[R] Error in CSV file

[Shivi82]

Hello All,
This is an easy fix but I am not able to find the root cause of the error. I
am trying to upload a csv file but it is throwing an error.
Have done a lot of research on google and some tutorial but cant find a
solution hence please advice:-
Syntax is :-   aaa<-read.csv(file ="VehicleData.csv",Header=TRUE)

Error:- Error in read.table(file = file, header = header, sep = sep, quote =
quote,  : 
  unused argument (Header = TRUE)

Snapshot of the file:-
Weight  Hours   PROCESS Month   Weekday Day
6828 13 INBOUND Mar   Fri   13
2504 16 INBOUND Mar   Fri   27
20   16 INBOUND Mar   Fri   27
1026216 INBOUND Mar   Fri   27
2500 17 INBOUND Mar   Fri   13

Kindly help. 




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Re: [R] Error in CSV file

[Ivan Calandra]

Hi Shivi,

R is case sensitive and the error message that the argument "Header" is 
unused (because unrecognized). Try with "header" (lower case "h") and it 
should work.


HTH,
Ivan

--
Ivan Calandra, ATER
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26 77 36 89
ivan.calan...@univ-reims.fr
https://www.researchgate.net/profile/Ivan_Calandra

Le 29/05/15 10:41, Shivi82 a écrit :

Hello All,
This is an easy fix but I am not able to find the root cause of the error. I
am trying to upload a csv file but it is throwing an error.
Have done a lot of research on google and some tutorial but cant find a
solution hence please advice:-
Syntax is :-   aaa<-read.csv(file ="VehicleData.csv",Header=TRUE)

Error:- Error in read.table(file = file, header = header, sep = sep, quote =
quote,  :
   unused argument (Header = TRUE)

Snapshot of the file:-
Weight  Hours   PROCESS Month   Weekday Day
6828 13 INBOUND Mar   Fri   13
2504 16 INBOUND Mar   Fri   27
20   16 INBOUND Mar   Fri   27
1026216 INBOUND Mar   Fri   27
2500 17 INBOUND Mar   Fri   13

Kindly help.




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Re: [R] Error in CSV file

[Rainer M Krug]

Shivi82  writes:

> Hello All,
> This is an easy fix but I am not able to find the root cause of the error. I
> am trying to upload a csv file but it is throwing an error.
> Have done a lot of research on google and some tutorial but cant find a
> solution hence please advice:-
> Syntax is :-   aaa<-read.csv(file ="VehicleData.csv",Header=TRUE)
>
> Error:- Error in read.table(file = file, header = header, sep = sep, quote =
> quote,  : 
>   unused argument (Header = TRUE)
 ^^

use "header = TRUE" instead of "Header = TRUE". R is case sensitive.

Cheers,

Rainer

>
> Snapshot of the file:-
> WeightHours   PROCESS Month   Weekday Day
> 6828   13 INBOUND Mar   Fri   13
> 2504   16 INBOUND Mar   Fri   27
> 20 16 INBOUND Mar   Fri   27
> 10262  16 INBOUND Mar   Fri   27
> 2500   17 INBOUND Mar   Fri   13
>
> Kindly help. 
>
>
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Error-in-CSV-file-tp4707879.html
> Sent from the R help mailing list archive at Nabble.com.
>

-- 
Rainer M. Krug
email: Rainerkrugsde
PGP: 0x0F52F982


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Re: [R] Error in CSV file

[Shivi82]

This ate my head like for 2 hours. God thanks for the help. 



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Re: [R] analysis of variance test

[Michael Dewey]

Dear Nezahat
In future it would be helpful if you

1 - gave us the data so we can reproduce what you are doing
2 - told us what the error was in case we cannot replicate ti
3 - did not post in HTML as it messes up everything in your post

What did you think x1 <- numeric was going to do?
Try
x1 <- numeric
str(x1)


On 28/05/2015 22:16, Nezahat HUnter wrote:


Let's say I have 12 observation of 5 variables and my first variable is categorical (with 
4 different levels). I am trying to find out statistical significance difference between 
these categorical levels for each variable, but my  function is not working! Please note 
that my data "x" are in data.frame format.
Any suggestion would be helpful.Many thanks.

function(x)
{
 x1 <- numeric
 x2 <- numeric
 for(i in 2:length(x)) {
 x1[i] <- summary(aov(x[, i] ~ factor(x[, 1])))
 x2[i] <- x1[i]$Pr[1]  #Pr is the probability values
 if(x2[i] < 0.06)
 x2[i] <- 1
 else x2[i] <- 0
 }
 x2
}




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Michael
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Re: [R] analysis of variance test

[Jim Lemon]

Hi Nezahat,
First, you are storing the code of the function "numeric" in x1 and
x2. You probably want to use:

x1<-numeric()
x2<-numeric()

Second, you are then storing the output of your aov summary (a list)
in x1, which requires a bit of analysis to get the information you
want (i.e. p value). The following will work for your example, but is
not a general solution.

nh_fun<-function(x) {
pvals <-numeric()
for(i in 2:length(x))
pvals[i-1]<-unlist(summary(aov(x[,i] ~
factor(x[,1])))[[1]][5])[1] <= 0.05
return(pvals)
}

nh_fun(x)

As you probably want to get the conventional <=0.05, I have changed
the criterion. If you want to understand why the mess of extractors
appears after the "summary" call, use the "str" function successively
on the return value from "summary"

Jim


On Fri, May 29, 2015 at 7:16 AM, Nezahat HUnter
 wrote:
>
> Let's say I have 12 observation of 5 variables and my first variable is 
> categorical (with 4 different levels). I am trying to find out statistical 
> significance difference between these categorical levels for each variable, 
> but my  function is not working! Please note that my data "x" are in 
> data.frame format.
> Any suggestion would be helpful.Many thanks.
>
> function(x)
> {
> x1 <- numeric
> x2 <- numeric
> for(i in 2:length(x)) {
> x1[i] <- summary(aov(x[, i] ~ factor(x[, 1])))
> x2[i] <- x1[i]$Pr[1]  #Pr is the probability values
> if(x2[i] < 0.06)
> x2[i] <- 1
> else x2[i] <- 0
> }
> x2
> }
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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] How to make new predictions from a GAM with a spline forced through the origin

[Gavan McGrath]

Hi,

I’m followed an example to fit a GAM with a spline forced through a point, i.e. 
(0,0). This works fine from one of Simon’s examples however when it comes to 
making a prediction from a new set of x values I’m a bit stumped.

In the example below a smooth term is constructed and the basis and penalties 
at x=0 are removed then the gam is fitted to a spline basis matrix X using 
spline penalties.

Can someone suggest a way that I can make predictions at new  x  values based 
on the gam b below.


Here is Simon Wood's example:

library(mgcv)
set.seed(0)
n <- 100
x <- runif(n)*4-1;x <- sort(x);
f <- exp(4*x)/(1+exp(4*x));y <- f+rnorm(100)*0.1;plot(x,y)
dat <- data.frame(x=x,y=y)

## Create a spline basis and penalty, making sure there is a knot
## at the constraint point, (0 here, but could be anywhere)
knots <- data.frame(x=seq(-1,3,length=9)) ## create knots
## set up smoother...
sm <- smoothCon(s(x,k=9,bs="cr"),dat,knots=knots)[[1]]

## 3rd parameter is value of spline at knot location 0,
## set it to 0 by dropping...
X <- sm$X[,-3]## spline basis
S <- sm$S[[1]][-3,-3] ## spline penalty
off <- y*0 + .6   ## offset term to force curve through (0, .6)

## fit spline constrained through (0, .6)...
b <- gam(y ~ X - 1 + offset(off),paraPen=list(X=list(S)))
lines(x,predict(b))



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[R] Help on R Functionality & Histogram

[Shivi82]

Hello Experts, 
I have couple of questions on the analysis I am creating.
1) How does R adopt to changes. The case I have here is that the excel I
have started initially had to be modified because the data I had was on
hourly basis ranging from 0 to 23 hours. After Changes 0 was modified to 24
in hours. Now do I need to recall this excel again in R using read.csv
syntax or is there another way to do so i.e. a kind of reload option
2) I am creating a histogram. I need on x axis 24 hours to be displayed
separately as 0,1,2, and thereon. However it only shows till 20 which makes
the look awkward. Also all l need to resize the labels and if possible
inside the bars. It used the below code, axis fonts have changed but labels
give an error with this code

Code:- hist(aaa$Hours,main="Hourly Weight",xlab = "Time",breaks = 25,col =
"yellow",ylim = c(0,9000),
 labels=TRUE, cex.axis=0.6,cex.label=0.6)

Kindly advice on the both the questions. Thanks. 

Histogram.png   



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[R] Help on R Functionality & Histogram

[Shivi82]

Hello Experts, 
I have couple of questions on the analysis I am creating.
1) How does R adopt to changes. The case I have here is that the excel I
have started initially had to be modified because the data I had was on
hourly basis ranging from 0 to 23 hours. After Changes 0 was modified to 24
in hours. Now do I need to recall this excel again in R using read.csv
syntax or is there another way to do so i.e. a kind of reload option
2) I am creating a histogram. I need on x axis 24 hours to be displayed
separately as 0,1,2, and thereon. However it only shows till 20 which makes
the look awkward. Also all l need to resize the labels and if possible
inside the bars. It used the below code, axis fonts have changed but labels
give an error with this code

Code:- hist(aaa$Hours,main="Hourly Weight",xlab = "Time",breaks = 25,col =
"yellow",ylim = c(0,9000),
 labels=TRUE, cex.axis=0.6,cex.label=0.6)

Kindly advice on the both the questions. Thanks. 






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Re: [R] Help on R Functionality & Histogram

[Sarah Goslee]

On Fri, May 29, 2015 at 7:53 AM, Shivi82  wrote:
> Hello Experts,
> I have couple of questions on the analysis I am creating.
> 1) How does R adopt to changes. The case I have here is that the excel I
> have started initially had to be modified because the data I had was on
> hourly basis ranging from 0 to 23 hours. After Changes 0 was modified to 24
> in hours. Now do I need to recall this excel again in R using read.csv
> syntax or is there another way to do so i.e. a kind of reload option

Using read.csv() is the reload option. R has no automatic interface to
external files.


> 2) I am creating a histogram. I need on x axis 24 hours to be displayed
> separately as 0,1,2, and thereon. However it only shows till 20 which makes
> the look awkward. Also all l need to resize the labels and if possible
> inside the bars. It used the below code, axis fonts have changed but labels
> give an error with this code
>
> Code:- hist(aaa$Hours,main="Hourly Weight",xlab = "Time",breaks = 25,col =
> "yellow",ylim = c(0,9000),
>  labels=TRUE, cex.axis=0.6,cex.label=0.6)

The most understandable approach is to break it down into chunks:
Create the histogram.
Add a custom axis.
Add custom labels.

# using fake data
aaa <- data.frame(Hours = sample(1:24, 1, replace=TRUE))

aaa.hist <- hist(aaa$Hours, main="Hourly Weight", xlab = "Time",
breaks = seq(0, 24), col = "yellow", ylim = c(0,9000), cex.axis=0.6,
xaxt="n")
axis(1, (0:23)+.5, 1:24, cex.axis=.6)
text((0:23)+.5, aaa.hist$counts-150, aaa.hist$counts, cex=.6)

Sarah

-- 
Sarah Goslee
http://www.functionaldiversity.org

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Re: [R] Help on R Functionality & Histogram

[Shivi82]

Thanks Sarah. This is magical. 
Thanks for explaining in such a length. 



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[R] An Odd Request

[Josh Grant]

Hello R-Users

I apologize in advance if my post is inappropriate. I read the entire
posting guide and found nothing to say so, but you never know. I am seeking
a knowledgable R-user that might be interested (for whatever reason) in
helping out on what I hope would be considered a worthy project.

I am a research scientist, albeit one with little programming ability. I
recently started a website which allows patients of different sorts to
suggest research studies. Everything is completely free and anonymous. When
several members express interest in a particular idea I attempt to build it
so they can actually run through the study. Clearly there are limits but we
currently we have 4 communities, chronic fatigue syndrome, fibromyalgia,
multiple sclerosis and pernicious anaemia and there are several active
studies in which people are submitting data every day. It's quite exciting
and I think it has great potential to help people, particularly with
disorders that have defied explanation.

I'm currently using google spreadsheets/forms to create symptom trackers
and interactive dashboards of the results which (most of the time) show
group results by default but which can show individual results if an ID is
entered. Unfortunately google spreadsheets is a little limited and I now
require the use of more complicated stats such as linear mixed models.

I know that I need to move to R, I understand the basics of running
statistical tests with packages such as LMER, but I have no clue how to go
about integrating such analyses into a website. I could certainly learn
how, would love to, and ultimately will, but if someone was interested in
joining me in this endeavour much more could be accomplished.

If you're interested in knowing more let me know.

Josh

[[alternative HTML version deleted]]

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[R] Problems with nls

[Abolfazl Saghafi]

Can some help me with a question on this bass model, please

As I read some articles on this topic, I understand that
1. the bass formula is
N(t) = pm + (q-p) N(t-1) - (q/m) (N(t-1))^2
2. which is a difference equation with the solution
N(t) = m (1 − exp(−(p+q)t)) / (1 + (q/p)exp(−(p+q)t))
3. So, using a linear regression would give us some some initial
estimations for the parameters m, p, q
4. we then can put the initial estimations into a NLS to get the better
estimations

Am I right?

Now the question is,
why is that I see people use cumulative data and try to fit it into a pdf as
M * ( ((P+Q)^2 / P) * exp(-(P+Q) * T79) ) / (1+(Q/P)*exp(-(P+Q)*T79))^2,

why not using the cumulative data and fit directly the N(t)

[[alternative HTML version deleted]]

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Re: [R] An Odd Request

[Charles Determan]

If you are primarily interested in making your R analyses in to a website
you should look in to the 'Shiny' package.  It makes generating web pages
very easy.  Here is a link to the Shiny Gallery providing some examples (
http://shiny.rstudio.com/gallery/).

Regards,
Charles

On Fri, May 29, 2015 at 7:48 AM, Josh Grant  wrote:

> Hello R-Users
>
> I apologize in advance if my post is inappropriate. I read the entire
> posting guide and found nothing to say so, but you never know. I am seeking
> a knowledgable R-user that might be interested (for whatever reason) in
> helping out on what I hope would be considered a worthy project.
>
> I am a research scientist, albeit one with little programming ability. I
> recently started a website which allows patients of different sorts to
> suggest research studies. Everything is completely free and anonymous. When
> several members express interest in a particular idea I attempt to build it
> so they can actually run through the study. Clearly there are limits but we
> currently we have 4 communities, chronic fatigue syndrome, fibromyalgia,
> multiple sclerosis and pernicious anaemia and there are several active
> studies in which people are submitting data every day. It's quite exciting
> and I think it has great potential to help people, particularly with
> disorders that have defied explanation.
>
> I'm currently using google spreadsheets/forms to create symptom trackers
> and interactive dashboards of the results which (most of the time) show
> group results by default but which can show individual results if an ID is
> entered. Unfortunately google spreadsheets is a little limited and I now
> require the use of more complicated stats such as linear mixed models.
>
> I know that I need to move to R, I understand the basics of running
> statistical tests with packages such as LMER, but I have no clue how to go
> about integrating such analyses into a website. I could certainly learn
> how, would love to, and ultimately will, but if someone was interested in
> joining me in this endeavour much more could be accomplished.
>
> If you're interested in knowing more let me know.
>
> Josh
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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] Problems with nls

[Bert Gunter]

AFAICS this has essentially nothing to do with R. Please post elsewhere,
e.g. on a statistics list like stats.stackexchange.com.

