[R] Help
I am writing my master thesis in which I compared two cultures . So for my statistics I need to compare Age,Sex,Culture as well as have a look at the tasks scores . Anyone familiar with this ? I’d love to share my script so you guide me where I did wrong . Regards __ 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] Calculating a P for trend
Hi all, I have a question about calculating a P for trend on my data. Let�s give an example that is similar to my own situation first: I have a continuous outcome, namely BMI. I want to investigate the effect of a specific medicine, let�s call it MedA on BMI. MedA is a variable that is categorical, coded as yes/no use of the medication. A also have the duration of use of the MedA, divided in three categories: use of MedA for 1-30 days, use of MedA for 31-60 days and use of MedA for 61-120 days (categories based on literature). I have performed a linear regression analyses and it seems like there is some kind of trend: the longer the use of MedA, the higher the BMI will be (the betas increase with time of use). So an exemplary table: Outcome: BMI Beta MedA use duration Use for 1-30 days 0.060 Use for 31-60 days 0.074 Use for 61-120 da 0.081 So, I have created three variables and I modelled them in Rstudio (on a multiple imputed dataset using MICE): mod1 <- with(imp, lm(BMI ~ MedA_1to30)) pool_ mod1 <- pool(mod1) summary(pool_ mod1, conf.int = TRUE) mod2 <- with(imp, lm(BMI ~ MedA_31to60)) pool_ mod2 <- pool(mod2) summary(pool_ mod2, conf.int = TRUE) mod3 <- with(imp, lm(BMI ~ MedA_61to120)) pool_ mod3 <- pool(mod3) summary(pool_ mod3, conf.int = TRUE) Now that I have done this, I want to calculate a p for trend. I do know what a P for trend measures, but I do not know how to calculate this myself. I read something about the partial.cor.trend.test() function from the trend package, but I do not know what I should fill in. Because I can only fill in an x and y, but I have three time variables. So I do not know how to solve this. Can somebody help me? If more information is necessary, I am happy to give it to you! [[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.
Re: [R] Question Mixed-Design Anova in R
Dear John and Peter,
Thank you both for your answers. I am going to try the solutions you gave me!
Thanks again,
Lisa
From: Fox, John
Sent: 23 November 2018 16:54:49
To: Lisa van der Burgh
Cc: r-help@R-project.org; peter dalgaard
Subject: RE: [R] Question Mixed-Design Anova in R
Dear Lisa,
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of peter
> dalgaard
> Sent: Friday, November 23, 2018 10:16 AM
> To: Lisa van der Burgh <40760...@student.eur.nl>
> Cc: r-help@R-project.org
> Subject: Re: [R] Question Mixed-Design Anova in R
>
> You seem to be bringing in a ton of stuff without looking at features in base
> R...
>
> Check
>
> help(mauchly.test)
> help(anova.mlm)
>
> and examples therein. There are also options in the "car" package.
With respect to the latter, see in particular the O'Brien-Kaiser example in
?Anova.
I hope this helps,
John
-
John Fox
Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: https://socialsciences.mcmaster.ca/jfox/
>
> -pd
>
> > On 23 Nov 2018, at 11:43 , Lisa van der Burgh <40760...@student.eur.nl>
> wrote:
> >
> > Hi Everyone,
> >
> >
> >
> > I have a question about Mixed-Design Anova in R. I want to obtain Mauchly s
> test of Sphericity and the Greenhouse-Geisser correction. I have managed to
> do it in SPSS:
> >
> >
> >
> > GLM Measure1 Measure2 Measure3 Measure4 Measure5 Measure6 BY
> Grouping
> >
> > /WSFACTOR=Measure 6 Polynomial
> >
> > /METHOD=SSTYPE(3)
> >
> > /PLOT=PROFILE(Measure*Grouping)
> >
> > /CRITERIA=ALPHA(.05)
> >
> > /WSDESIGN=Measure
> >
> > /DESIGN=Grouping.
> >
> >
> >
> > I have tried to replicate this in R:
> >
> > library("dplyr")
> >
> > library("tidyr")
> >
> > library("ggplot2")
> >
> > library("ez")
> >
> >
> >
> > PatientID <- c(1:10)
> >
> > Measure1 <- c(3,5,7,4,NA,7,4,4,7,2)
> >
> > Measure2 <- c(1,2,5,6,8,9,5,NA,6,7)
> >
> > Measure3 <- c(3,3,5,7,NA,4,5,7,8,1)
> >
> > Measure4 <- c(1,2,5,NA,3,NA,6,7,3,6)
> >
> > Measure5 <- c(2,3,NA,8,3,5,6,3,6,4)
> >
> > Measure6 <- c(1,2,4,6,8,3,5,6,NA,4)
> >
> > Grouping <- c(1,0,1,1,1,0,0,1,1,0)
> >
> > dataframe <- data.frame(PatientID, Measure1, Measure2, Measure3,
> > Measure4, Measure5, Measure6, Grouping)
> >
> > dataframe$Grouping <- as.factor(dataframe$Grouping)
> >
> > dataframe
> >
> >
> >
> > ezPrecis(dataframe)
> >
> > glimpse(dataframe)
> >
> >
> >
> > dataframe %>% count(PatientID)
> >
> >
> >
> > dataframe %>% count(PatientID, Grouping, Measure1, Measure2,
> Measure3,
> > Measure4, Measure5, Measure6) %>%
> >
> > filter(PatientID %in% c(1:243)) %>%
> >
> > print(n = 10)
> >
> >
> >
> > # So, we have a mixed design with one between factor (Grouping) and 6
> within factors (Measure 1 to 6).
