.9902072 > 34exact 0.02 70 0.9867702 > 35exact 0.02 75 0.9826010 > 36exact 0.02 80 0.9776446 > 37exact 0.02 85 0.9927058 > 38exact 0.02 90 0.9904482 > 39exact 0.02 95 0.9877327 > 40exact 0.02 100 0.9845164 > 41exact 0.03 5 0.9915279 > 42exact 0.03 10 0.9972351 > 43exact 0.03 15 0.9906286 > 44exact 0.03 20 0.9789916 > 45exact 0.03 25 0.9938142 > 46exact 0.03 30 0.9880954 > 47exact 0.03 35 0.9797802 > 48exact 0.03 40 0.9933299 > 49exact 0.03 45 0.9890462 > 50exact 0.03 50 0.9831894 > 51exact 0.03 55 0.9755598 > 52exact 0.03 60 0.9908560 > 53exact 0.03 65 0.9866943 > 54exact 0.03 70 0.9813629 > 55exact 0.03 75 0.9926775 > 56exact 0.03 80 0.9896911 > 57exact 0.03 85 0.9859049 > 58exact 0.03 90 0.9812172 > 59exact 0.03 95 0.9755343 > 60exact 0.03 100 0.9893762 > 61exact 0.04 5 0.9852420 > 62exact 0.04 10 0.9937863 > 63exact 0.04 15 0.9797082 > 64exact 0.04 20 0.9925871 > 65exact 0.04 25 0.9834784 > 66exact 0.04 30 0.9936800 > 67exact 0.04 35 0.9877867 > 68exact 0.04 40 0.9789777 > 69exact 0.04 45 0.9912599 > 70exact 0.04 50 0.9855896 > 71exact 0.04 55 0.9777638 > 72exact 0.04 60 0.9901122 > 73exact 0.04 65 0.9849824 > 74exact 0.04 70 0.9781965 > 75exact 0.04 75 0.9897956 > 76exact 0.04 80 0.9852643 > 77exact 0.04 85 0.9794261 > 78exact 0.04 90 0.9899813 > 79exact 0.04 95 0.9653302 > 80exact 0.04 100 0.9641378 > 81exact 0.05 5 0.9774075 > 82exact 0.05 10 0.9884964 > 83exact 0.05 15 0.9945327 > 84exact 0.05 20 0.9840985 > 85exact 0.05 25 0.9928351 > 86exact 0.05 30 0.9843645 > 87exact 0.05 35 0.9927483 > 88exact 0.05 40 0.9861231 > 89exact 0.05 45 0.9761385 > 90exact 0.05 50 0.9882136 > 91exact 0.05 55 0.9806825 > 92exact 0.05 60 0.9902109 > 93exact 0.05 65 0.9844774 > 94exact 0.05 70 0.9766393 > 95exact 0.05 75 0.9662306 > 96exact 0.05 80 0.9650815 > 97exact 0.05 85 0.9772934 > 98exact 0.05 90 0.9755923 > 99exact 0.05 95 0.9718140 > 100 exact 0.05 100 0.9826071 > 101 exact 0.06 5 0.9980297 > 102 exact 0.06 10 0.9811622 > 103 exact 0.06 15 0.9896401 > 104 exact 0.06 20 0.9943659 > 105 exact 0.06 25 0.9849507 > 106 exact 0.06 30 0.9920548 > 107 exact 0.06 35 0.9831689 > 108 exact 0.06 40 0.9909419 > 109 exact 0.06 45 0.9829932 > 110 exact 0.06 50 0.9906217 > 111 exact 0.06 55 0.9836566 > 112 exact 0.06 60 0.9663670 > 113 exact 0.06 65 0.9668145 > 114 exact 0.06 70 0.9630279 > 115 exact 0.06 75 0.9763348 > 116 exact 0.06 80 0.9716289 > 117 exact 0.06 85 0.9820840 > 118 exact 0.06 90 0.9772655 > 119 exact 0.06 95 0.9687703 > 120 exact 0.06 100 0.9680765 > 121 exact 0.07 5 0.9969201 > 122 exact 0.07 10 0.9964239 > 123 exact 0.07 15 0.9824673 > 124 exact 0.07 20 0.9892932 > 125 exact 0.07 25 0.9934691 > 126 exact 0.07 30 0.9837683 > 127 exact 0.07 35 0.9902956 > 128 exact 0.07 40 0.9801496 > 129 exact 0.07 45 0.9879752 > 130 exact 0.07 50 0.9779901 > 131 exact 0.07 55 0.9679391 >
Re: [R] Errors in if statement
On Sep 28, 2012, at 1:16 PM, JiangZhengyu wrote:
>
> Hi guys, I have many rows (>1000) and columns (>30) of "geno" matrix. I use
> the following loop and condition statement (adapted from someone else code).
> I always have an error below. I was wondering if anyone knows what's the
> problem & how to fix it.
Boy, it surely looks like missing values are the problem. Have you read:
?sum
--
David.
> Thanks,Zhengyu ### geno matrix P1 P2 P3 P4
> 1 2 2 3 2
> 2 2 2 1 1
> 1 2 1 2 NANA 2 3 4 5 ###
> for(i in 1:4) {
> cat(i,"")
> if(sum(geno[i,]!=2)>3 && sum(geno[i,]==1)>=1 && sum(geno[i,]==3)>=1){
> tmp = 1
> }
> } ### 1 2 Error in if (sum(geno[i, ] != 2) > 3 && sum(geno[i, ] == 1)
> >= 1 && sum(geno[i, :
> missing value where TRUE/FALSE needed
>
> [[alternative HTML version deleted]]
David Winsemius, MD
Alameda, CA, USA
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