Cheers,
Bert



On Fri, May 29, 2015 at 6:44 AM, Abolfazl Saghafi <
abolfazl.sagh...@gmail.com> wrote:

> Can some help me with a question on this bass model, please
>
> As I read some articles on this topic, I understand that
> 1. the bass formula is
> N(t) = pm + (q-p) N(t-1) - (q/m) (N(t-1))^2
> 2. which is a difference equation with the solution
> N(t) = m (1 − exp(−(p+q)t)) / (1 + (q/p)exp(−(p+q)t))
> 3. So, using a linear regression would give us some some initial
> estimations for the parameters m, p, q
> 4. we then can put the initial estimations into a NLS to get the better
> estimations
>
> Am I right?
>
> Now the question is,
> why is that I see people use cumulative data and try to fit it into a pdf
> as
> M * ( ((P+Q)^2 / P) * exp(-(P+Q) * T79) ) / (1+(Q/P)*exp(-(P+Q)*T79))^2,
>
> why not using the cumulative data and fit directly the N(t)
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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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Re: [R] best way to handle database connections from within a package

[Mark Sharp]

I would simply separate the database connect and disconnect functions from the 
query functions. 

Mark
R. Mark Sharp, Ph.D.
msh...@txbiomed.org





> On May 28, 2015, at 12:18 PM, Luca Cerone  wrote:
> 
> Dear all,
> I am writing a package that is a collection of queries to be run
> against a postgresql database,
> so that the users do not have to worry about the structure of the database.
> 
> In my package I import dbDriver, dbUnloadDriver, dbConnect,
> dbDisconnect from the package DBI
> and dbGetQuery from the package RPostgreSQL.
> 
> All the function in a function in my package have the same structure:
> 
> getFancyData <- function( from, to) {
>on.exit( dbDisconnect(con), add=TRUE)
>on.exit( dbUnloadDriver(drv), add=TRUE)
>drv <- dbDriver("PostgreSQL")
>con <- dbConnect(drv,
> user=pkguser,
> host=pkghost,
> password=pkgpassword,
> port = pkgport)
> 
>query <- sprintf("select * from fancyTable where dt between '%s'
> and '%s'", from, to)
>res <- dbGetQuery(con,query)
>return(res)
> }
> 
> The various access details are read from an encrypted profile that the
> user has to
> create when she installs the package.
> 
> Such functions work perfectly fine, but I have to replicate a lot of
> times loading and unloading the driver and connecting and
> disconnecting from the database.
> 
> I am wondering if there is a better way to do this job, like loading
> the driver and opening the connection only once when the package is
> loaded. However I have to make sure that
> if R crashes or the code where the function is called contains an
> error then the connection
> with the database is closed. How would you implement this?
> 
> 
> Also how would you write a functional that would at least allow me to
> avoid replicating
> the boilerplate code to load and unload the drivers?
> 
> I am thinking something on the lines of:
> 
> querybuild <- function(query, )
>on.exit( dbDisconnect(con), add=TRUE)
>on.exit( dbUnloadDriver(drv), add=TRUE)
>query <- sprintf(query, ... )
>res <- dbSendQuery(query)
>return(res)
> }
> 
> and then define
> 
> getFancyData <- function(from, to) querybuild("select * from
> fancyTable where dt between '%s' and '%s'", from, to)
> 
> Do you see a better way?
> 
> Thanks a lot in advance for your help and advice on this!
> 
> Cheers,
> Luca
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Why I am not able to load library(R.matlab)? Other packages are fine.

[C W]

Hi Henrik,

I don't quite get what I should do here.  I am not familiar with
R.methodS3.  Can you tell me what command exactly do I need to do?

Thanks,

Mike

On Thu, May 28, 2015 at 3:30 PM, Henrik Bengtsson  wrote:

> For some unknown reason, you've managed to install R.matlab without
> the dependency R.methodsS3 (cf.
> http://cran.r-project.org/web/packages/R.matlab/) or it happened due
> to some other glitch somewhere.
>
> Try to reinstall R.matlab.  If that doesn't help, explicitly install
> R.methodsS3 and retry.  If you get the same error with the other
> dependencies (R.oo and R.utils), do the same.
>
> /Henrik
>
>
>
> On Thu, May 28, 2015 at 11:47 AM, C W  wrote:
> > Dear R list,
> >
> > I am trying to do use the R.matlab library, I did the following, but it
> > does not work.
> >
> >> library(R.matlab)
> > Error in loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()),
> versionCheck
> > = vI[[j]]) :
> >   there is no package called ‘R.methodsS3’
> > Error: package or namespace load failed for ‘R.matlab’
> >
> > This is my session info.
> >
> >> sessionInfo()
> > R version 3.2.0 (2015-04-16)
> > Platform: x86_64-apple-darwin13.4.0 (64-bit)
> > Running under: OS X 10.10.3 (Yosemite)
> >
> > 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
> >
> > My R is up-to-date, R 3.2.0.  Why is this happening?  Is it because I
> > installed the new R version, instead of updating it?  Maybe things are
> in a
> > different directory?
> >
> > Thanks so much,
> >
> > Mike
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > 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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[R] alternatives to KS test applicable to K-samples

[Wensui Liu]

Good morning, All
I have a stat question not specifically related to the the programming language.
To compare distributional consistency / discrepancy between two
samples, we usually use kolmogorov-smirnov test, which is implemented
in R with ks.test() or in SAS with "pro npar1way edf".
I am wondering if there is any alternative to KS test that could be
generalized to K-samples.

Thanks and have a nice weekend.

wensui

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Re: [R] Problem with comparing multiple data sets

[Mohammad Alimohammadi]

Hi everyone.

I tried the (modeest) package on my initial test data and it worked.
However, it doesn't work on the entire data set. I saved one of the
protions that gives error. (Not for all of the values but for some of
them). For example: lines 36 and 37 and 39 correctly show the mode value
but 38 and 40 are not correct. Such error is repeated for many of the
values.

[36,] 2
[37,] 2
[38,] Numeric,3
[39,] 1
[40,] Numeric,3



#This is what I did:
> df<- read.csv(file="Part1-modif.csv", head=TRUE, sep=",")
> Out<- apply(df[,2:length(df)],1, mfv)
> t(t(Out))


#This is the data set

structure(list(terms = structure(c(2L, 4L, 4L, 4L, 3L, 1L, 5L,
5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label =
c("#authentication,access control",
"#privacy,personal data", "#security,malicious,security", "data
controller",
"id management,security", "password,recovery"), class = "factor"),
class.1 = c(2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L,
2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L), class.2 = c(2L, 2L, 2L,
0L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L,
2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L,
2L, 2L), class.3 = c(2L, 0L, 2L, 2L, 1L, 1L, 0L, 0L, 0L,
2L, 2L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("terms",
"class.1", "class.2", "class.3"), class = "data.frame", row.names = c(NA,
-50L))



also when I try to include the terms to the result it gives me an error:

> mode.names<- data.frame (df[,1],Out)
Error in data.frame(df[, 1], Out) :
arguments imply differing number of rows: 50, 3







On Thu, May 28, 2015 at 9:24 AM, Mohammad Alimohammadi <
mxalimoha...@ualr.edu> wrote:

> Thank you David for your help !
>
> On Wed, May 27, 2015 at 7:31 PM, David L Carlson 
> wrote:
>
>>  cat(paste0("[", 1:length(Out), "] #dac ", Out), sep="\n")
>>
>>  David
>>
>> *From:* Mohammad Alimohammadi [mailto:mxalimoha...@ualr.edu]
>> *Sent:* Wednesday, May 27, 2015 2:29 PM
>> *To:* David L Carlson; r-help@r-project.org
>>
>> *Subject:* Re: [R] Problem with comparing multiple data sets
>>
>>
>>
>> Thanks David it worked !
>>
>>
>>
>> One more thing. I hope it's not complicated. Is it also possible to
>> display the terms for each row next to it?
>>
>>
>>
>> for example:
>>
>>
>>
>> [1] #dac2
>>
>> [2] #dac0
>>
>> [3] #dac1
>>
>> ...
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> On Wed, May 27, 2015 at 2:18 PM, David L Carlson 
>> wrote:
>>
>> Save the result of the apply() function:
>>
>> Out <- apply(df[ ,2:length(df)], 1, mfv)
>>
>> Then there are several options:
>>
>> Approximately what you asked for
>> data.frame(Out)
>> t(t(Out))
>>
>> More typing but exactly what you asked for
>> cat(paste0("[", 1:length(Out), "] ", Out), sep="\n")
>>
>>
>> David L. Carlson
>> Department of Anthropology
>> Texas A&M University
>>
>>
>>
>> -Original Message-
>> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Mohammad
>> Alimohammadi
>> Sent: Wednesday, May 27, 2015 1:47 PM
>> To: John Kane; r-help@r-project.org
>> Subject: Re: [R] Problem with comparing multiple data sets
>>
>> Ok. so I read about the ("modeest") package that gives the results that I
>> am looking for (most repeated value).
>>
>> I modified the data frame a little and moved the text to the first column.
>> This is the data frame with all 3 possible classes for each term.
>>
>> =
>> structure(list(terms = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
>> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, 4L,
>> 4L, 4L, 4L, 3L, 3L, 3L, 3L, 2L, 2L, 2L), .Label = c("#dac",
>> "#mac,#security",
>> "accountability,anonymous", "data security,encryption,security"
>> ), class = "factor"), class.1 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L,
>> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
>> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 2L, 1L,
>> 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L), class.2 = c(2L, 2L,
>> 2L, 2L, 0L, 0L, 2L, 0L, 0L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 0L,
>> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 2L,
>> 0L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 0L, 0L, 0L, 0L, 1L, 1L, 1L),
>> class.3 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
>> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
>> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 2L, 1L, 1L, 1L, 1L,
>> 0L, 0L, 0L, 0L, 2L, 1L, 2L)), .Names = c("terms", "class

[R] Converting unique strings to unique numbers

[Kate Ignatius]

I have a pedigree file as so:

X0001 BYX859  0  0  2  1 BYX859
X0001 BYX894  0  0  1  1 BYX894
X0001 BYX862 BYX894 BYX859  2  2 BYX862
X0001 BYX863 BYX894 BYX859  2  2 BYX863
X0001 BYX864 BYX894 BYX859  2  2 BYX864
X0001 BYX865 BYX894 BYX859  2  2 BYX865

And I was hoping to change all unique string values to numbers.

That is:

BYX859 = 1
BYX894 = 2
BYX862 = 3
BYX863 = 4
BYX864 = 5
BYX865 = 6

But only in columns 2 - 4.  Essentially I would like the data to look like this:

X0001 1 0 0  2  1 BYX859
X0001 2 0 0  1  1 BYX894
X0001 3 2 1  2  2 BYX862
X0001 4 2 1  2  2 BYX863
X0001 5 2 1  2  2 BYX864
X0001 6 2 1  2  2 BYX865

Is this possible with factors?

Thanks!

K.

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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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] alternatives to KS test applicable to K-samples

[Cade, Brian]

Wensui:  There are the multi-response permutation procedures (MRPP) that
readily test the omnibus hypothesis of no distributional differences among
multiple samples for univariate or multivariate responses.  There also are
empirical coverage tests that test a similar hypothesis among multiple
samples but only for univariate responses.  Both are included in the USGS
Blossom package for R linked here:
https://www.fort.usgs.gov/products/23735 (not
yet distributed via CRAN).  The MRPP may also be available in other R
packages on CRAN (vegan ?).

Brian

Brian S. Cade, PhD

U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO  80526-8818

email:  ca...@usgs.gov 
tel:  970 226-9326


On Fri, May 29, 2015 at 10:31 AM, Wensui Liu  wrote:

> Good morning, All
> I have a stat question not specifically related to the the programming
> language.
> To compare distributional consistency / discrepancy between two
> samples, we usually use kolmogorov-smirnov test, which is implemented
> in R with ks.test() or in SAS with "pro npar1way edf".
> I am wondering if there is any alternative to KS test that could be
> generalized to K-samples.
>
> Thanks and have a nice weekend.
>
> wensui
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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] Converting unique strings to unique numbers

[MacQueen, Don]

Here is an example to get you started:

mycol <- c('b','a','d','d','b','c')
as.numeric(factor(mycol))

-Don

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 5/29/15, 9:58 AM, "Kate Ignatius"  wrote:

>I have a pedigree file as so:
>
>X0001 BYX859  0  0  2  1 BYX859
>X0001 BYX894  0  0  1  1 BYX894
>X0001 BYX862 BYX894 BYX859  2  2 BYX862
>X0001 BYX863 BYX894 BYX859  2  2 BYX863
>X0001 BYX864 BYX894 BYX859  2  2 BYX864
>X0001 BYX865 BYX894 BYX859  2  2 BYX865
>
>And I was hoping to change all unique string values to numbers.
>
>That is:
>
>BYX859 = 1
>BYX894 = 2
>BYX862 = 3
>BYX863 = 4
>BYX864 = 5
>BYX865 = 6
>
>But only in columns 2 - 4.  Essentially I would like the data to look
>like this:
>
>X0001 1 0 0  2  1 BYX859
>X0001 2 0 0  1  1 BYX894
>X0001 3 2 1  2  2 BYX862
>X0001 4 2 1  2  2 BYX863
>X0001 5 2 1  2  2 BYX864
>X0001 6 2 1  2  2 BYX865
>
>Is this possible with factors?
>
>Thanks!
>
>K.
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Converting unique strings to unique numbers

[Jeff Newmiller]

Of course, but I would not recommend it. A factor is a vector of integers with 
an attribute containing the labels that those integers correspond to. You seem 
to be asking for a factor that has lost the definitions part. But hey, 
newvector <- as.integer(factor(oldvector)) should get you what you asked for 
one column at a time.
---
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.

On May 29, 2015 9:58:22 AM PDT, Kate Ignatius  wrote:
>I have a pedigree file as so:
>
>X0001 BYX859  0  0  2  1 BYX859
>X0001 BYX894  0  0  1  1 BYX894
>X0001 BYX862 BYX894 BYX859  2  2 BYX862
>X0001 BYX863 BYX894 BYX859  2  2 BYX863
>X0001 BYX864 BYX894 BYX859  2  2 BYX864
>X0001 BYX865 BYX894 BYX859  2  2 BYX865
>
>And I was hoping to change all unique string values to numbers.
>
>That is:
>
>BYX859 = 1
>BYX894 = 2
>BYX862 = 3
>BYX863 = 4
>BYX864 = 5
>BYX865 = 6
>
>But only in columns 2 - 4.  Essentially I would like the data to look
>like this:
>
>X0001 1 0 0  2  1 BYX859
>X0001 2 0 0  1  1 BYX894
>X0001 3 2 1  2  2 BYX862
>X0001 4 2 1  2  2 BYX863
>X0001 5 2 1  2  2 BYX864
>X0001 6 2 1  2  2 BYX865
>
>Is this possible with factors?
>
>Thanks!
>
>K.
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>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 -- To UNSUBSCRIBE and more, see
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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] Converting unique strings to unique numbers

[William Dunlap]

match() will do what you want.  E.g., run your data through
the following function.

f <- function (data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L)
}
data
}



Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Fri, May 29, 2015 at 9:58 AM, Kate Ignatius 
wrote:

> I have a pedigree file as so:
>
> X0001 BYX859  0  0  2  1 BYX859
> X0001 BYX894  0  0  1  1 BYX894
> X0001 BYX862 BYX894 BYX859  2  2 BYX862
> X0001 BYX863 BYX894 BYX859  2  2 BYX863
> X0001 BYX864 BYX894 BYX859  2  2 BYX864
> X0001 BYX865 BYX894 BYX859  2  2 BYX865
>
> And I was hoping to change all unique string values to numbers.
>
> That is:
>
> BYX859 = 1
> BYX894 = 2
> BYX862 = 3
> BYX863 = 4
> BYX864 = 5
> BYX865 = 6
>
> But only in columns 2 - 4.  Essentially I would like the data to look like
> this:
>
> X0001 1 0 0  2  1 BYX859
> X0001 2 0 0  1  1 BYX894
> X0001 3 2 1  2  2 BYX862
> X0001 4 2 1  2  2 BYX863
> X0001 5 2 1  2  2 BYX864
> X0001 6 2 1  2  2 BYX865
>
> Is this possible with factors?
>
> Thanks!
>
> K.
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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] Why I am not able to load library(R.matlab)? Other packages are fine.