> >
> >
> >
> > dat_means <- dataframe %>%
> >
> > group_by(Grouping, Measure1, Measure2, Measure3, Measure4,
> Measure5,
> > Measure6) %>%
> >
> > summarise(mRT = mean(c(Measure1, Measure2, Measure3, Measure4,
> > Measure5, Measure6))) %>% ungroup()
> >
> > View(dat_means)
> >
> >
> >
> > ggplot(dat_means, aes(c(Measure1, Measure2, Measure3, Measure4,
> > Measure5, Measure6), mRT, colour = Grouping)) +
> >
> > geom_line(aes(group = Grouping)) +
> >
> > geom_point(aes(shape = Grouping), size = 3) +
> >
> > facet_wrap(~group)
> >
> >
> >
> > ANOVA <- ezANOVA(dat, x, PatientID, within = .( c(Measure1, Measure2,
> > Measure3, Measure4, Measure5, Measure6)),
> >
> >between = Grouping, type = 3)
> >
> >
> >
> > print(ANOVA)
> >
> >
> >
> >
> >
> > However, this does not work. I know I am probably doing it completely
> wrong, but I do not know how to solve it. Besides that, I do not know what to
> fill in at the x .
> >
> > Can somebody help me?
> >
> >
> >
> > Thank you in advance.
> >
> > Lisa
> >
> >
> > [[alternative HTML version deleted]]
> >
> > __
[R] Question Mixed-Design Anova in R
Hi Everyone,
I have a question about Mixed-Design Anova in R. I want to obtain Mauchly�s
test of Sphericity and the Greenhouse-Geisser correction. I have managed to do
it in SPSS:
GLM Measure1 Measure2 Measure3 Measure4 Measure5 Measure6 BY Grouping
/WSFACTOR=Measure 6 Polynomial
/METHOD=SSTYPE(3)
/PLOT=PROFILE(Measure*Grouping)
/CRITERIA=ALPHA(.05)
/WSDESIGN=Measure
/DESIGN=Grouping.
I have tried to replicate this in R:
library("dplyr")
library("tidyr")
library("ggplot2")
library("ez")
PatientID <- c(1:10)
Measure1 <- c(3,5,7,4,NA,7,4,4,7,2)
Measure2 <- c(1,2,5,6,8,9,5,NA,6,7)
Measure3 <- c(3,3,5,7,NA,4,5,7,8,1)
Measure4 <- c(1,2,5,NA,3,NA,6,7,3,6)
Measure5 <- c(2,3,NA,8,3,5,6,3,6,4)
Measure6 <- c(1,2,4,6,8,3,5,6,NA,4)
Grouping <- c(1,0,1,1,1,0,0,1,1,0)
dataframe <- data.frame(PatientID, Measure1, Measure2, Measure3, Measure4,
Measure5, Measure6, Grouping)
dataframe$Grouping <- as.factor(dataframe$Grouping)
dataframe
ezPrecis(dataframe)
glimpse(dataframe)
dataframe %>% count(PatientID)
dataframe %>% count(PatientID, Grouping, Measure1, Measure2, Measure3,
Measure4, Measure5, Measure6) %>%
filter(PatientID %in% c(1:243)) %>%
print(n = 10)
# So, we have a mixed design with one between factor (Grouping) and 6 within
factors (Measure 1 to 6).
dat_means <- dataframe %>%
group_by(Grouping, Measure1, Measure2, Measure3, Measure4, Measure5,
Measure6) %>%
summarise(mRT = mean(c(Measure1, Measure2, Measure3, Measure4, Measure5,
Measure6))) %>% ungroup()
View(dat_means)
ggplot(dat_means, aes(c(Measure1, Measure2, Measure3, Measure4, Measure5,
Measure6), mRT, colour = Grouping)) +
geom_line(aes(group = Grouping)) +
geom_point(aes(shape = Grouping), size = 3) +
facet_wrap(~group)
ANOVA <- ezANOVA(dat, x, PatientID, within = .( c(Measure1, Measure2, Measure3,
Measure4, Measure5, Measure6)),
between = Grouping, type = 3)
print(ANOVA)
However, this does not work. I know I am probably doing it completely wrong,
but I do not know how to solve it. Besides that, I do not know what to fill in
at the �x�.
Can somebody help me?
Thank you in advance.
Lisa
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Help plotting recurrence matrix
Hi everyone,
I am trying to plot a recurrence matrix (package: nonlinearTseries).
This is my command, which gives the plot as found in the attached image
(example 1):
Team1 <- read.delim("Team1.txt", header =TRUE)
Team1_plot <- rqa(time.series = Team1$Kat5, embedding.dim = 1, time.lag = 1,
radius = 0.1,lmin =2, vmin =2, distanceToBorder = 2, save.RM
= TRUE,
do.plot = FALSE )
RP <- Team1_plot$recurrence.matrix
image (RP)
My first problem
1. I cannot change the labeling of the axis limits
While this command helps rotate the datapoints WITHIN the plot as desired
(example 2):
image(t(RP[,ncol(RP):1])) # rotate matrix by 90° and plot
the labeling of the axes is still wrong (the y-axis still starts with 223).
Adding parameters such as ylim or ymax has not helped.
2. I am trying to convert it into a colour-coded image. I have
categories ranging from numbers 1 to 5 (as well as random unique larger
numbers referring to nonrecurrent values). With some help, I have managed to
assign my numeric values in my data to colours using this function:
RP2 <- as.matrix(RP)*1
TS <- Team1$Kat5
for (i in seq(1,dim(RP2)[1])) {
for (j in seq(1,dim(RP2)[1])) {
if (RP2[i,j] == 1 & TS[i] < 6) { # test if recurrence is present in RP
AND corresponding value in the time-series is < 6, in which case that
Recurrent is coded according to the value in the time series
RP2[i,j] <- TS[i]
} else if (RP2[i,j] == 1 & TS[i] < 6) { # if the above is not the case,
test if recurrence is present in RP AND corresponding value in the
time-series is > 5, in which case that recurrence is deleted from the RP
RP2[i,j] <- 0
}
}
}
image.scale(RP2, zlim=c(1,5),col= c("yellow", "red", "blue", "green",
"orange")
My problem is that the image I receive from doing this (see attachment) is
scaled from 0 to 1 and I seem to be incapable of understanding why this is
and how to change it. Once again, I have tried using y and x lim as well as
max functions, but have not managed to attain any results.
I realize these are amateur questions but I have been struggling with this
for quite some time and would really appreciate any help I can get.
Thanks and best regards,
Lisa
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Re: [R] Recoding variables in R
Thank you all for your help, it worked! Op 23 mei 2018 om 19:27 heeft marta azores mailto:martazo...@gmail.com>> het volgende geschreven: Try that code NewDF<-DF[!DF$Anxiolytics==1,] 2018-05-23 10:14 GMT+00:00 Lisa van der Burgh mailto:lisavdbu...@hotmail.com>>: Hi all, I have a very general question and I think for you maybe very easy, but I am not able to solve it. I have a dataset and that dataset contains the variable Anxiolytics. This variable is coded as 0, 1, 2. The variable looks as follows: > summary(DF$Anxiolytics) 01 2 NA's 1102 0 20440 You can see that the variable is coded as 0, 1, 2, but group 1 is 'empty'. So I want to remove this group, such that I get: > summary(DF$Anxiolytics) 0 2 NA's 1102 20 440 And then I want to recode the variable in a way that 2 becomes 1, such as: > summary(DF$Anxiolytics) 0 1 NA's 1102 20 440 Can you help me with the code for doing this? Thank you in advance! Best, Lisa [[alternative HTML version deleted]] __ R-help@r-project.org<mailto:R-help@r-project.org> mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Recoding variables in R
Hi all, I have a very general question and I think for you maybe very easy, but I am not able to solve it. I have a dataset and that dataset contains the variable Anxiolytics. This variable is coded as 0, 1, 2. The variable looks as follows: > summary(DF$Anxiolytics) 01 2 NA's 1102 0 20440 You can see that the variable is coded as 0, 1, 2, but group 1 is 'empty'. So I want to remove this group, such that I get: > summary(DF$Anxiolytics) 0 2 NA's 1102 20 440 And then I want to recode the variable in a way that 2 becomes 1, such as: > summary(DF$Anxiolytics) 0 1 NA's 1102 20 440 Can you help me with the code for doing this? Thank you in advance! Best, Lisa [[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.
[R] R-Studio+Commander
> Hello, > > I would like to Know if i can install r on my iPad Air 2? Is this possible? > > Regards Lisa-Marie Kindler Von meinem iPhone gesendet [[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.