[Ben Bolker]

C W  gmail.com> writes:

> 
> Hi Henrik,
> 
> I don't quite get what I should do here.  I am not familiar with
> R.methodS3.  Can you tell me what command exactly do I need to do?
> 
> Thanks,
> 
> Mike

install.packages("R.methodsS3")
install.packages("R.matlab")
library("R.matlab")



  [snip snip snip]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] alternatives to KS test applicable to K-samples

[Wensui Liu]

Very nice, Brian

Sincerely appreciate your assistance!

On Friday, May 29, 2015, Cade, Brian  wrote:

> Wensui:  There are the multi-response permutation procedures (MRPP) that
> readily test the omnibus hypothesis of no distributional differences among
> multiple samples for univariate or multivariate responses.  There also are
> empirical coverage tests that test a similar hypothesis among multiple
> samples but only for univariate responses.  Both are included in the USGS
> Blossom package for R linked here:
> https://www.fort.usgs.gov/products/23735 (not yet distributed via CRAN).
> The MRPP may also be available in other R packages on CRAN (vegan ?).
>
> Brian
>
> Brian S. Cade, PhD
>
> U. S. Geological Survey
> Fort Collins Science Center
> 2150 Centre Ave., Bldg. C
> Fort Collins, CO  80526-8818
>
> email:  ca...@usgs.gov
> 
> tel:  970 226-9326
>
>
> On Fri, May 29, 2015 at 10:31 AM, Wensui Liu  > wrote:
>
>> Good morning, All
>> I have a stat question not specifically related to the the programming
>> language.
>> To compare distributional consistency / discrepancy between two
>> samples, we usually use kolmogorov-smirnov test, which is implemented
>> in R with ks.test() or in SAS with "pro npar1way edf".
>> I am wondering if there is any alternative to KS test that could be
>> generalized to K-samples.
>>
>> Thanks and have a nice weekend.
>>
>> wensui
>>
>> __
>> R-help@r-project.org
>>  mailing list --
>> To UNSUBSCRIBE and more, see
>> 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.
>>
>
>

-- 
==
WenSui Liu
Credit Risk Manager, 53 Bancorp
wensui@53.com
513-295-4370
==

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Re: [R] Converting unique strings to unique numbers

[Hervé Pagès]

Hi Kate,

I found that matching the character vector to itself is a very
effective way to do this:

  x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
 "is", "of", "little", "interest")
  ids <- match(x, x)
  ids
  # [1]  1  2  3  4  5  6  7  8  3 10 11

By using this trick, many manipulations on character vectors can
be replaced by manipulations on integer vectors, which are sometimes
way more efficient.

Cheers,
H.


On 05/29/2015 09:58 AM, Kate Ignatius wrote:

I have a pedigree file as so:

X0001 BYX859  0  0  2  1 BYX859
X0001 BYX894  0  0  1  1 BYX894
X0001 BYX862 BYX894 BYX859  2  2 BYX862
X0001 BYX863 BYX894 BYX859  2  2 BYX863
X0001 BYX864 BYX894 BYX859  2  2 BYX864
X0001 BYX865 BYX894 BYX859  2  2 BYX865

And I was hoping to change all unique string values to numbers.

That is:

BYX859 = 1
BYX894 = 2
BYX862 = 3
BYX863 = 4
BYX864 = 5
BYX865 = 6

But only in columns 2 - 4.  Essentially I would like the data to look like this:

X0001 1 0 0  2  1 BYX859
X0001 2 0 0  1  1 BYX894
X0001 3 2 1  2  2 BYX862
X0001 4 2 1  2  2 BYX863
X0001 5 2 1  2  2 BYX864
X0001 6 2 1  2  2 BYX865

Is this possible with factors?

Thanks!

K.

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--
Hervé Pagès

Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024

E-mail: hpa...@fredhutch.org
Phone:  (206) 667-5791
Fax:(206) 667-1319

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Re: [R] alternatives to KS test applicable to K-samples

[David Winsemius]

On May 29, 2015, at 9:31 AM, Wensui Liu wrote:

> Good morning, All
> I have a stat question not specifically related to the the programming 
> language.
> To compare distributional consistency / discrepancy between two
> samples, we usually use kolmogorov-smirnov test, which is implemented
> in R with ks.test() or in SAS with "pro npar1way edf".
> I am wondering if there is any alternative to KS test that could be
> generalized to K-samples.

The 'coin' package (Hothorn, Hornick, van de Weil, and Zeileis) presents a 
variety of permutation and rank-based tests that would probably be more 
powerful than any multi-group variant of the KS test. The multi-group variant 
of the Wilcoxon Rank Sum Test presented in the examples for the help page: 
?wilcox_test is the Nemenyi-Damico-Wolfe-Dunn test.

-- 

David Winsemius
Alameda, CA, USA

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Re: [R] Converting unique strings to unique numbers

[Kate Ignatius]

I found this helpful.  However - the second to forth columns come out
all zero - was this the intention?

That is:

X0001 0 0 0  2  1 BYX859
X0001 0 0 0  1  1 BYX894
X0001 0 0 0  2  2 BYX862
X0001 0 0 0  2  2 BYX863
X0001 0 0 0  2  2 BYX864
X0001 0 0 0  2  2 BYX865

On Fri, May 29, 2015 at 1:31 PM, William Dunlap  wrote:
> match() will do what you want.  E.g., run your data through
> the following function.
>
> f <- function (data)
> {
> uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
> uniqStrings <- setdiff(uniqStrings, "0")
> for (j in 2:4) {
> data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L)
> }
> data
> }
>
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Fri, May 29, 2015 at 9:58 AM, Kate Ignatius 
> wrote:
>>
>> I have a pedigree file as so:
>>
>> X0001 BYX859  0  0  2  1 BYX859
>> X0001 BYX894  0  0  1  1 BYX894
>> X0001 BYX862 BYX894 BYX859  2  2 BYX862
>> X0001 BYX863 BYX894 BYX859  2  2 BYX863
>> X0001 BYX864 BYX894 BYX859  2  2 BYX864
>> X0001 BYX865 BYX894 BYX859  2  2 BYX865
>>
>> And I was hoping to change all unique string values to numbers.
>>
>> That is:
>>
>> BYX859 = 1
>> BYX894 = 2
>> BYX862 = 3
>> BYX863 = 4
>> BYX864 = 5
>> BYX865 = 6
>>
>> But only in columns 2 - 4.  Essentially I would like the data to look like
>> this:
>>
>> X0001 1 0 0  2  1 BYX859
>> X0001 2 0 0  1  1 BYX894
>> X0001 3 2 1  2  2 BYX862
>> X0001 4 2 1  2  2 BYX863
>> X0001 5 2 1  2  2 BYX864
>> X0001 6 2 1  2  2 BYX865
>>
>> Is this possible with factors?
>>
>> Thanks!
>>
>> K.
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> 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] Result differences in 32-bit vs. 64-bit point.in.polygon?

[Lensing, Shelly Y]

Is anyone aware of point.in.polygon giving different results for 32-bit vs. 
64-bit R? Our OS is 64-bit Windows 7 Enterprise. I'm working with someone 
else's extensive R program and the final results are close but not exactly 
matching. We're thinking it might be something with the point.in.polygon 
function (one of many possibilities, including leaps).

Thanks much,

Shelly Lensing
Biostatistics / University of Arkansas for Medical Sciences
4301 W. Markham St. #781 / Little Rock, AR  72205
V: 501.686.8203 / F: 501-526-6729 / COPH 3236

--
Confidentiality Notice: This e-mail message, including a...{{dropped:10}}

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Re: [R] Converting unique strings to unique numbers

[Sarah Goslee]

On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès  wrote:
> Hi Kate,
>
> I found that matching the character vector to itself is a very
> effective way to do this:
>
>   x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
>  "is", "of", "little", "interest")
>   ids <- match(x, x)
>   ids
>   # [1]  1  2  3  4  5  6  7  8  3 10 11
>
> By using this trick, many manipulations on character vectors can
> be replaced by manipulations on integer vectors, which are sometimes
> way more efficient.

Hm. I hadn't thought of that approach - I use the
as.numeric(factor(...)) approach.

So I was curious, and compared the two:


set.seed(43)
x <- sample(letters, 1, replace=TRUE)

system.time({
  for(i in seq_len(2)) {
  ids1 <- match(x, x)
}})

#   user  system elapsed
#  9.657   0.000   9.657

system.time({
  for(i in seq_len(2)) {
  ids2 <- as.numeric(factor(x, levels=letters))
}})

#   user  system elapsed
#   6.160.006.16

Using factor() is faster. More importantly, using factor() lets you
set the order of the indices in an expected fashion, where match()
assigns them in the order of occurrence.

head(data.frame(x, ids1, ids2))

  x ids1 ids2
1 m1   13
2 x2   24
3 b32
4 s4   19
5 i59
6 o6   15

In a problem like Kate's where there are several columns for which the
same ordering of indices is desired, that becomes really important.

If you take Bill Dunlap's modification of the match() approach, it
resolves both problems: matching against the pooled unique values is
both faster than the factor() version and gives the same result:


On Fri, May 29, 2015 at 1:31 PM, William Dunlap  wrote:
> match() will do what you want.  E.g., run your data through
> the following function.
>
f <- function (data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L)
}
data
}

##

y <- data.frame(id = 1:5000, v1 = sample(letters, 5000, replace=TRUE),
v2 = sample(letters, 5000, replace=TRUE), v3 = sample(letters, 5000,
replace=TRUE), stringsAsFactors=FALSE)


system.time({
  for(i in seq_len(2)) {
ids3 <- f(data.frame(y))
}})

#   user  system elapsed
# 22.515   0.000  22.518



ff <- function(data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- as.numeric(factor(data[[j]], levels=uniqStrings))
}
data
}

system.time({
  for(i in seq_len(2)) {
ids4 <- ff(data.frame(y))
}})

#user  system elapsed
#  26.083   0.002  26.090

head(ids3)

  id v1 v2 v3
1  1  1  2  8
2  2  2 19 22
3  3  3 21 16
4  4  4 10 17
5  5  1  8 18
6  6  1 12 26

head(ids4)

  id v1 v2 v3
1  1  1  2  8
2  2  2 19 22
3  3  3 21 16
4  4  4 10 17
5  5  1  8 18
6  6  1 12 26

Kate, if you're getting all zeros, check str(yourdataframe) - it's
likely that when you imported your data into R the strings were
already converted to factors, which is not what you want (ask me how I
know this!).

Sarah



> On 05/29/2015 09:58 AM, Kate Ignatius wrote:
>>
>> I have a pedigree file as so:
>>
>> X0001 BYX859  0  0  2  1 BYX859
>> X0001 BYX894  0  0  1  1 BYX894
>> X0001 BYX862 BYX894 BYX859  2  2 BYX862
>> X0001 BYX863 BYX894 BYX859  2  2 BYX863
>> X0001 BYX864 BYX894 BYX859  2  2 BYX864
>> X0001 BYX865 BYX894 BYX859  2  2 BYX865
>>
>> And I was hoping to change all unique string values to numbers.
>>
>> That is:
>>
>> BYX859 = 1
>> BYX894 = 2
>> BYX862 = 3
>> BYX863 = 4
>> BYX864 = 5
>> BYX865 = 6
>>
>> But only in columns 2 - 4.  Essentially I would like the data to look like
>> this:
>>
>> X0001 1 0 0  2  1 BYX859
>> X0001 2 0 0  1  1 BYX894
>> X0001 3 2 1  2  2 BYX862
>> X0001 4 2 1  2  2 BYX863
>> X0001 5 2 1  2  2 BYX864
>> X0001 6 2 1  2  2 BYX865
>>
>> Is this possible with factors?
>>
>> Thanks!
>>
>> K.
>>


-- 
Sarah Goslee
http://www.functionaldiversity.org

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[R] Automatically updating a plot from a regularly updated data file

[Sam Albers]

Hi all,

I have a question about using R in a way that may not be correct but I
thought I would ask anyway.

I have an instrument that outputs a text file with comma separated data. A
new line is added to the file each time the instrument takes a new reading.
Is there any way to configure R such that a script to generate a plot from
said text file is re-run each time the file is modified (i.e. a new line is
added). So basically update an exported plot each time a new line of data
is collected.

Is this type of thing possible in R? If not can anyone recommend some
Windows (or Linux if need be) tools that could help me accomplish this
preferably still utilizing R's plotting capabilites? I know that there are
other tools that can do this all but nothing makes figures as nicely as R.

I suppose more generally this is a question about way to automate processes
with R to take advantage of R's functionality.

Thanks in advance.

Sam

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[R] about transforming a data.frame

[Bogdan Tanasa]

Dear all,

I would appreciate a suggestion on the following : I am working with a
data.frame (below) :

  EXPCT   row_names   col_names
1   test -5B4:B5:B6B1:B2:B3
2   test -2B7:B8:B9B1:B2:B3
3   test -2D4:D5:D6H4:H5:H6
4   test -2D10:D11:D12 F10:F11:F12
5   test -2D10:D11:D12H1:H2:H3
6   test -2E10:E11:E12G7:G8:G9
7   test -4 A1:A2:A3D1:D2:D3
8   test -4   B10:B11:B12B1:B2:B3

what would be the easiest way to consider UNIQUE elements in the ROW_NAMES
or the UNIQUE elements in the COL_NAMES and :

print how many times these UNIQUE ELEMENTS associate with the numbers -5,
-2, or -4 (these numbers are on the column names CT) ..

thanks,

bogdan

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Re: [R] Converting unique strings to unique numbers

[Hervé Pagès]

Hi Sarah,

On 05/29/2015 12:04 PM, Sarah Goslee wrote:

On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès  wrote:

Hi Kate,

I found that matching the character vector to itself is a very
effective way to do this:

   x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
  "is", "of", "little", "interest")
   ids <- match(x, x)
   ids
   # [1]  1  2  3  4  5  6  7  8  3 10 11

By using this trick, many manipulations on character vectors can
be replaced by manipulations on integer vectors, which are sometimes
way more efficient.


Hm. I hadn't thought of that approach - I use the
as.numeric(factor(...)) approach.

So I was curious, and compared the two:


set.seed(43)
x <- sample(letters, 1, replace=TRUE)

system.time({
   for(i in seq_len(2)) {
   ids1 <- match(x, x)
}})

#   user  system elapsed
#  9.657   0.000   9.657

system.time({
   for(i in seq_len(2)) {
   ids2 <- as.numeric(factor(x, levels=letters))
}})

#   user  system elapsed
#   6.160.006.16

Using factor() is faster.


That's an unfair comparison, because you already know what the levels
are so you can supply them to your call to factor(). Most of the time
you don't know what the levels are so either you just do factor(x) and
let the factor() constructor compute the levels for you, or you compute
them yourself upfront with something like factor(x, levels=unique(x)).

  library(microbenchmark)

  microbenchmark(
{ids1 <- match(x, x)},
{ids2 <- as.integer(factor(x, levels=letters))},
{ids3 <- as.integer(factor(x))},
{ids4 <- as.integer(factor(x, levels=unique(x)))}
  )
  Unit: microseconds
  expr min   lq
   { ids1 <- match(x, x) } 245.979 262.2390
   { ids2 <- as.integer(factor(x, levels = letters)) } 214.115 219.2320
 { ids3 <- as.integer(factor(x)) } 380.782 388.7295
 { ids4 <- as.integer(factor(x, levels = unique(x))) } 332.250 342.6630
   mean   median  uq max neval
   267.3210 264.4845 268.348 293.894   100
   226.9913 220.9870 226.147 314.875   100
   402.2242 394.7165 412.075 481.410   100
   349.7405 345.3090 353.162 383.002   100


More importantly, using factor() lets you
set the order of the indices in an expected fashion, where match()
assigns them in the order of occurrence.

head(data.frame(x, ids1, ids2))

   x ids1 ids2
1 m1   13
2 x2   24
3 b32
4 s4   19
5 i59
6 o6   15

In a problem like Kate's where there are several columns for which the
same ordering of indices is desired, that becomes really important.


I'm not sure why which particular ID gets assigned to each string would
matter but maybe I'm missing something. What really matters is that each
string receives a unique ID. match(x, x) does that.

In Kate's problem, where the strings are in more than one column,
and you want the ID to be unique across the columns, you need to do
match(x, x) where 'x' contains the strings from all the columns
that you want to replace:

  m <- matrix(c(
"X0001", "BYX859",0,0,  2,  1, "BYX859",
"X0001", "BYX894",0,0,  1,  1, "BYX894",
"X0001", "BYX862", "BYX894", "BYX859",  2,  2, "BYX862",
"X0001", "BYX863", "BYX894", "BYX859",  2,  2, "BYX863",
"X0001", "BYX864", "BYX894", "BYX859",  2,  2, "BYX864",
"X0001", "BYX865", "BYX894", "BYX859",  2,  2, "BYX865"
  ), ncol=7, byrow=TRUE)

  x <- m[ , 2:4]
  id <- match(x, x, nomatch=0, incomparables="0")
  m[ , 2:4] <- id

No factor needed. No loop needed. ;-)

Cheers,
H.



If you take Bill Dunlap's modification of the match() approach, it
resolves both problems: matching against the pooled unique values is
both faster than the factor() version and gives the same result:


On Fri, May 29, 2015 at 1:31 PM, William Dunlap  wrote:

match() will do what you want.  E.g., run your data through
the following function.


f <- function (data)
{
 uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
 uniqStrings <- setdiff(uniqStrings, "0")
 for (j in 2:4) {
 data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L)
 }
 data
}

##

y <- data.frame(id = 1:5000, v1 = sample(letters, 5000, replace=TRUE),
v2 = sample(letters, 5000, replace=TRUE), v3 = sample(letters, 5000,
replace=TRUE), stringsAsFactors=FALSE)


system.time({
   for(i in seq_len(2)) {
 ids3 <- f(data.frame(y))
}})

#   user  system elapsed
# 22.515   0.000  22.518



ff <- function(data)
{
 uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
 uniqStrings <- setdiff(uniqStrings, "0")
 for (j in 2:4) {
 data[[j]] <- as.numeric(factor(data[[j]], levels=uniqStrings))
 }
 data
}

system.time({
   for(i in seq_len(2)) {
 ids4 <- ff(data.frame(y))
}})

#user  system elapsed
#  26.083   0.002  26.090

head(ids3)

   id v1 v2 v3
1  1  1  2  8
2  2  2 19 22
3  3  3 21 16
4  4  4 10 17
5  5  1  8

Re: [R] about transforming a data.frame

[Sarah Goslee]

Hi,

Please use dput() to provide your data, as it can get somewhat mangled
by copy and pasting, especially if you post in HTML (as you are asked
not to do in the posting guide).

What is a unique element? is "B4:B5:B6" an element, or are "B4" and
"B5" each elements? That is, what is the result you expect to obtain
for the sample data you provided?

What code have you tried? I would think table() might be involved, and
possibly strsplit(), but will refrain from putting more time into this
until you provide a reproducible dataset with dput() and some clearer
idea of your intent.

Sarah

On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa  wrote:
> Dear all,
>
> I would appreciate a suggestion on the following : I am working with a
> data.frame (below) :
>
>   EXPCT   row_names   col_names
> 1   test -5B4:B5:B6B1:B2:B3
> 2   test -2B7:B8:B9B1:B2:B3
> 3   test -2D4:D5:D6H4:H5:H6
> 4   test -2D10:D11:D12 F10:F11:F12
> 5   test -2D10:D11:D12H1:H2:H3
> 6   test -2E10:E11:E12G7:G8:G9
> 7   test -4 A1:A2:A3D1:D2:D3
> 8   test -4   B10:B11:B12B1:B2:B3
>
> what would be the easiest way to consider UNIQUE elements in the ROW_NAMES
> or the UNIQUE elements in the COL_NAMES and :
>
> print how many times these UNIQUE ELEMENTS associate with the numbers -5,
> -2, or -4 (these numbers are on the column names CT) ..
>
> thanks,
>
> bogdan
>
-- 
Sarah Goslee
http://www.functionaldiversity.org

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Re: [R] Converting unique strings to unique numbers

[William Dunlap]

>I'm not sure why which particular ID gets assigned to each string would
>matter but maybe I'm missing something. What really matters is that each
>string receives a unique ID. match(x, x) does that.

I think each row of the OP's dataset represented an individual (column 2)
followed by its mother and father (columns 3 and 4).  I assume that the
marker "0" means that a parent is not in the dataset.  If you match against
the strings in column 2 only, in their original order, then the resulting
numbers
give the row number of an individual, making it straightforward to look up
information regarding the ancestors of an individual.  Hence the choice of
numeric ID's may be important.



Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Fri, May 29, 2015 at 1:29 PM, Hervé Pagès  wrote:

> Hi Sarah,
>
> On 05/29/2015 12:04 PM, Sarah Goslee wrote:
>
>> On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès 
>> wrote:
>>
>>> Hi Kate,
>>>
>>> I found that matching the character vector to itself is a very
>>> effective way to do this:
>>>
>>>x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
>>>   "is", "of", "little", "interest")
>>>ids <- match(x, x)
>>>ids
>>># [1]  1  2  3  4  5  6  7  8  3 10 11
>>>
>>> By using this trick, many manipulations on character vectors can
>>> be replaced by manipulations on integer vectors, which are sometimes
>>> way more efficient.
>>>
>>
>> Hm. I hadn't thought of that approach - I use the
>> as.numeric(factor(...)) approach.
>>
>> So I was curious, and compared the two:
>>
>>
>> set.seed(43)
>> x <- sample(letters, 1, replace=TRUE)
>>
>> system.time({
>>for(i in seq_len(2)) {
>>ids1 <- match(x, x)
>> }})
>>
>> #   user  system elapsed
>> #  9.657   0.000   9.657
>>
>> system.time({
>>for(i in seq_len(2)) {
>>ids2 <- as.numeric(factor(x, levels=letters))
>> }})
>>
>> #   user  system elapsed
>> #   6.160.006.16
>>
>> Using factor() is faster.
>>
>
> That's an unfair comparison, because you already know what the levels
> are so you can supply them to your call to factor(). Most of the time
> you don't know what the levels are so either you just do factor(x) and
> let the factor() constructor compute the levels for you, or you compute
> them yourself upfront with something like factor(x, levels=unique(x)).
>
>   library(microbenchmark)
>
>   microbenchmark(
> {ids1 <- match(x, x)},
> {ids2 <- as.integer(factor(x, levels=letters))},
> {ids3 <- as.integer(factor(x))},
> {ids4 <- as.integer(factor(x, levels=unique(x)))}
>   )
>   Unit: microseconds
>   expr min   lq
>{ ids1 <- match(x, x) } 245.979 262.2390
>{ ids2 <- as.integer(factor(x, levels = letters)) } 214.115 219.2320
>  { ids3 <- as.integer(factor(x)) } 380.782 388.7295
>  { ids4 <- as.integer(factor(x, levels = unique(x))) } 332.250 342.6630
>mean   median  uq max neval
>267.3210 264.4845 268.348 293.894   100
>226.9913 220.9870 226.147 314.875   100
>402.2242 394.7165 412.075 481.410   100
>349.7405 345.3090 353.162 383.002   100
>
>  More importantly, using factor() lets you
>> set the order of the indices in an expected fashion, where match()
>> assigns them in the order of occurrence.
>>
>> head(data.frame(x, ids1, ids2))
>>
>>x ids1 ids2
>> 1 m1   13
>> 2 x2   24
>> 3 b32
>> 4 s4   19
>> 5 i59
>> 6 o6   15
>>
>> In a problem like Kate's where there are several columns for which the
>> same ordering of indices is desired, that becomes really important.
>>
>
> I'm not sure why which particular ID gets assigned to each string would
> matter but maybe I'm missing something. What really matters is that each
> string receives a unique ID. match(x, x) does that.
>
> In Kate's problem, where the strings are in more than one column,
> and you want the ID to be unique across the columns, you need to do
> match(x, x) where 'x' contains the strings from all the columns
> that you want to replace:
>
>   m <- matrix(c(
> "X0001", "BYX859",0,0,  2,  1, "BYX859",
> "X0001", "BYX894",0,0,  1,  1, "BYX894",
> "X0001", "BYX862", "BYX894", "BYX859",  2,  2, "BYX862",
> "X0001", "BYX863", "BYX894", "BYX859",  2,  2, "BYX863",
> "X0001", "BYX864", "BYX894", "BYX859",  2,  2, "BYX864",
> "X0001", "BYX865", "BYX894", "BYX859",  2,  2, "BYX865"
>   ), ncol=7, byrow=TRUE)
>
>   x <- m[ , 2:4]
>   id <- match(x, x, nomatch=0, incomparables="0")
>   m[ , 2:4] <- id
>
> No factor needed. No loop needed. ;-)
>
> Cheers,
> H.
>
>
>> If you take Bill Dunlap's modification of the match() approach, it
>> resolves both problems: matching against the pooled unique values is
>> both faster than the factor() version and gives the same result:
>>
>>
>> On Fri, May 29, 2015 at 1:31 PM, William Dunlap 
>> wrote:
>>
>>> match() will 

Re: [R] about transforming a data.frame

[Bogdan Tanasa]

Hi Sarah,

thank you for your help. I have simplified the example, by reading the
elements in a data frame, eg :

df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
"D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
col_names = c
("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
CT = c(5,2,2,2,2,2,4,4) )

I have used the the count() in the plyr package :

count_row_names <- count(df$row_names)
count_col_names <- count(df$col_names)

however, I would need to correlate these UNIQUE ELEMENTS in the columns
"row_names" or "col_names" with the numbers they associate in the  CT
columns, eg :

""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
associate with "2" (in the CT column).

thank you very much,

bogdan




On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
wrote:

> Hi,
>
> Please use dput() to provide your data, as it can get somewhat mangled
> by copy and pasting, especially if you post in HTML (as you are asked
> not to do in the posting guide).
>
> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
> "B5" each elements? That is, what is the result you expect to obtain
> for the sample data you provided?
>
> What code have you tried? I would think table() might be involved, and
> possibly strsplit(), but will refrain from putting more time into this
> until you provide a reproducible dataset with dput() and some clearer
> idea of your intent.
>
> Sarah
>
> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa  wrote:
> > Dear all,
> >
> > I would appreciate a suggestion on the following : I am working with a
> > data.frame (below) :
> >
> >   EXPCT   row_names   col_names
> > 1   test -5B4:B5:B6B1:B2:B3
> > 2   test -2B7:B8:B9B1:B2:B3
> > 3   test -2D4:D5:D6H4:H5:H6
> > 4   test -2D10:D11:D12 F10:F11:F12
> > 5   test -2D10:D11:D12H1:H2:H3
> > 6   test -2E10:E11:E12G7:G8:G9
> > 7   test -4 A1:A2:A3D1:D2:D3
> > 8   test -4   B10:B11:B12B1:B2:B3
> >
> > what would be the easiest way to consider UNIQUE elements in the
> ROW_NAMES
> > or the UNIQUE elements in the COL_NAMES and :
> >
> > print how many times these UNIQUE ELEMENTS associate with the numbers -5,
> > -2, or -4 (these numbers are on the column names CT) ..
> >
> > thanks,
> >
> > bogdan
> >
> --
> Sarah Goslee
> http://www.functionaldiversity.org
>

[[alternative HTML version deleted]]

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Re: [R] Why I am not able to load library(R.matlab)? Other packages are fine.

[C W]

Wow, thanks Ben.  That worked very well.

I guess I didn't have R.methodS3?  But that doesn't make sense, because I
was using R.matlab few weeks ago.  I believe I was on R 3.1.

Maybe it's in R 3.1 folder?  I am using a Mac, btw.

Cheers,

-M

On Fri, May 29, 2015 at 1:55 PM, Ben Bolker  wrote:

> C W  gmail.com> writes:
>
> >
> > Hi Henrik,
> >
> > I don't quite get what I should do here.  I am not familiar with
> > R.methodS3.  Can you tell me what command exactly do I need to do?
> >
> > Thanks,
> >
> > Mike
>
> install.packages("R.methodsS3")
> install.packages("R.matlab")
> library("R.matlab")
>
>
>
>   [snip snip snip]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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] about transforming a data.frame

[John Kane]

Bogdan, the request was for data in dput() format. 

Type ?dput for more information.

Do dput(myfile) copy the ouput and paste into the email

You should get something like: 
structure(list(c1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 
5L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 
8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L), .Label = c("(0.509,0.614]", 
"(0.614,0.718]", "(0.718,0.822]", "(0.822,0.926]", "(0.926,1.03]", 
"(1.03,1.13]", "(1.13,1.24]", "(1.24,1.34]", "(1.34,1.45]", "(1.45,1.55]"
), class = "factor"), s1 = c(0.51, 0.52, 0.58, 0.58, 0.59, 0.6, 
0.63, 0.65, 0.68, 0.74, 0.74, 0.75, 0.77, 0.77, 0.77, 0.78, 0.79, 
0.84, 0.84, 0.85, 0.87, 0.93, 0.93, 0.95, 0.99, 1.04, 1.09, 1.11, 
1.13, 1.14, 1.14, 1.14, 1.17, 1.18, 1.19, 1.22, 1.22, 1.23, 1.28, 
1.29, 1.3, 1.32, 1.37, 1.38, 1.38, 1.4, 1.43, 1.47, 1.52, 1.55
)), .Names = c("c1", "s1"), row.names = c(NA, -50L), class = "data.frame")

Data in duput() format is the preferred way to get data in R-help since it 
provides a perfect copy of what you have on your machine.  Any other way of 
providing data risks the recipients reading it into R differently than it is on 
your machine.

John Kane
Kingston ON Canada


> -Original Message-
> From: tan...@gmail.com
> Sent: Fri, 29 May 2015 13:58:20 -0700
> To: sarah.gos...@gmail.com
> Subject: Re: [R] about transforming a data.frame
> 
> Hi Sarah,
> 
> thank you for your help. I have simplified the example, by reading the
> elements in a data frame, eg :
> 
> df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
> col_names = c
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> CT = c(5,2,2,2,2,2,4,4) )
> 
> I have used the the count() in the plyr package :
> 
> count_row_names <- count(df$row_names)
> count_col_names <- count(df$col_names)
> 
> however, I would need to correlate these UNIQUE ELEMENTS in the columns
> "row_names" or "col_names" with the numbers they associate in the  CT
> columns, eg :
> 
> ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
> associate with "2" (in the CT column).
> 
> thank you very much,
> 
> bogdan
> 
> 
> 
> 
> On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
> wrote:
> 
>> Hi,
>> 
>> Please use dput() to provide your data, as it can get somewhat mangled
>> by copy and pasting, especially if you post in HTML (as you are asked
>> not to do in the posting guide).
>> 
>> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
>> "B5" each elements? That is, what is the result you expect to obtain
>> for the sample data you provided?
>> 
>> What code have you tried? I would think table() might be involved, and
>> possibly strsplit(), but will refrain from putting more time into this
>> until you provide a reproducible dataset with dput() and some clearer
>> idea of your intent.
>> 
>> Sarah
>> 
>> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa  wrote:
>>> Dear all,
>>> 
>>> I would appreciate a suggestion on the following : I am working with a
>>> data.frame (below) :
>>> 
>>>   EXPCT   row_names   col_names
>>> 1   test -5B4:B5:B6B1:B2:B3
>>> 2   test -2B7:B8:B9B1:B2:B3
>>> 3   test -2D4:D5:D6H4:H5:H6
>>> 4   test -2D10:D11:D12 F10:F11:F12
>>> 5   test -2D10:D11:D12H1:H2:H3
>>> 6   test -2E10:E11:E12G7:G8:G9
>>> 7   test -4 A1:A2:A3D1:D2:D3
>>> 8   test -4   B10:B11:B12B1:B2:B3
>>> 
>>> what would be the easiest way to consider UNIQUE elements in the
>> ROW_NAMES
>>> or the UNIQUE elements in the COL_NAMES and :
>>> 
>>> print how many times these UNIQUE ELEMENTS associate with the numbers
>>> -5,
>>> -2, or -4 (these numbers are on the column names CT) ..
>>> 
>>> thanks,
>>> 
>>> bogdan
>>> 
>> --
>> Sarah Goslee
>> http://www.functionaldiversity.org
>> 
> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.