Re: [R] error in data.farme--duplicate row.names error
Don, Thank you for your helpful response. At this point, I do believe it is a package error and I have contacted the developer. Thanks, Lisa -Original Message- From: MacQueen, Don [mailto:macque...@llnl.gov] Sent: Thursday, May 19, 2016 4:27 PM To: Rees, Lisa Marie (MU-Student); r-help@r-project.org Subject: Re: [R] error in data.farme--duplicate row.names error You will probably have to contact the maintainer of the package, since the error appears to be generated inside the package's function. Immediately after the error, type traceback() The results might give you a clue. Or they might not! There might be some requirements on the second argument of the DefineGame() function. Check the help page for DefineGame and see if the object you supplied, value, meets those requirements. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 5/19/16, 7:16 AM, "R-help on behalf of Rees, Lisa Marie (MU-Student)" wrote: >I'm using the "GameTheory" package--- DefineGame(14,values) and values >is equal to 16,383 observations. > >I keep getting the following error- >[Error in data.frame(rep(0, 2^n - 1), row.names = Bmat) : > duplicate row.names: 1, 11, 111, 12, 112, 1112, 2, 13, 113, 1113, >3, 1213, 11213, 111213, 213, 14, 114, 1114, 4, 1214, 11214, >111214, 214, 1314, 11314, 111314, 314, 121314, 1121314, >11121314, 21314] > >What can I do to fix this issue? I would greatly appreciate any help. > >Thank you. > > > > [[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. __ 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] error in data.farme--duplicate row.names error
Michael, Thanks for your response. I tried table(table(Bmat)) and it gave me the following error: [ Error in table(Bmat) : object 'Bmat' not found] FYI-- "values" contains 16,383 observations ranging from 0 to less than 1. Lisa -Original Message- From: Michael Dewey [mailto:li...@dewey.myzen.co.uk] Sent: Thursday, May 19, 2016 11:05 AM To: Rees, Lisa Marie (MU-Student); r-help@r-project.org Subject: Re: [R] error in data.farme--duplicate row.names error Dear Lisa What does Bmat contain? Perhaps try table(table(Bmat)) and see if any entries are greater than unity. On 19/05/2016 15:16, Rees, Lisa Marie (MU-Student) wrote: > I'm using the "GameTheory" package--- DefineGame(14,values) and values is > equal to 16,383 observations. > > I keep getting the following error- > [Error in data.frame(rep(0, 2^n - 1), row.names = Bmat) : > duplicate row.names: 1, 11, 111, 12, 112, 1112, 2, 13, 113, > 1113, 3, 1213, 11213, 111213, 213, 14, 114, 1114, 4, 1214, > 11214, 111214, 214, 1314, 11314, 111314, 314, 121314, 1121314, > 11121314, 21314] > > What can I do to fix this issue? I would greatly appreciate any help. > > Thank you. > > > > [[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. > -- Michael http://www.dewey.myzen.co.uk/home.html __ 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] error in data.farme--duplicate row.names error
I'm using the "GameTheory" package--- DefineGame(14,values) and values is equal to 16,383 observations. I keep getting the following error- [Error in data.frame(rep(0, 2^n - 1), row.names = Bmat) : duplicate row.names: 1, 11, 111, 12, 112, 1112, 2, 13, 113, 1113, 3, 1213, 11213, 111213, 213, 14, 114, 1114, 4, 1214, 11214, 111214, 214, 1314, 11314, 111314, 314, 121314, 1121314, 11121314, 21314] What can I do to fix this issue? I would greatly appreciate any help. Thank you. [[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.
[R] Help with RUSBoost
I am considering using RUSBoost (https://github.com/SteveOhh/RUSBoost) and was wondering if anyone has used this package, and could give me some insight and help. The help would be on more of the machine learning side, I just have a few questions about implementation. [[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.
[R] Tomorrow Webinar: Enter a KDD Cup or Kaggle Competition without being an expert!
Webinar: Enter a KDD Cup or Kaggle Competition, You don't need to be an expert! Tomorrow, January 13, 2015 from 10am-11am PDT * If inconvenient time, please register and we will send you a recording. Click to Register: http://hubs.ly/y0q_jf0 Abstract: * Quickly achieve a place in the top 5: We will show you how Salford's TreeNet gradient boosting can be used for the 2009 KDD Cup competition to quickly achieve a place in the top 5. * Takeaway: At the end of this webinar, our goal is that you will be able to build a TreeNet model that can bring you within decimal places of a winning solution. * Starting Point for Kaggle, KDD and other data science competitions: Use this information as a starting point for Kaggle competitions and other KDD Cup competitions. * 30-day software access: All attendees receive 30-day access to TreeNet gradient boosting, and other Salford Predictive Modeler technology. Click to Register: http://hubs.ly/y0q_jf0 [[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.