FREE ONLINE PHOTOSHARING - Share your photos online with your friends and 
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Re: [R] Why I am not able to load library(R.matlab)? Other packages are fine.

[Ben Bolker]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

 I think Henrik's point (which I merely clarified) was that something
funky (we'll probably never know what, and it's not worth figuring out
unless it happens again/to other people) had gone wrong and that the
easiest thing to do was just to reinstall.

References:
* https://www.youtube.com/watch?v=t2F1rFmyQmY
*
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.208.9970&rep=rep1&type=pdf


On 15-05-29 05:11 PM, C W wrote:
> Wow, thanks Ben.  That worked very well.
> 
> I guess I didn't have R.methodS3?  But that doesn't make sense,
> because I was using R.matlab few weeks ago.  I believe I was on R
> 3.1.
> 
> Maybe it's in R 3.1 folder?  I am using a Mac, btw.
> 
> Cheers,
> 
> -M
> 
> On Fri, May 29, 2015 at 1:55 PM, Ben Bolker 
> wrote:
> 
>> C W  gmail.com> writes:
>> 
>>> 
>>> Hi Henrik,
>>> 
>>> I don't quite get what I should do here.  I am not familiar
>>> with R.methodS3.  Can you tell me what command exactly do I
>>> need to do?
>>> 
>>> Thanks,
>>> 
>>> Mike
>> 
>> install.packages("R.methodsS3") install.packages("R.matlab") 
>> library("R.matlab")
>> 
>> 
>> 
>> [snip snip snip]
>> 
>> __ 
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more,
>> see 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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Re: [R] Automatically updating a plot from a regularly updated data file

[MacQueen, Don]

A lot will depend on how frequently data is added to the file, how big the
file gets, and how important it is to see updated plots quickly.

I have R doing exactly what you describe, and have found logic like this
(which might be described as crude) to be sufficient

while( {some condition} ) {
  {read the data file}
  {make the plot}
  Sys.sleep( {some number of seconds} )
}

Of course this is not actually noticing that the file has changed and
responding, it is just updating at regular intervals. But that might be
good enough.


A slightly more sophisticated approach would be to set up a loop like the
above, and have the sleep time short, but within the loop use

  file.info({the csv file})

and when the modification time is later than the previous modification
time, read the data and update the plot.


If the file gets really big, you might not want to reload the entire file
each time. That might lead you into things like keeping track of how many
lines the file has, and only reading the new lines -- if you need your
plots to be cumulative. In that situation you might end up using the
pipe() function to create your connection to the file, and pass the OS's
'tail' command (Linux or Mac, not sure about Win) to pipe.

If you only need to plot the last, say, X hours of data, then you may not
need to keep track of the number of lines, just read the last N lines
(hopefully not too hard to figure out what N should be).

If you don't want an R process running indefinitely, as is the case for
the above, you can, on Linux and Mac, set up a cron job to run an R script
as often as once per minute. I have at least one such task where it
happens every 2 minutes, and makes plots of the current data. In this
case, we have 16 measurement devices each sending data to a MySQL database
once per minute; the R script pulls the data from the database every 2
minutes and plots, and the system works well for our needs. Windows will
have some equivalent to cron, I just don't know what it is.

FWIW, all of the above write png files which are viewed via a webserver.

-Don

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 5/29/15, 12:51 PM, "Sam Albers"  wrote:

>Hi all,
>
>I have a question about using R in a way that may not be correct but I
>thought I would ask anyway.
>
>I have an instrument that outputs a text file with comma separated data. A
>new line is added to the file each time the instrument takes a new
>reading.
>Is there any way to configure R such that a script to generate a plot from
>said text file is re-run each time the file is modified (i.e. a new line
>is
>added). So basically update an exported plot each time a new line of data
>is collected.
>
>Is this type of thing possible in R? If not can anyone recommend some
>Windows (or Linux if need be) tools that could help me accomplish this
>preferably still utilizing R's plotting capabilites? I know that there are
>other tools that can do this all but nothing makes figures as nicely as R.
>
>I suppose more generally this is a question about way to automate
>processes
>with R to take advantage of R's functionality.
>
>Thanks in advance.
>
>Sam
>
>   [[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>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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Re: [R] about transforming a data.frame

[Bogdan Tanasa]

Hi John,

thanks for clarifications, yes, of course, the dput() output is the
following :

dput(dataframe_matches_ddCT)

structure(list(FIGURE = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label
= "test", class = "factor"), ddCT = c(-5.4595, -2.7467,
-2.7467, -2.7467, -2.7467, -2.7467, -4.5927, -4.5927), row_names =
structure(c(1L, 2L, 3L, 4L, 4L, 5L, 6L, 7L), .Label = c("B4:B5:B6",
"B7:B8:B9", "D4:D5:D6", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3",
"B10:B11:B12"
), class = "factor"), col_names = structure(c(1L, 1L, 2L, 3L, 4L, 5L, 6L,
1L), .Label = c("B1:B2:B3", "H4:H5:H6", "F10:F11:F12",
"H1:H2:H3", "G7:G8:G9", "D1:D2:D3"), class = "factor"),
CTaverage_MATRIX_SUBSTRACTIONS = c(-5.4595413208,
-2.7467829387, -2.74099286393334, -2.7433134714, -2.7480595907,
-2.755259196, -4.59402211506667, -4.5927206675)), .Names = c("FIGURE",
"ddCT", "row_names", "col_names", "CTaverage_MATRIX_SUBSTRACTIONS"
), row.names = c(NA, 8L), class = "data.frame")

thanks again for your input,

-- bogdan

On Fri, May 29, 2015 at 2:11 PM, John Kane  wrote:

> Bogdan, the request was for data in dput() format.
>
> Type ?dput for more information.
>
> Do dput(myfile) copy the ouput and paste into the email
>
> You should get something like:
> structure(list(c1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
> 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L,
> 5L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L,
> 8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L), .Label = c("(0.509,0.614]",
> "(0.614,0.718]", "(0.718,0.822]", "(0.822,0.926]", "(0.926,1.03]",
> "(1.03,1.13]", "(1.13,1.24]", "(1.24,1.34]", "(1.34,1.45]", "(1.45,1.55]"
> ), class = "factor"), s1 = c(0.51, 0.52, 0.58, 0.58, 0.59, 0.6,
> 0.63, 0.65, 0.68, 0.74, 0.74, 0.75, 0.77, 0.77, 0.77, 0.78, 0.79,
> 0.84, 0.84, 0.85, 0.87, 0.93, 0.93, 0.95, 0.99, 1.04, 1.09, 1.11,
> 1.13, 1.14, 1.14, 1.14, 1.17, 1.18, 1.19, 1.22, 1.22, 1.23, 1.28,
> 1.29, 1.3, 1.32, 1.37, 1.38, 1.38, 1.4, 1.43, 1.47, 1.52, 1.55
> )), .Names = c("c1", "s1"), row.names = c(NA, -50L), class = "data.frame")
>
> Data in duput() format is the preferred way to get data in R-help since it
> provides a perfect copy of what you have on your machine.  Any other way of
> providing data risks the recipients reading it into R differently than it
> is on your machine.
>
> John Kane
> Kingston ON Canada
>
>
> > -Original Message-
> > From: tan...@gmail.com
> > Sent: Fri, 29 May 2015 13:58:20 -0700
> > To: sarah.gos...@gmail.com
> > Subject: Re: [R] about transforming a data.frame
> >
> > Hi Sarah,
> >
> > thank you for your help. I have simplified the example, by reading the
> > elements in a data frame, eg :
> >
> > df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> > "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
> > col_names = c
> >
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> > CT = c(5,2,2,2,2,2,4,4) )
> >
> > I have used the the count() in the plyr package :
> >
> > count_row_names <- count(df$row_names)
> > count_col_names <- count(df$col_names)
> >
> > however, I would need to correlate these UNIQUE ELEMENTS in the columns
> > "row_names" or "col_names" with the numbers they associate in the  CT
> > columns, eg :
> >
> > ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
> > associate with "2" (in the CT column).
> >
> > thank you very much,
> >
> > bogdan
> >
> >
> >
> >
> > On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
> > wrote:
> >
> >> Hi,
> >>
> >> Please use dput() to provide your data, as it can get somewhat mangled
> >> by copy and pasting, especially if you post in HTML (as you are asked
> >> not to do in the posting guide).
> >>
> >> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
> >> "B5" each elements? That is, what is the result you expect to obtain
> >> for the sample data you provided?
> >>
> >> What code have you tried? I would think table() might be involved, and
> >> possibly strsplit(), but will refrain from putting more time into this
> >> until you provide a reproducible dataset with dput() and some clearer
> >> idea of your intent.
> >>
> >> Sarah
> >>
> >> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa 
> wrote:
> >>> Dear all,
> >>>
> >>> I would appreciate a suggestion on the following : I am working with a
> >>> data.frame (below) :
> >>>
> >>>   EXPCT   row_names   col_names
> >>> 1   test -5B4:B5:B6B1:B2:B3
> >>> 2   test -2B7:B8:B9B1:B2:B3
> >>> 3   test -2D4:D5:D6H4:H5:H6
> >>> 4   test -2D10:D11:D12 F10:F11:F12
> >>> 5   test -2D10:D11:D12H1:H2:H3
> >>> 6   test -2E10:E11:E12G7:G8:G9
> >>> 7   test -4 A1:A2:A3D1:D2:D3
> >>> 8   test -4   B10:B11:B12B1:B2:B3
> >>>
> >>> what would be the easiest way to consider UNIQUE elements in the
> >> ROW_NAMES
> >>> or the UNIQUE elements in the COL_NAMES and :
> >>>
> >>> print how many times these UNIQUE E

Re: [R] about transforming a data.frame

[Bogdan Tanasa]

Hi Jim,

yes, thank you, that is the desired output. one more question please :
after using the dataframe :

df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
"D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
col_names = c
("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
CT = c(5,2,2,2,2,2,4,4) )

and :

table(df$row_names,df$CT)
table(df$col_names,df$CT)

how could I quickly verify that "B1:B2:B3" (for example) hits the CT values
of 2,4,5  at least one time ? an example is in

table(df$col_names,df$CT) ?

thank you very much,

-- bogdan



On Fri, May 29, 2015 at 2:40 PM, Jim Lemon  wrote:

> Hi Bogdan,
> Sarah has already suggested this, but doesn't:
>
> table(df$row_names,df$CT)
> table(df$col_names,df$CT)
>
> give you what you want?
>
> Jim
>
>
> On Sat, May 30, 2015 at 7:11 AM, John Kane  wrote:
> > Bogdan, the request was for data in dput() format.
> >
> > Type ?dput for more information.
> >
> > Do dput(myfile) copy the ouput and paste into the email
> >
> > You should get something like:
> > structure(list(c1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
> > 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L,
> > 5L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L,
> > 8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L), .Label = c("(0.509,0.614]",
> > "(0.614,0.718]", "(0.718,0.822]", "(0.822,0.926]", "(0.926,1.03]",
> > "(1.03,1.13]", "(1.13,1.24]", "(1.24,1.34]", "(1.34,1.45]", "(1.45,1.55]"
> > ), class = "factor"), s1 = c(0.51, 0.52, 0.58, 0.58, 0.59, 0.6,
> > 0.63, 0.65, 0.68, 0.74, 0.74, 0.75, 0.77, 0.77, 0.77, 0.78, 0.79,
> > 0.84, 0.84, 0.85, 0.87, 0.93, 0.93, 0.95, 0.99, 1.04, 1.09, 1.11,
> > 1.13, 1.14, 1.14, 1.14, 1.17, 1.18, 1.19, 1.22, 1.22, 1.23, 1.28,
> > 1.29, 1.3, 1.32, 1.37, 1.38, 1.38, 1.4, 1.43, 1.47, 1.52, 1.55
> > )), .Names = c("c1", "s1"), row.names = c(NA, -50L), class =
> "data.frame")
> >
> > Data in duput() format is the preferred way to get data in R-help since
> it provides a perfect copy of what you have on your machine.  Any other way
> of providing data risks the recipients reading it into R differently than
> it is on your machine.
> >
> > John Kane
> > Kingston ON Canada
> >
> >
> >> -Original Message-
> >> From: tan...@gmail.com
> >> Sent: Fri, 29 May 2015 13:58:20 -0700
> >> To: sarah.gos...@gmail.com
> >> Subject: Re: [R] about transforming a data.frame
> >>
> >> Hi Sarah,
> >>
> >> thank you for your help. I have simplified the example, by reading the
> >> elements in a data frame, eg :
> >>
> >> df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> >> "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
> >> col_names = c
> >>
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> >> CT = c(5,2,2,2,2,2,4,4) )
> >>
> >> I have used the the count() in the plyr package :
> >>
> >> count_row_names <- count(df$row_names)
> >> count_col_names <- count(df$col_names)
> >>
> >> however, I would need to correlate these UNIQUE ELEMENTS in the columns
> >> "row_names" or "col_names" with the numbers they associate in the  CT
> >> columns, eg :
> >>
> >> ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
> >> associate with "2" (in the CT column).
> >>
> >> thank you very much,
> >>
> >> bogdan
> >>
> >>
> >>
> >>
> >> On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
> >> wrote:
> >>
> >>> Hi,
> >>>
> >>> Please use dput() to provide your data, as it can get somewhat mangled
> >>> by copy and pasting, especially if you post in HTML (as you are asked
> >>> not to do in the posting guide).
> >>>
> >>> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
> >>> "B5" each elements? That is, what is the result you expect to obtain
> >>> for the sample data you provided?
> >>>
> >>> What code have you tried? I would think table() might be involved, and
> >>> possibly strsplit(), but will refrain from putting more time into this
> >>> until you provide a reproducible dataset with dput() and some clearer
> >>> idea of your intent.
> >>>
> >>> Sarah
> >>>
> >>> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa 
> wrote:
>  Dear all,
> 
>  I would appreciate a suggestion on the following : I am working with a
>  data.frame (below) :
> 
>    EXPCT   row_names   col_names
>  1   test -5B4:B5:B6B1:B2:B3
>  2   test -2B7:B8:B9B1:B2:B3
>  3   test -2D4:D5:D6H4:H5:H6
>  4   test -2D10:D11:D12 F10:F11:F12
>  5   test -2D10:D11:D12H1:H2:H3
>  6   test -2E10:E11:E12G7:G8:G9
>  7   test -4 A1:A2:A3D1:D2:D3
>  8   test -4   B10:B11:B12B1:B2:B3
> 
>  what would be the easiest way to consider UNIQUE elements in the
> >>> ROW_NAMES
>  or the UNIQUE elements in the COL_NAMES and :
> 
>  print how many times these UNIQUE ELEMENTS associate with the num

Re: [R] about transforming a data.frame

[Sarah Goslee]

I'm still not really clear on what you need (format, etc), but this
may help you get started:

> with(df, table(CT, row_names))
   row_names
CT  A1:A2:A3 B10:B11:B12 B4:B5:B6 B7:B8:B9 D10:D11:D12 D4:D5:D6 E10:E11:E12
  20   001   21   1
  41   100   00   0
  50   010   00   0
> with(df, table(CT, col_names))
   col_names
CT  B1:B2:B3 D1:D2:D3 F10:F11:F12 G7:G8:G9 H1:H2:H3 H4:H5:H6
  210   1111
  411   0000
  510   0000
>

On Fri, May 29, 2015 at 4:58 PM, Bogdan Tanasa  wrote:
> Hi Sarah,
>
> thank you for your help. I have simplified the example, by reading the
> elements in a data frame, eg :
>
> df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
> col_names = c
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> CT = c(5,2,2,2,2,2,4,4) )
>
> I have used the the count() in the plyr package :
>
> count_row_names <- count(df$row_names)
> count_col_names <- count(df$col_names)
>
> however, I would need to correlate these UNIQUE ELEMENTS in the columns
> "row_names" or "col_names" with the numbers they associate in the  CT
> columns, eg :
>
> ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
> associate with "2" (in the CT column).
>
> thank you very much,
>
> bogdan
>
>
>
>
> On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
> wrote:
>>
>> Hi,
>>
>> Please use dput() to provide your data, as it can get somewhat mangled
>> by copy and pasting, especially if you post in HTML (as you are asked
>> not to do in the posting guide).
>>
>> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
>> "B5" each elements? That is, what is the result you expect to obtain
>> for the sample data you provided?
>>
>> What code have you tried? I would think table() might be involved, and
>> possibly strsplit(), but will refrain from putting more time into this
>> until you provide a reproducible dataset with dput() and some clearer
>> idea of your intent.
>>
>> Sarah
>>
>> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa  wrote:
>> > Dear all,
>> >
>> > I would appreciate a suggestion on the following : I am working with a
>> > data.frame (below) :
>> >
>> >   EXPCT   row_names   col_names
>> > 1   test -5B4:B5:B6B1:B2:B3
>> > 2   test -2B7:B8:B9B1:B2:B3
>> > 3   test -2D4:D5:D6H4:H5:H6
>> > 4   test -2D10:D11:D12 F10:F11:F12
>> > 5   test -2D10:D11:D12H1:H2:H3
>> > 6   test -2E10:E11:E12G7:G8:G9
>> > 7   test -4 A1:A2:A3D1:D2:D3
>> > 8   test -4   B10:B11:B12B1:B2:B3
>> >
>> > what would be the easiest way to consider UNIQUE elements in the
>> > ROW_NAMES
>> > or the UNIQUE elements in the COL_NAMES and :
>> >
>> > print how many times these UNIQUE ELEMENTS associate with the numbers
>> > -5,
>> > -2, or -4 (these numbers are on the column names CT) ..
>> >
>> > thanks,
>> >
>> > bogdan
>> >

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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] about transforming a data.frame

[Bogdan Tanasa]

Dear Sarah,

thank you very much, it is very helpful. please may I ask one more question
about a quick and easy tutorial about the loading multiple files (from a
folder) in R, and processing one file at a time ?  thanks very much again,

bogdan

On Fri, May 29, 2015 at 2:55 PM, Sarah Goslee 
wrote:

> I'm still not really clear on what you need (format, etc), but this
> may help you get started:
>
> > with(df, table(CT, row_names))
>row_names
> CT  A1:A2:A3 B10:B11:B12 B4:B5:B6 B7:B8:B9 D10:D11:D12 D4:D5:D6 E10:E11:E12
>   20   001   21   1
>   41   100   00   0
>   50   010   00   0
> > with(df, table(CT, col_names))
>col_names
> CT  B1:B2:B3 D1:D2:D3 F10:F11:F12 G7:G8:G9 H1:H2:H3 H4:H5:H6
>   210   1111
>   411   0000
>   510   0000
> >
>
> On Fri, May 29, 2015 at 4:58 PM, Bogdan Tanasa  wrote:
> > Hi Sarah,
> >
> > thank you for your help. I have simplified the example, by reading the
> > elements in a data frame, eg :
> >
> > df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> > "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
> > col_names = c
> >
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> > CT = c(5,2,2,2,2,2,4,4) )
> >
> > I have used the the count() in the plyr package :
> >
> > count_row_names <- count(df$row_names)
> > count_col_names <- count(df$col_names)
> >
> > however, I would need to correlate these UNIQUE ELEMENTS in the columns
> > "row_names" or "col_names" with the numbers they associate in the  CT
> > columns, eg :
> >
> > ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
> > associate with "2" (in the CT column).
> >
> > thank you very much,
> >
> > bogdan
> >
> >
> >
> >
> > On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
> > wrote:
> >>
> >> Hi,
> >>
> >> Please use dput() to provide your data, as it can get somewhat mangled
> >> by copy and pasting, especially if you post in HTML (as you are asked
> >> not to do in the posting guide).
> >>
> >> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
> >> "B5" each elements? That is, what is the result you expect to obtain
> >> for the sample data you provided?
> >>
> >> What code have you tried? I would think table() might be involved, and
> >> possibly strsplit(), but will refrain from putting more time into this
> >> until you provide a reproducible dataset with dput() and some clearer
> >> idea of your intent.
> >>
> >> Sarah
> >>
> >> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa 
> wrote:
> >> > Dear all,
> >> >
> >> > I would appreciate a suggestion on the following : I am working with a
> >> > data.frame (below) :
> >> >
> >> >   EXPCT   row_names   col_names
> >> > 1   test -5B4:B5:B6B1:B2:B3
> >> > 2   test -2B7:B8:B9B1:B2:B3
> >> > 3   test -2D4:D5:D6H4:H5:H6
> >> > 4   test -2D10:D11:D12 F10:F11:F12
> >> > 5   test -2D10:D11:D12H1:H2:H3
> >> > 6   test -2E10:E11:E12G7:G8:G9
> >> > 7   test -4 A1:A2:A3D1:D2:D3
> >> > 8   test -4   B10:B11:B12B1:B2:B3
> >> >
> >> > what would be the easiest way to consider UNIQUE elements in the
> >> > ROW_NAMES
> >> > or the UNIQUE elements in the COL_NAMES and :
> >> >
> >> > print how many times these UNIQUE ELEMENTS associate with the numbers
> >> > -5,
> >> > -2, or -4 (these numbers are on the column names CT) ..
> >> >
> >> > thanks,
> >> >
> >> > bogdan
> >> >
>

[[alternative HTML version deleted]]

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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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] Why I am not able to load library(R.matlab)? Other packages are fine.

[C W]

Hi Ben,

Thanks for the fun clip.  I love it.  Have a wonderful day!

-M

On Fri, May 29, 2015 at 5:10 PM, Ben Bolker  wrote:

> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
>  I think Henrik's point (which I merely clarified) was that something
> funky (we'll probably never know what, and it's not worth figuring out
> unless it happens again/to other people) had gone wrong and that the
> easiest thing to do was just to reinstall.
>
> References:
> * https://www.youtube.com/watch?v=t2F1rFmyQmY
> *
>
> http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.208.9970&rep=rep1&type=pdf
>
>
> On 15-05-29 05:11 PM, C W wrote:
> > Wow, thanks Ben.  That worked very well.
> >
> > I guess I didn't have R.methodS3?  But that doesn't make sense,
> > because I was using R.matlab few weeks ago.  I believe I was on R
> > 3.1.
> >
> > Maybe it's in R 3.1 folder?  I am using a Mac, btw.
> >
> > Cheers,
> >
> > -M
> >
> > On Fri, May 29, 2015 at 1:55 PM, Ben Bolker 
> > wrote:
> >
> >> C W  gmail.com> writes:
> >>
> >>>
> >>> Hi Henrik,
> >>>
> >>> I don't quite get what I should do here.  I am not familiar
> >>> with R.methodS3.  Can you tell me what command exactly do I
> >>> need to do?
> >>>
> >>> Thanks,
> >>>
> >>> Mike
> >>
> >> install.packages("R.methodsS3") install.packages("R.matlab")
> >> library("R.matlab")
> >>
> >>
> >>
> >> [snip snip snip]
> >>
> >> __
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more,
> >> see 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.
> >>
> >
>
> -BEGIN PGP SIGNATURE-
> Version: GnuPG v1.4.11 (GNU/Linux)
>
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> Qo0q9ed2csPRLbLLrpX2FPKbxLg/g6MSxmIQ118tbWhkzKfRoyxCZHLcT+U2xLuR
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> =nO+b
> -END PGP SIGNATURE-
>

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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Help on R Functionality & Histogram

[Boris Steipe]

Don't use Nabble when posting to the R-Help forum.

Responses inline.

On May 29, 2015, at 7:54 AM, Shivi82  wrote:

> Hello Experts, 
> I have couple of questions on the analysis I am creating.
> 1) How does R adopt to changes. The case I have here is that the excel I
> have started initially had to be modified because the data I had was on
> hourly basis ranging from 0 to 23 hours. After Changes 0 was modified to 24
> in hours. Now do I need to recall this excel again in R using read.csv
> syntax or is there another way to do so i.e. a kind of reload option

No. Reload the data by rerunning your script.


> 2) I am creating a histogram. I need on x axis 24 hours to be displayed
> separately as 0,1,2, and thereon. However it only shows till 20 which makes
> the look awkward. Also all l need to resize the labels and if possible
> inside the bars. It used the below code, axis fonts have changed but labels
> give an error with this code
> 
> Code:- hist(aaa$Hours,main="Hourly Weight",xlab = "Time",breaks = 25,col =
> "yellow",ylim = c(0,9000),
> labels=TRUE, cex.axis=0.6,cex.label=0.6)

The very understandable warning message you must have got with that call tells 
you that there is no such argument "cex.label".

hist() calls plot.histogram() which internally calls text() to write the 
labels. text() has an argument "cex", but even if you supply it to hist(), it 
is not passed to text() via the function body of plot.histogram(). You could 
modify plot.histogram but the more immediate solution is to set labels = FALSE, 
and explicitly use text() to write your labels. Try something like

x <- hist(aaa$Hours,
 main="Hourly Weight",
 xlab = "Time",
 breaks = 25,
 col = "yellow",
 ylim = c(0,9000),
 labels=FALSE,
 cex.axis=0.6)
 
text(x$mids, x$counts * 1.05, labels = x$counts, cex=0.5)



B.


> 
> Kindly advice on the both the questions. Thanks. 
> 
> 
> 
> 
> 
> 
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Help-on-R-Functionality-Histogram-tp4707887.html
> Sent from the R help mailing list archive at Nabble.com.
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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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Re: [R] about transforming a data.frame

[Sarah Goslee]

LMGTFY: 
http://stackoverflow.com/questions/11433432/importing-multiple-csv-files-into-r

On Fri, May 29, 2015 at 5:58 PM, Bogdan Tanasa  wrote:
> Dear Sarah,
>
> thank you very much, it is very helpful. please may I ask one more question
> about a quick and easy tutorial about the loading multiple files (from a
> folder) in R, and processing one file at a time ?  thanks very much again,
>
> bogdan
>
> On Fri, May 29, 2015 at 2:55 PM, Sarah Goslee 
> wrote:
>>
>> I'm still not really clear on what you need (format, etc), but this
>> may help you get started:
>>
>> > with(df, table(CT, row_names))
>>row_names
>> CT  A1:A2:A3 B10:B11:B12 B4:B5:B6 B7:B8:B9 D10:D11:D12 D4:D5:D6
>> E10:E11:E12
>>   20   001   21
>> 1
>>   41   100   00
>> 0
>>   50   010   00
>> 0
>> > with(df, table(CT, col_names))
>>col_names
>> CT  B1:B2:B3 D1:D2:D3 F10:F11:F12 G7:G8:G9 H1:H2:H3 H4:H5:H6
>>   210   1111
>>   411   0000
>>   510   0000
>> >
>>
>> On Fri, May 29, 2015 at 4:58 PM, Bogdan Tanasa  wrote:
>> > Hi Sarah,
>> >
>> > thank you for your help. I have simplified the example, by reading the
>> > elements in a data frame, eg :
>> >
>> > df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
>> > "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
>> > col_names = c
>> >
>> > ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
>> > CT = c(5,2,2,2,2,2,4,4) )
>> >
>> > I have used the the count() in the plyr package :
>> >
>> > count_row_names <- count(df$row_names)
>> > count_col_names <- count(df$col_names)
>> >
>> > however, I would need to correlate these UNIQUE ELEMENTS in the columns
>> > "row_names" or "col_names" with the numbers they associate in the  CT
>> > columns, eg :
>> >
>> > ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
>> > associate with "2" (in the CT column).
>> >
>> > thank you very much,
>> >
>> > bogdan
>> >
>> >
>> >
>> >
>> > On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
>> > wrote:
>> >>
>> >> Hi,
>> >>
>> >> Please use dput() to provide your data, as it can get somewhat mangled
>> >> by copy and pasting, especially if you post in HTML (as you are asked
>> >> not to do in the posting guide).
>> >>
>> >> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
>> >> "B5" each elements? That is, what is the result you expect to obtain
>> >> for the sample data you provided?
>> >>
>> >> What code have you tried? I would think table() might be involved, and
>> >> possibly strsplit(), but will refrain from putting more time into this
>> >> until you provide a reproducible dataset with dput() and some clearer
>> >> idea of your intent.
>> >>
>> >> Sarah
>> >>
>> >> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa 
>> >> wrote:
>> >> > Dear all,
>> >> >
>> >> > I would appreciate a suggestion on the following : I am working with
>> >> > a
>> >> > data.frame (below) :
>> >> >
>> >> >   EXPCT   row_names   col_names
>> >> > 1   test -5B4:B5:B6B1:B2:B3
>> >> > 2   test -2B7:B8:B9B1:B2:B3
>> >> > 3   test -2D4:D5:D6H4:H5:H6
>> >> > 4   test -2D10:D11:D12 F10:F11:F12
>> >> > 5   test -2D10:D11:D12H1:H2:H3
>> >> > 6   test -2E10:E11:E12G7:G8:G9
>> >> > 7   test -4 A1:A2:A3D1:D2:D3
>> >> > 8   test -4   B10:B11:B12B1:B2:B3
>> >> >
>> >> > what would be the easiest way to consider UNIQUE elements in the
>> >> > ROW_NAMES
>> >> > or the UNIQUE elements in the COL_NAMES and :
>> >> >
>> >> > print how many times these UNIQUE ELEMENTS associate with the numbers
>> >> > -5,
>> >> > -2, or -4 (these numbers are on the column names CT) ..
>> >> >
>> >> > thanks,
>> >> >
>> >> > bogdan
>> >> >
>
>

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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] Result differences in 32-bit vs. 64-bit point.in.polygon?

[Duncan Murdoch]

On 29/05/2015 2:36 PM, Lensing, Shelly Y wrote:
> Is anyone aware of point.in.polygon giving different results for 32-bit vs. 
> 64-bit R? Our OS is 64-bit Windows 7 Enterprise. I'm working with someone 
> else's extensive R program and the final results are close but not exactly 
> matching. We're thinking it might be something with the point.in.polygon 
> function (one of many possibilities, including leaps).

Often 32 bit R does calculations slightly more accurately than 64 bit R
does.  This is because the 64 bit compiler is more likely to do
calculations in 64 bit precision when the 32 bit compiler does them in
80 bit precision.  Of course, individual calculations being more
accurate doesn't mean the final answer is, but small numeric differences
in floating point calculations are to be expected.

Duncan Murdoch

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Problem with comparing multiple data sets

[John Kane]

Hi Mohammad,
I have no idea what is happening but for some reason your new data (renamed df1 
since df is a reserved word in R) is outputting a list whereas dff1 (your 
original test data) is giving a vector as you wanted.

It may be obvious but I don't see why df1 is giving us a list.  As far as I can 
tell the two data sets are structually the same.