[R] Help with Caret
*My code is below:*
library('RMySQL')
library('DMwR')
library('tm')
library('Snowball')
library('SnowballC')
con <- dbConnect(MySQL(), user="root", password="stuff0645",
dbname="TwitterCelebs", host="localhost")
rt_outlier <- dbGetQuery(con, "SELECT *,tweet_text from
outlier_info,tweets where outlier_info.tweet_id=tweets.tweet_id limit
500")
rt_not_outlier <- dbGetQuery(con,"Select *, tweet_text from
not_outlier_info,tweets where
not_outlier_info.tweet_id=tweets.tweet_id limit 500");
dbDisconnect(con)
all_tweets = rbind(rt_outlier,rt_not_outlier)
all_tweets[,"tweet_text"] <- iconv(all_tweets[,"tweet_text"], to = "utf-8")
corpus = Corpus(VectorSource(all_tweets[,"tweet_text"]))
corpus = tm_map(corpus,removePunctuation)
corpus <- tm_map(corpus, tolower)
corpus <- tm_map(corpus, stemDocument)
corpus <- tm_map(corpus,removeWords,stopwords("english"))
corpus <- tm_map(corpus,stripWhitespace)
corpus <- tm_map(corpus,removeNumbers)
mydata.dtm <- TermDocumentMatrix(corpus,control=list(weighting=weightTfIdf,
minWordLength=2, findFreqTerms=5))
dataframe <- as.data.frame(inspect(mydata.dtm))
d=as.data.frame(t(dataframe))
classData = c(rep(0,500),rep(1,500))
classData = as.factor(classData)
library('caret')
ctrl = trainControl(method = "repeatedcv", repeats = 3,)
set.seed(2)
mymodel <- train(d, classData,trControl=ctrl,method="J48",model=FALSE)
Basically, what is happening here is that I keep getting the error and
warnings:
Error in train.default(d, classData, method = "J48", model = FALSE) :
final tuning parameters could not be determined
In addition: Warning messages:1: In train.default(d, classData, method
= "J48", model = FALSE) :
Models using Weka will not work with parallel processing with
multicore/doMC2: In nominalTrainWorkflow(x = x, y = y, wts = weights,
info = trainInfo, :
There were missing values in resampled performance measures.3: In
train.default(d, classData, method = "J48", model = FALSE) :
missing values found in aggregated results
What am I doing wrong? Also note I'm using model=FALSE in training to
conserve memory, as this has been a problem
--
Lisa Gandy, PhD
Assistant Professor
Central Michigan University
Computer Science Dept
Pearce 119
989-774-3696
gand...@cmich.edu
https://intlab.cps.cmich.edu/lgandy/
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Chicago, Data Mining Training (Hands-On, $35), Chicago, May 23rd
Join us for a hands-on data mining training in Chicago, IL on May 23, 2014: * Cost: $35 * Registration: http://hub.am/1ftOoR0 * Learn data mining quickly using Salford Systems' Commercial Data Mining Software (No programming skills needed). Agenda: 8:30am-9:00am Breakfast (provided) * 9:00am-12:00pm * Introduction to Data Mining * Case Study Examples 12:00pm-1:00pm Lunch (provided) * 1:00pm-3:00pm * Build and Score Predictive Models * Optimization for Predictive Accuracy * Create Reports: Translating insights into actionable results Why you should attend: * Get step-by-step instruction for the most popular data mining techniques used in predictive analytics including decision trees, classification, segmentation, non-linear regression, ensemble methods, boosted decision trees, etc. * Walk away with everything you will need to start your own data mining projects. * Be ready to apply your new data mining knowledge at your organization to create immediate value. * All attendees receive 90-day access to the SPM Salford Predictive Modeler technology. We hope to see you in Chicago! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] match from a data.frame in dependence of an ID
Try xtabs() >> df2 = data.frame(ID = c(10,10,10,10,10,11,11,12),Group = c(1,2,3,4,5,3,4,4),Value = c(10,20,30,40,50,60,70,80)) >> xtabs(df2$Value~df2$ID + df2$Group) I think this is exactly what you want. On Fri, Mar 28, 2014 at 4:51 PM, Mat wrote: > Hello togehter, > > i have a litte problem. I have an output data.frame which look like this > one: > ID > 1 10 > 2 11 > 3 12 > > Now I have another data.frame with more than one line for each ID: > IDGroupValue > 1 10110 > 2 10220 > 3 10330 > 4 10440 > 5 10550 > 6 11360 > 7 11470 > 8 12480 > > How can i match these two data.frame for the following result: > > ID1 2 3 4 5 > 1 1010 2030 40 50 > 2 11 60 70 > 3 12 80 > > > > -- > View this message in context: > http://r.789695.n4.nabble.com/match-from-a-data-frame-in-dependence-of-an-ID-tp4687745.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Merge two vectors into one
He wants a[1] b[1] a[2] b[2] a[3] b[3] I think you can do: x = as.vector(rbind(a, b)) On Mon, Mar 24, 2014 at 2:39 PM, Frans Marcelissen < fransiepansiekever...@gmail.com> wrote: > Why not simply > > > a<-1:3 > > b<-4:5 > > c(a,b) > [1] 1 2 3 4 5 > > > 2014-03-22 23:22 GMT+01:00 Tham Tran : > > > Dear R users, > > > > Given two vectors x and y > > a=1 2 3 > > b=4 5 6 > > > > i want to combine them into a single vector z as 1 4 2 5 3 6 > > > > Thanks for your help > > > > Tham > > > > > > > > -- > > View this message in context: > > http://r.789695.n4.nabble.com/Merge-two-vectors-into-one-tp4687361.html > > Sent from the R help mailing list archive at Nabble.com. > > > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Import multiple files into R
But I don't think a list can contain that much data - we dont know how big
each txt file is.