The two data sets are below the program.  
## =
library(modeest)

# Original test data 
str(dff2)
head(dff2)

# sample of new data
str(d1)
head(df1)

Out.dff2  <- apply(dff2[ ,2:length(dff2)], 1, mfv)
str(Out.dff2)

Out.df1  <-  apply(df1[ , 2:length(df1)], 1, mfv)
str(Out.df1)


## =
## New data set 
df1  <- structure(list(terms = structure(c(2L, 4L, 4L, 4L, 3L, 1L, 5L,
5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label =
c("#authentication,access control",
"#privacy,personal data", "#security,malicious,security", "data controller",
"id management,security", "password,recovery"), class = "factor"),
class.1 = c(2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L,
2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L), class.2 = c(2L, 2L, 2L,
0L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L,
2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L,
2L, 2L), class.3 = c(2L, 0L, 2L, 2L, 1L, 1L, 0L, 0L, 0L,
2L, 2L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("terms",
"class.1", "class.2", "class.3"), class = "data.frame", row.names = c(NA,
-50L))

## Original test data set

dff2  <-   structure(list(terms = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, 4L,
 4L, 4L, 4L, 3L, 3L, 3L, 3L, 2L, 2L, 2L), .Label = c("#dac",
 "#mac,#security",
 "accountability,anonymous", "data security,encryption,security"
 ), class = "factor"), class.1 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L,
 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 2L, 1L,
 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L), class.2 = c(2L, 2L,
 2L, 2L, 0L, 0L, 2L, 0L, 0L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 0L,
 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 2L,
 0L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 0L, 0L, 0L, 0L, 1L, 1L, 1L),
 class.3 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 2L, 1L, 1L, 1L, 1L,
 0L, 0L, 0L, 0L, 2L, 1L, 2L)), .Names = c("terms", "class.1",
 "class.2", "class.3"), class = "data.frame", row.names = c(NA,
 -49L))

##=



John Kane
Kingston ON Canada

-Original Message-
From: mxalimoha...@ualr.edu
Sent: Fri, 29 May 2015 11:40:41 -0500
To: dcarl...@tamu.edu, drjimle...@gmail.com, jrkrid...@inbox.com, 
r-help@r-project.org
Subject: Re: [R] Problem with comparing multiple data sets

Hi everyone.

I tried the (modeest) package on my initial test data and it worked. However, 
it doesn't work on the entire data set. I saved one of the protions that gives 
error. (Not for all of the values but for some of them). For example: lines 36 
and 37 and 39 correctly show the mode value but 38 and 40 are not correct. Such 
error is repeated for many of the values.

[36,] 2        

[37,] 2        

[38,] Numeric,3

[39,] 1        

[40,] Numeric,3



#This is what I did:

> df<- read.csv(file="Part1-modif.csv", head=TRUE, sep=",")

> Out<- apply(df[,2:length(df)],1, mfv)

> t(t(Out))

#This is the data set 

structure(list(terms = structure(c(2L, 4L, 4L, 4L, 3L, 1L, 5L, 

5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 

6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 

6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = c("#authentication,access 
control", 

"#privacy,personal data", "#security,malicious,security", "data controller", 

"id management,security", "password,recovery"), class = "factor"), 

    class.1 = c(2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 

    2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 

    1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 

    2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L), class.2 = c(2L, 2L, 2L, 

    0L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 

    2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 

    2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 

    2L, 2L), class.3 = c(

Re: [R] about transforming a data.frame

[Bogdan Tanasa]

thanks a lot Sarah, very much appreciate it !

On Fri, May 29, 2015 at 3:18 PM, Sarah Goslee 
wrote:

> LMGTFY:
> http://stackoverflow.com/questions/11433432/importing-multiple-csv-files-into-r
>
> On Fri, May 29, 2015 at 5:58 PM, Bogdan Tanasa  wrote:
> > Dear Sarah,
> >
> > thank you very much, it is very helpful. please may I ask one more
> question
> > about a quick and easy tutorial about the loading multiple files (from a
> > folder) in R, and processing one file at a time ?  thanks very much
> again,
> >
> > bogdan
> >
> > On Fri, May 29, 2015 at 2:55 PM, Sarah Goslee 
> > wrote:
> >>
> >> I'm still not really clear on what you need (format, etc), but this
> >> may help you get started:
> >>
> >> > with(df, table(CT, row_names))
> >>row_names
> >> CT  A1:A2:A3 B10:B11:B12 B4:B5:B6 B7:B8:B9 D10:D11:D12 D4:D5:D6
> >> E10:E11:E12
> >>   20   001   21
> >> 1
> >>   41   100   00
> >> 0
> >>   50   010   00
> >> 0
> >> > with(df, table(CT, col_names))
> >>col_names
> >> CT  B1:B2:B3 D1:D2:D3 F10:F11:F12 G7:G8:G9 H1:H2:H3 H4:H5:H6
> >>   210   1111
> >>   411   0000
> >>   510   0000
> >> >
> >>
> >> On Fri, May 29, 2015 at 4:58 PM, Bogdan Tanasa 
> wrote:
> >> > Hi Sarah,
> >> >
> >> > thank you for your help. I have simplified the example, by reading the
> >> > elements in a data frame, eg :
> >> >
> >> > df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> >> > "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3",
> "B10:B11:B12"),
> >> > col_names = c
> >> >
> >> >
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> >> > CT = c(5,2,2,2,2,2,4,4) )
> >> >
> >> > I have used the the count() in the plyr package :
> >> >
> >> > count_row_names <- count(df$row_names)
> >> > count_col_names <- count(df$col_names)
> >> >
> >> > however, I would need to correlate these UNIQUE ELEMENTS in the
> columns
> >> > "row_names" or "col_names" with the numbers they associate in the  CT
> >> > columns, eg :
> >> >
> >> > ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
> >> > associate with "2" (in the CT column).
> >> >
> >> > thank you very much,
> >> >
> >> > bogdan
> >> >
> >> >
> >> >
> >> >
> >> > On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee  >
> >> > wrote:
> >> >>
> >> >> Hi,
> >> >>
> >> >> Please use dput() to provide your data, as it can get somewhat
> mangled
> >> >> by copy and pasting, especially if you post in HTML (as you are asked
> >> >> not to do in the posting guide).
> >> >>
> >> >> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
> >> >> "B5" each elements? That is, what is the result you expect to obtain
> >> >> for the sample data you provided?
> >> >>
> >> >> What code have you tried? I would think table() might be involved,
> and
> >> >> possibly strsplit(), but will refrain from putting more time into
> this
> >> >> until you provide a reproducible dataset with dput() and some clearer
> >> >> idea of your intent.
> >> >>
> >> >> Sarah
> >> >>
> >> >> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa 
> >> >> wrote:
> >> >> > Dear all,
> >> >> >
> >> >> > I would appreciate a suggestion on the following : I am working
> with
> >> >> > a
> >> >> > data.frame (below) :
> >> >> >
> >> >> >   EXPCT   row_names   col_names
> >> >> > 1   test -5B4:B5:B6B1:B2:B3
> >> >> > 2   test -2B7:B8:B9B1:B2:B3
> >> >> > 3   test -2D4:D5:D6H4:H5:H6
> >> >> > 4   test -2D10:D11:D12 F10:F11:F12
> >> >> > 5   test -2D10:D11:D12H1:H2:H3
> >> >> > 6   test -2E10:E11:E12G7:G8:G9
> >> >> > 7   test -4 A1:A2:A3D1:D2:D3
> >> >> > 8   test -4   B10:B11:B12B1:B2:B3
> >> >> >
> >> >> > what would be the easiest way to consider UNIQUE elements in the
> >> >> > ROW_NAMES
> >> >> > or the UNIQUE elements in the COL_NAMES and :
> >> >> >
> >> >> > print how many times these UNIQUE ELEMENTS associate with the
> numbers
> >> >> > -5,
> >> >> > -2, or -4 (these numbers are on the column names CT) ..
> >> >> >
> >> >> > thanks,
> >> >> >
> >> >> > bogdan
> >> >> >
> >
> >
>

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] about transforming a data.frame

[Jim Lemon]

Hi Bogdan,
Sarah has already suggested this, but doesn't:

table(df$row_names,df$CT)
table(df$col_names,df$CT)

give you what you want?

Jim


On Sat, May 30, 2015 at 7:11 AM, John Kane  wrote:
> Bogdan, the request was for data in dput() format.
>
> Type ?dput for more information.
>
> Do dput(myfile) copy the ouput and paste into the email
>
> You should get something like:
> structure(list(c1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
> 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L,
> 5L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L,
> 8L, 8L, 9L, 9L, 9L, 9L, 9L, 10L, 10L, 10L), .Label = c("(0.509,0.614]",
> "(0.614,0.718]", "(0.718,0.822]", "(0.822,0.926]", "(0.926,1.03]",
> "(1.03,1.13]", "(1.13,1.24]", "(1.24,1.34]", "(1.34,1.45]", "(1.45,1.55]"
> ), class = "factor"), s1 = c(0.51, 0.52, 0.58, 0.58, 0.59, 0.6,
> 0.63, 0.65, 0.68, 0.74, 0.74, 0.75, 0.77, 0.77, 0.77, 0.78, 0.79,
> 0.84, 0.84, 0.85, 0.87, 0.93, 0.93, 0.95, 0.99, 1.04, 1.09, 1.11,
> 1.13, 1.14, 1.14, 1.14, 1.17, 1.18, 1.19, 1.22, 1.22, 1.23, 1.28,
> 1.29, 1.3, 1.32, 1.37, 1.38, 1.38, 1.4, 1.43, 1.47, 1.52, 1.55
> )), .Names = c("c1", "s1"), row.names = c(NA, -50L), class = "data.frame")
>
> Data in duput() format is the preferred way to get data in R-help since it 
> provides a perfect copy of what you have on your machine.  Any other way of 
> providing data risks the recipients reading it into R differently than it is 
> on your machine.
>
> John Kane
> Kingston ON Canada
>
>
>> -Original Message-
>> From: tan...@gmail.com
>> Sent: Fri, 29 May 2015 13:58:20 -0700
>> To: sarah.gos...@gmail.com
>> Subject: Re: [R] about transforming a data.frame
>>
>> Hi Sarah,
>>
>> thank you for your help. I have simplified the example, by reading the
>> elements in a data frame, eg :
>>
>> df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
>> "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
>> col_names = c
>> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
>> CT = c(5,2,2,2,2,2,4,4) )
>>
>> I have used the the count() in the plyr package :
>>
>> count_row_names <- count(df$row_names)
>> count_col_names <- count(df$col_names)
>>
>> however, I would need to correlate these UNIQUE ELEMENTS in the columns
>> "row_names" or "col_names" with the numbers they associate in the  CT
>> columns, eg :
>>
>> ""B1:B2:B3" associate with "5, 2, 4" (in CT column), or "D10:D11:D12"
>> associate with "2" (in the CT column).
>>
>> thank you very much,
>>
>> bogdan
>>
>>
>>
>>
>> On Fri, May 29, 2015 at 1:32 PM, Sarah Goslee 
>> wrote:
>>
>>> Hi,
>>>
>>> Please use dput() to provide your data, as it can get somewhat mangled
>>> by copy and pasting, especially if you post in HTML (as you are asked
>>> not to do in the posting guide).
>>>
>>> What is a unique element? is "B4:B5:B6" an element, or are "B4" and
>>> "B5" each elements? That is, what is the result you expect to obtain
>>> for the sample data you provided?
>>>
>>> What code have you tried? I would think table() might be involved, and
>>> possibly strsplit(), but will refrain from putting more time into this
>>> until you provide a reproducible dataset with dput() and some clearer
>>> idea of your intent.
>>>
>>> Sarah
>>>
>>> On Fri, May 29, 2015 at 4:19 PM, Bogdan Tanasa  wrote:
 Dear all,

 I would appreciate a suggestion on the following : I am working with a
 data.frame (below) :

   EXPCT   row_names   col_names
 1   test -5B4:B5:B6B1:B2:B3
 2   test -2B7:B8:B9B1:B2:B3
 3   test -2D4:D5:D6H4:H5:H6
 4   test -2D10:D11:D12 F10:F11:F12
 5   test -2D10:D11:D12H1:H2:H3
 6   test -2E10:E11:E12G7:G8:G9
 7   test -4 A1:A2:A3D1:D2:D3
 8   test -4   B10:B11:B12B1:B2:B3

 what would be the easiest way to consider UNIQUE elements in the
>>> ROW_NAMES
 or the UNIQUE elements in the COL_NAMES and :

 print how many times these UNIQUE ELEMENTS associate with the numbers
 -5,
 -2, or -4 (these numbers are on the column names CT) ..

 thanks,

 bogdan

>>> --
>>> Sarah Goslee
>>> http://www.functionaldiversity.org
>>>
>>
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>>
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Re: [R] TWS and R

[Austin Trombley]

Has anyone found a solution to this?  I am having the same issue?  thanks! 

On Thursday, November 15, 2012 at 10:35:48 PM UTC-8, abcd1234 wrote:
>
> Hi all, 
>
>  The TWS on my system is unable to connect to my R session. Here is the 
> error that I'm getting: 
>
> /> tws<-twsConnect() 
> Error in socketConnection(host = host, port = port, open = "ab", blocking 
> = 
> blocking) : 
> cannot open the connection 
> In addition: Warning message: 
> In socketConnection(host = host, port = port, open = "ab", blocking = 
> blocking) : 
> localhost:7496 cannot be opened/ 
>
> Here is the session info for the R session: 
> / 
> R version 2.15.1 (2012-06-22) 
>  Platform: x86_64-pc-linux-gnu (64-bit) 
> locale: 
>  [1] LC_CTYPE=en_IN.UTF-8LC_NUMERIC=C 
>
>  [3] LC_TIME=en_IN.UTF-8LC_COLLATE=en_IN.UTF-8 
>   
> [5] LC_MONETARY=en_IN.UTF-8LC_MESSAGES=en_IN.UTF-8 
>   
> [7] LC_PAPER=CLC_NAME=C 
>
>  [9] LC_ADDRESS=CLC_TELEPHONE=C 
>
>  [11] LC_MEASUREMENT=en_IN.UTF-8 LC_IDENTIFICATION=C 
>
>  attached base packages: 
> [1] statsgraphicsgrDevices utilsdatasets 
> [6] methodsbase 
>   
> other attached packages: 
>  [1] IBrokers_0.9-10 xts_0.8-6zoo_1.7-8 
>
>  loaded via a namespace (and not attached): 
>  [1] grid_2.15.1lattice_0.20-0 tools_2.15.1/ 
>
> I have checked the "Enable Activex and Socket clients"  but it hasn't 
> helped. Since I'm running on an Ubuntu machine, I even tried changing the 
> parameter blocking in the command twsConnect() to 
>
> 1. blocking = FALSE 
>
> 2. According to the one mentioned here 
>
> http://code.google.com/p/ibrokers/source/detail?r=84&path=/trunk/R/twsConnect.R
>  
>   
> but nothing has helped. 
>
> I have also added 127.0.0.1 to the "Trusted IP" option. 
>
>  Please let me know what I should do. 
>
> Thanks. 
>
>
>
>
> -- 
> View this message in context: 
> http://r.789695.n4.nabble.com/TWS-and-R-tp4649699.html 
> Sent from the R help mailing list archive at Nabble.com. 
>
> __ 
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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. 
>
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[R] vectorized code

[zeynab jibril]

HI

I was working on online example, where virus is spread through a graph. The
example is sufficient for small graph i.e. small number of edges and nodes.
But I tried it on very large graph i.e. 1 nodes and 2 edges, but
the below function is not sufficient for large graph because its slow.