So I'd like to do:
city.set <- c('Asheville','Charlotte')
month.set <- c('Dec', 'Jan', 'Feb')
comb.set <- expand.grid(city.set,month.set)
comb.set$filename = paste(comb.set[,1],comb.set[,2],sep = '_')
for i in 1:nrow(comb.set){
db <- read.table(comb.set$filename[i] + '.txt',header = TRUE)
save(db,filename + '.dat')
rm(db)
}
So that each file is saved in a .dat file as a data frame, and she only
need to load and remove whenever she uses the file.
On Thu, Mar 20, 2014 at 5:14 PM, PIKAL Petr wrote:
> Hi
>
> Best way is to put those files in one directory and start R from this
> directory or set this directory as working one.
>
> ?set.wd
> Then you can use list.files() to get file names.
> ?list.files
>
> myfiles<-list.files()
>
> Than strip txt. E.g. by
>
> filenames<-strsplit(list.files(), "\\."))
>
> After that you can use for cycle to read all files, however I would prefer
> list structure.
>
> files<-vector(mode="list", length(myfiles))
>
> for (i in 1:length(myfiles)) {
> files[[i]]<-read.table(myfiles[i])
> names(files)[[i]]<-filenames[[i]][1]
> }
>
> Instead of 75 objects you will get one list object and you can easily use
> interactively or programmatically its parts.
>
> Regards
> Petr
>
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.org] On Behalf Of ajaykumar
> > Sent: Thursday, March 20, 2014 6:15 AM
> > To: r-help@r-project.org
> > Subject: [R] Import multiple files into R
> >
> > Hi I want to import around 75 files. Each file has a name and a time
> > and is comma separated.
> > For example some of my file names are
> > Asheville_Dec.txt
> > Asheville_Jan.txt
> > Asheville_Feb.txt
> > Charlotte_Dec.txt
> > Chapelhill_Jan.txt
> >
> > The time months are only Dec Jan and Feb. The locations are different.
> > I have data for these 3 months at 25 cities which I want to import in R
> > I don't want to use read.table 75 times and keep changing names.
> >
> > Is there any way I can keep all the city names in an object and the
> > three months in another object and then use a loop over read.table of
> > location_time.txt I would also like to view them as 75 different data
> > sets in R So the file Asheville_Dec must be in a dataset called
> > Asheville_Dec in R
> >
> > Can this be done.
> > Please help
> >
> >
> >
> > --
> > View this message in context: http://r.789695.n4.nabble.com/Import-
> > multiple-files-into-R-tp4687178.html
> > Sent from the R help mailing list archive at Nabble.com.
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-
> > guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
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Re: [R] computation
In this calculation: (dg0%*%t(dx0)) is [,1] [,2] [1,]0 -0.75 [2,]0 0.50 So: exp2 = NaN It is extremely easy to find out the issue if you go back line by line, this is a basic procedure of debugging. Please do it yourself next time. On Mon, Jan 13, 2014 at 3:09 PM, IZHAK shabsogh wrote: > kindly help me fine what is the mistake with following: > my aim is to compute those steps and obtain a vector with values (3,5) > but i am geting (NAN,NAN) > > Q<-matrix(c(5,-3,-3,2),2,2) > b<-rbind(0,1) > H0<-diag(2) > x0<-rbind(0,0) > d0<-b > g0<--b > a0<--(t(g0)%*%d0)/(t(d0)%*%Q%*%d0) > x1<-x0+a0[,1]*d0 > dx0<-x1-x0 > g1<-Q%*%x1-b > dg0<-g1-g0 > exp1<-(t(dg0)%*%H0%*%dg0)/(t(dg0)%*%dx0) > exp2<-(dx0%*%t(dx0))/(dg0%*%t(dx0)) > exp3<-(dx0%*%t(dg0)%*%H0 +H0%*%dg0%*%t(dx0))/(dx0%*%t(dg0)) > H1<-H0+(1+exp1[,1])*exp2-exp3 > d1<--H1%*%g1 > a2<-(t(g1)%*%d1)/(t(d1)%*%Q%*%d1) > x2<-x1+a2[,1]*d1 > > [[alternative HTML version deleted]] > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] X11 font -adobe-helvetica size 15 could not be loaded
HI All,
I got this Error message, when I tried to plot
X11 font -adobe-helvetica-%s-%s-*-*-%d-*-*-*-*-*-*-*, face 5 at size 15 could
not be loaded
sessionInfo()
R version 3.0.0 (2013-04-03)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_CA.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_CA.UTF-8 LC_COLLATE=en_CA.UTF-8
[5] LC_MONETARY=en_CA.UTF-8 LC_MESSAGES=en_CA.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] stringr_0.6.2 reshape2_1.2.2
loaded via a namespace (and not attached):
[1] plyr_1.8 tools_3.0.0
I looked online for a solution.and found this
(https://stat.ethz.ch/pipermail/r-help/2002-June/022116.html). But,
/etc/X11 $ ls
app-defaults rgb.txt xkb Xresources Xsession.options
default-display-manager X Xreset Xsession XvMCConfig
fonts xinit Xreset.d Xsession.d Xwrapper.config
doesn't have XF86Config
capabilities("X11")
X11
TRUE
names(X11Fonts())
[1] "serif" "sans" "mono" "Times" "Helvetica"
[6] "CyrTimes" "CyrHelvetica" "Arial" "Mincho"
Then, I installed xfonts-100dpi, xfonts-75dpi (Ubuntu-based).
xset -q #output
Font Path:
/usr/share/fonts/X11/misc,/usr/share/fonts/X11/Type1,built-ins
Regards,
Lisa.
__
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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] Hands-on Webinar: Advances in Regression: Modern Ensemble and Data Mining Approaches (no charge)
Hands-on Webinar (no charge) Advances in Regression: Modern Ensemble and Data Mining Approaches **Part of the series: The Evolution of Regression from Classical Linear Regression to Modern Ensembles Register Now for Parts 3, 4: https://www1.gotomeeting.com/register/500959705 **All registrants will automatically receive access to recordings of Parts 1 & 2. Course Abstract: Overcoming Linear Regression Limitations Regression is one of the most popular modeling methods, but the classical approach has significant problems. This webinar series addresses these problems. Are you working with larger datasets? Is your data challenging? Does your data include missing values, nonlinear relationships, local patterns and interactions? This webinar series is for you! In our March 29th session (Part 3), we will focus on modern ensemble and data mining approaches. These methods dramatically improve the performance of weak learners such as regression trees. The techniques discussed here enhance the performance of regression trees considerably. These methods inherit the good features of trees (variable selection, missing data, mixed predictors) and improve on the weak features such as prediction performance. Did you miss parts 1 and 2? With your registration, you will receive links to the recordings of Part 1 and 2. Covered in part 1 and 2 are improvements to conventional and logistic regression, as well as a discussion of classical, regularized, and nonlinear regression from both a theoretical and hands-on point of view. The hands-on component includes a step-by-step demonstration with instructions for reproducing the demo at your leisure. Especially for the dedicated student: after watching this recording, you will be able to apply these methods to your own data. Part 3: March 29, 10-11am PST - Regression methods discussed: * Nonlinear Ensemble Approaches: o TreeNet Gradient Boosting o Random Forests o Gradient Boosting incorporating Random Forests * Ensemble Post-Processing: o ISLE Importance Sampled Learning Ensembles o RuleLearner rule based learning ensembles Part 4: April 12, 10-11am PST - Hands-on demonstration of concepts discussed in Part 3 * Step-by-step demonstration * Datasets and software available for download * Instructions for reproducing demo at your leisure * For the dedicated student: apply these methods to your own data (optional) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] try/tryCatch
here is the error:
> aa<-metatrialstry(beta_5_50)
Error in asMethod(object) : matrix is not symmetric [1,2]
metatrials, the function that i am attempting to convert with try/tryCatch,
gives me back a matrix with as many rows are there are simulations (z) in
the aray with dim(x,y,z). with the data i attached, x is 500(number of
patients), y is 9 (these are covariates), and z is 500.
metatrials<-function(mydata){
a<-matrix(data=NA, nrow=dim(mydata)[3], ncol=5)
colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se")
for(ii in 1:dim(mydata)[3]){
tmp<-mydata[,,ii]
tmp1<-as.data.frame(tmp)
names(tmp1)=c("persons", "d1", "tp", "fn", "fp", "fn", "detect", "d0",
"outcome")
lm1<-lmer(outcome~0+d1+d0+(0+d1+d0 | persons), family=binomial,
data=tmp1, nAGQ=3)
a[ii,1]=lm1@fixef[1]
a[ii,2]=lm1@fixef[2]
a[ii,3]=vcov(lm1)[1,2]/prod(sqrt(diag(vcov(lm1
a[ii,4:5]=sqrt(diag(vcov(lm1)))
}
return(a)
}
##
what i want is for the function to go on to the next data set in the array
and simply return an NA for that line in the metatrials results. so
basically, just keep going.
thanks so much for your help!