My question is how can the below function be converted to Vectorized code
can be optimized for large graphs?

spreadVirus <- function(G,Vinitial,Activation_probability){



# Precompute all outgoing graph adjacencies



G$AdjList = get.adjlist(G,mode="out")



# Initialize various graph attributes

V(G)$color= "blue"

E(G)$color= "black"

V(G)[Vinitial]$color<- "yellow"



# List to store the incremental graphs (for plotting later)

Glist <- list(G)

count <- 1



# Spread the infection

active <- Vinitial



while(length(active)>0){

new_infected <- NULL

E(G)$color = "black"



for(v in active){

# spread through the daily contacts of vertex v



daily_contacts <- G$AdjList[[v]]



E(G)[v %->% daily_contacts]$color <- "red"



for(v1 in daily_contacts){



if(V(G)[v1]$color == "blue" & new_color=="red") {



V(G)[v1]$color <- "red"



new_infected <- c(new_infected,v1)



 }

}

}

# the next active set

#this needed for updating



active <- new_infected



# Add graph to list

# optional dependening on if i want to graph

count <- count + 1

Glist[[count]] <- G

}

return(Glist)

}

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Re: [R] Converting unique strings to unique numbers

[Hervé Pagès]

Hi Bill,

On 05/29/2015 01:48 PM, William Dunlap wrote:

I'm not sure why which particular ID gets assigned to each string would
matter but maybe I'm missing something. What really matters is that each
string receives a unique ID. match(x, x) does that.


I think each row of the OP's dataset represented an individual (column 2)
followed by its mother and father (columns 3 and 4).  I assume that the
marker "0" means that a parent is not in the dataset.  If you match against
the strings in column 2 only, in their original order, then the
resulting numbers
give the row number of an individual,


Note that the code I gave happens to do exactly that (assuming that
column 2 contains no duplicates, but your code is also relying on that
assumption in order to have the ids match the row numbers).

We're discussing the merit of match(x, x) versus match(x, unique(x)).
All I'm trying to say is that the unique(x) step (which doubles the cost
of the whole operation, because it also uses hashing, like match() does)
is generally not needed. It doesn't seem to be needed in Kate's use
case.

H.


making it straightforward to look up
information regarding the ancestors of an individual.  Hence the choice of
numeric ID's may be important.

Bill Dunlap
TIBCO Software
wdunlap tibco.com 

On Fri, May 29, 2015 at 1:29 PM, Hervé Pagès mailto:hpa...@fredhutch.org>> wrote:

Hi Sarah,

On 05/29/2015 12:04 PM, Sarah Goslee wrote:

On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès
mailto:hpa...@fredhutch.org>> wrote:

Hi Kate,

I found that matching the character vector to itself is a very
effective way to do this:

x <- c("a", "bunch", "of", "strings", "whose", "exact",
"content",
   "is", "of", "little", "interest")
ids <- match(x, x)
ids
# [1]  1  2  3  4  5  6  7  8  3 10 11

By using this trick, many manipulations on character vectors can
be replaced by manipulations on integer vectors, which are
sometimes
way more efficient.


Hm. I hadn't thought of that approach - I use the
as.numeric(factor(...)) approach.

So I was curious, and compared the two:


set.seed(43)
x <- sample(letters, 1, replace=TRUE)

system.time({
for(i in seq_len(2)) {
ids1 <- match(x, x)
}})

#   user  system elapsed
#  9.657   0.000   9.657

system.time({
for(i in seq_len(2)) {
ids2 <- as.numeric(factor(x, levels=letters))
}})

#   user  system elapsed
#   6.160.006.16

Using factor() is faster.


That's an unfair comparison, because you already know what the levels
are so you can supply them to your call to factor(). Most of the time
you don't know what the levels are so either you just do factor(x) and
let the factor() constructor compute the levels for you, or you compute
them yourself upfront with something like factor(x, levels=unique(x)).

   library(microbenchmark)

   microbenchmark(
 {ids1 <- match(x, x)},
 {ids2 <- as.integer(factor(x, levels=letters))},
 {ids3 <- as.integer(factor(x))},
 {ids4 <- as.integer(factor(x, levels=unique(x)))}
   )
   Unit: microseconds
   expr min
  lq
{ ids1 <- match(x, x) } 245.979
262.2390
{ ids2 <- as.integer(factor(x, levels = letters)) } 214.115
219.2320
  { ids3 <- as.integer(factor(x)) } 380.782
388.7295
  { ids4 <- as.integer(factor(x, levels = unique(x))) } 332.250
342.6630
mean   median  uq max neval
267.3210 264.4845 268.348 293.894   100
226.9913 220.9870 226.147 314.875   100
402.2242 394.7165 412.075 481.410   100
349.7405 345.3090 353.162 383.002   100

More importantly, using factor() lets you
set the order of the indices in an expected fashion, where match()
assigns them in the order of occurrence.

head(data.frame(x, ids1, ids2))

x ids1 ids2
1 m1   13
2 x2   24
3 b32
4 s4   19
5 i59
6 o6   15

In a problem like Kate's where there are several columns for
which the
same ordering of indices is desired, that becomes really important.


I'm not sure why which particular ID gets assigned to each string would
matter but maybe I'm missing something. What really matters is that each
string receives a unique ID. match(x, x) does that.

In Kate's problem, where the strings are in more than one column,
and you want the ID to be unique across the columns, you need to do
 

Re: [R] about transforming a data.frame

[Jim Lemon]

Hi Bogdan,
If you mean "How can I verify that "B1:B2:B3" is paired with all of
the values 2, 4 and 5"

apply(table(df$col_names,df$CT),1,all)

and if you mean "How can I verify that "B1:B2:B3" is paired with at
least one of the values 2, 4 and 5"

apply(table(df$col_names,df$CT),1,any)

Jim


Hi Jim,

yes, thank you, that is the desired output. one more question please :
after using the dataframe :

df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
"D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3",
"B10:B11:B12"),  col_names = c
("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
CT = c(5,2,2,2,2,2,4,4) )

and :

table(df$row_names,df$CT)
table(df$col_names,df$CT)

how could I quickly verify that "B1:B2:B3" (for example) hits the CT
values of 2,4,5  at least one time ? an example is in

table(df$col_names,df$CT) ?

thank you very much,

-- bogdan

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Re: [R] about transforming a data.frame

[Bogdan Tanasa]

Thanks a lot Jim. If I may ask one more little question please,

shall I ask the question ""How can I verify that "B1:B2:B3" is paired with
ALL of the values 2, 4 and 5 ",

regardless of the pairing value (in our case, for the code below, the
"pairing value" for "B1:B2:B3" is 1, but it can be 2,3,4, etc BUT NOT
zero),

how could I test for that ? or this is the way that "apply" works for "all"
argument ?

a good documentation for "apply" function will help too . thanks, and happy
weekend !

-- bogdan


On Fri, May 29, 2015 at 4:21 PM, Jim Lemon  wrote:

> Hi Bogdan,
> If you mean "How can I verify that "B1:B2:B3" is paired with all of
> the values 2, 4 and 5"
>
> apply(table(df$col_names,df$CT),1,all)
>
> and if you mean "How can I verify that "B1:B2:B3" is paired with at
> least one of the values 2, 4 and 5"
>
> apply(table(df$col_names,df$CT),1,any)
>
> Jim
>
>
> Hi Jim,
>
> yes, thank you, that is the desired output. one more question please :
> after using the dataframe :
>
> df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
> "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3",
> "B10:B11:B12"),  col_names = c
>
> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
> CT = c(5,2,2,2,2,2,4,4) )
>
> and :
>
> table(df$row_names,df$CT)
> table(df$col_names,df$CT)
>
> how could I quickly verify that "B1:B2:B3" (for example) hits the CT
> values of 2,4,5  at least one time ? an example is in
>
> table(df$col_names,df$CT) ?
>
> thank you very much,
>
> -- bogdan
>

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Re: [R] about transforming a data.frame

[Bogdan Tanasa]

Hi Jim,

thanks again. now I see : the answer to my previous question seems to be
yes, as "all" functions works on logical vectors ... best wishes,

-- bogdan

On Fri, May 29, 2015 at 4:29 PM, Bogdan Tanasa  wrote:

> Thanks a lot Jim. If I may ask one more little question please,
>
> shall I ask the question ""How can I verify that "B1:B2:B3" is paired with
> ALL of the values 2, 4 and 5 ",
>
> regardless of the pairing value (in our case, for the code below, the
> "pairing value" for "B1:B2:B3" is 1, but it can be 2,3,4, etc BUT NOT
> zero),
>
> how could I test for that ? or this is the way that "apply" works for
> "all" argument ?
>
> a good documentation for "apply" function will help too . thanks, and
> happy weekend !
>
> -- bogdan
>
>
> On Fri, May 29, 2015 at 4:21 PM, Jim Lemon  wrote:
>
>> Hi Bogdan,
>> If you mean "How can I verify that "B1:B2:B3" is paired with all of
>> the values 2, 4 and 5"
>>
>> apply(table(df$col_names,df$CT),1,all)
>>
>> and if you mean "How can I verify that "B1:B2:B3" is paired with at
>> least one of the values 2, 4 and 5"
>>
>> apply(table(df$col_names,df$CT),1,any)
>>
>> Jim
>>
>>
>> Hi Jim,
>>
>> yes, thank you, that is the desired output. one more question please :
>> after using the dataframe :
>>
>> df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
>> "D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3",
>> "B10:B11:B12"),  col_names = c
>>
>> ("B1:B2:B3","B1:B2:B3","H4:H5:H6","F10:F11:F12","H1:H2:H3","G7:G8:G9","D1:D2:D3","B1:B2:B3"),
>> CT = c(5,2,2,2,2,2,4,4) )
>>
>> and :
>>
>> table(df$row_names,df$CT)
>> table(df$col_names,df$CT)
>>
>> how could I quickly verify that "B1:B2:B3" (for example) hits the CT
>> values of 2,4,5  at least one time ? an example is in
>>
>> table(df$col_names,df$CT) ?
>>
>> thank you very much,
>>
>> -- bogdan
>>
>
>

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[R] Toronto CRAN mirror 403 error?

[Mark Drummond]

I've been getting a 403 when I try pulling from the Toronto CRAN mirror
today.

http://cran.utstat.utoronto.ca/

Is there a contact list for mirror managers?

-- 
Cheers, Mark

*Mark Drummond*
m...@markdrummond.ca

When I get sad, I stop being sad and be Awesome instead. TRUE STORY.

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Re: [R] Problem with comparing multiple data sets

[Jim Lemon]

Hi Mohammad,
It looks like you are still having problems with this. Given your
latest data set, as below, here is something that might do what you
want. From David's message, I'm not sure whether you are operating on
a single data frame or a list.

# this is the data set as taken from your message below
madf<-structure(list(terms = structure(c(2L, 4L, 4L, 4L, 3L, 1L, 5L,
5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label =
c("#authentication,access control",
"#privacy,personal data", "#security,malicious,security", "data controller",
"id management,security", "password,recovery"), class = "factor"),
class.1 = c(2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L,
2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L,
1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L), class.2 = c(2L, 2L, 2L,
0L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L,
2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L,
2L, 2L), class.3 = c(2L, 0L, 2L, 2L, 1L, 1L, 0L, 0L, 0L,
2L, 2L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("terms",
"class.1", "class.2", "class.3"), class = "data.frame", row.names = c(NA,
-50L))

# define a function that extracts the value from one field
# selected by a value in another field
extract_by_value<-function(x,field1,value1,field2) {
 return(x[x[,field1]==value1,field2])
}

# define another function that equates all of the values
sub_value<-function(x,field1,value1,field2,value2) {
 x[x[,field1]==value1,field2]<-value2
 return(x)
}

# this now steps through every value in "key_field"
# and operates on every field listed in "change_fields"
conformity<-function(x,key_field,change_fields) {
 keys<-unique(x[,key_field])
 for(key in keys) {
  for(change_field in change_fields) {
   # get the most frequent value in change_field
   # for the desired value in key_field
   most_freq<-as.numeric(names(which.max(table(
extract_by_value(x,key_field,key,change_field)
   # now set all the values to the most frequent
   x<-sub_value(x,key_field,key,change_field,most_freq)
  }
 }
 return(x)
}

conformity(madf,"terms",c("class.1","class.2","class.3"))

Obviously you will want to save the return value of "conformity" into
your original data frame or create a new one.

Jim

> Hi everyone.
>
> I tried the (modeest) package on my initial test data and it worked. However, 
> it doesn't work on the entire data set. I saved one of the protions that 
> gives error. (Not for all of the values but for some of them). For example: 
> lines 36 and 37 and 39 correctly show the mode value but 38 and 40 are not 
> correct. Such error is repeated for many of the values.
>
> [36,] 2
>
> [37,] 2
>
> [38,] Numeric,3
>
> [39,] 1
>
> [40,] Numeric,3
>
> 
>
> #This is what I did:
>
>> df<- read.csv(file="Part1-modif.csv", head=TRUE, sep=",")
>
>> Out<- apply(df[,2:length(df)],1, mfv)
>
>> t(t(Out))
>
> #This is the data set
>
> structure(list(terms = structure(c(2L, 4L, 4L, 4L, 3L, 1L, 5L,
>
> 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
>
> 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
>
> 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = 
> c("#authentication,access control",
>
> "#privacy,personal data", "#security,malicious,security", "data controller",
>
> "id management,security", "password,recovery"), class = "factor"),
>
> class.1 = c(2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L,
>
> 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L,
>
> 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L,
>
> 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L), class.2 = c(2L, 2L, 2L,
>
> 0L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 2L,
>
> 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L,
>
> 2L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L,
>
> 2L, 2L), class.3 = c(2L, 0L, 2L, 2L, 1L, 1L, 0L, 0L, 0L,
>
> 2L, 2L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
>
> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
>
> 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("terms",
>
> "class.1", "class.2", "class.3"), class = "data.frame", row.names = c(NA,
>
> -50L))
>
> 
>
> also when I try to include the terms to the result it gives me an error:
>
>> mode.names<- data.frame (df[,1],Out)
>
> Error in data.frame(df[, 1], Out) :
>
> arguments imply differing number of rows: 50, 3
>

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Re: [R] Toronto CRAN mirror 403 error?

[Jeff Newmiller]

This is why there are mirrors. You don't have to wait for them or tell them to 
do their jobs.
---
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  Live:   OO#.. Dead: OO#..  Playing
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Sent from my phone. Please excuse my brevity.

On May 29, 2015 7:12:56 PM PDT, Mark Drummond  wrote:
>I've been getting a 403 when I try pulling from the Toronto CRAN mirror
>today.
>
>http://cran.utstat.utoronto.ca/
>
>Is there a contact list for mirror managers?

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Re: [R] Toronto CRAN mirror 403 error?

[David Winsemius]

On May 29, 2015, at 7:12 PM, Mark Drummond wrote:

> I've been getting a 403 when I try pulling from the Toronto CRAN mirror
> today.
> 
> http://cran.utstat.utoronto.ca/

Right. It's been out for the last 2.7 days:

http://cran.r-project.org/mirmon_report.html#ca

> 
> Is there a contact list for mirror managers?

Why do you care? Why not use another mirror? The 
http://lib.stat.cmu.edu/R/CRAN/ mirror should be fairly close if you are on 
that side of the continent.

-- 
David.

> 
> -- 
> Cheers, Mark
> 
> *Mark Drummond*
> m...@markdrummond.ca
> 
> When I get sad, I stop being sad and be Awesome instead. TRUE STORY.
> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

David Winsemius
Alameda, CA, USA

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Re: [R] Toronto CRAN mirror 403 error?

[Gabor Grothendieck]

On Fri, May 29, 2015 at 10:12 PM, Mark Drummond  wrote:
> I've been getting a 403 when I try pulling from the Toronto CRAN mirror
> today.
>
> http://cran.utstat.utoronto.ca/
>
> Is there a contact list for mirror managers?
>

See the cran_mirrors.csv file in
  R.home("doc")
of your R distribution.

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Re: [R] Help on R Functionality & Histogram

[Shivi82]

Thanks you Sarah. This was very impressive and really helped me out.




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