-Lisa
On Mon, Mar 18, 2013 at 4:24 PM, jim holtman wrote:
> It would help if you told us what type of error you are getting and to
> also provide sample data so that we could run it to see what happens. I
> use 'try' a lot to catch errors and have not had any problems with it.
>
> On Mon, Mar 18, 2013 at 6:11 AM, lisa wrote:
>
>> Hi All,
>>
>> I have tried every fix on my try or tryCatch that I have found on the
>> internet, but so far have not been able to get my R code to continue with
>> the "for loop" after the lmer model results in an error.
>>
>> Here is two attemps of my code, the input is a 3D array file, but really
>> any function would do
>>
>> metatrialstry<-function(mydata){
>>
>> a<-matrix(data=NA, nrow=dim(mydata)[3], ncol=5)
>> #colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se",
>> "counter")#colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se")
>> k=1
>> for(ii in 1:dim(mydata)[3]){
>> tmp<-mydata[,,ii]
>> tmp1<-as.data.frame(tmp)
>> names(tmp1)=c("persons", "d1", "tp", "fn", "fp", "fn", "detect", "d0",
>> "outcome")
>> lm1<-try(lmer(outcome~0+d1+d0+(0+d1+d0 | persons), family=binomial,
>> data=tmp1, nAGQ=3), silent=T)
>> if(class(lm1)[1]!='try-error'){
>> a[ii,1]=lm1@fixef[1]
>> a[ii,2]=lm1@fixef[2]
>> a[ii,3]=vcov(lm1)[1,2]/prod(sqrt(diag(vcov(lm1
>> a[ii,4:5]=sqrt(diag(vcov(lm1)))
>> }
>> }
>> #k=k+1
>> #a[ii,6]=k
>>
>> return(a)
>> }
>>
>> #
>> # try / try catch ###
>> #
>>
>>
>>
>> metatrialstry<-function(mydata){
>>
>> a<-matrix(data=NA, nrow=dim(mydata)[3], ncol=5)
>> #colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se", "counter")
>> colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se")
>> #a[,6]=rep(0, length(a[,6]))
>> for(ii in 1:dim(mydata)[3]){
>> tmp<-mydata[,,ii]
>> tmp1<-as.data.frame(tmp)
>> names(tmp1)=c("persons", "d1", "tp", "fn", "fp", "fn", "detect", "d0",
>> "outcome")
>> lm1<-tryCatch(lmer(outcome~0+d1+d0+(0+d1+d0 | persons),
>> family=binomial, data=tmp1, nAGQ=3), error=function(e) e)
>> a[ii,1]=lm1@fixef[1]
>> a[ii,2]=lm1@fixef[2]
>> a[ii,3]=vcov(lm1)[1,2]/prod(sqrt(diag(vcov(lm1
>> a[ii,4:5]=sqrt(diag(vcov(lm1)))
>> }
>> return(a)
>> }
>>
>>
>> Any guidance would be greatly appreciated...
>>
>> thanks!
>> Lisa
>>
>> --
>> Lisa Wang
>> email: wang.li...@gmail.com
>> cell: +49 -0176-87786557
>> Tübingen, Germany, 72070
>>
>> [[alternative HTML version deleted]]
>>
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
>
> --
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.
--
Lisa Wang
email: wang.li...@gmail.com
cell: +49 -0176-87786557
Tübingen, Germany, 72070
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] try/tryCatch
Hi All,
I have tried every fix on my try or tryCatch that I have found on the
internet, but so far have not been able to get my R code to continue with
the "for loop" after the lmer model results in an error.
Here is two attemps of my code, the input is a 3D array file, but really
any function would do
metatrialstry<-function(mydata){
a<-matrix(data=NA, nrow=dim(mydata)[3], ncol=5)
#colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se",
"counter")#colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se")
k=1
for(ii in 1:dim(mydata)[3]){
tmp<-mydata[,,ii]
tmp1<-as.data.frame(tmp)
names(tmp1)=c("persons", "d1", "tp", "fn", "fp", "fn", "detect", "d0",
"outcome")
lm1<-try(lmer(outcome~0+d1+d0+(0+d1+d0 | persons), family=binomial,
data=tmp1, nAGQ=3), silent=T)
if(class(lm1)[1]!='try-error'){
a[ii,1]=lm1@fixef[1]
a[ii,2]=lm1@fixef[2]
a[ii,3]=vcov(lm1)[1,2]/prod(sqrt(diag(vcov(lm1
a[ii,4:5]=sqrt(diag(vcov(lm1)))
}
}
#k=k+1
#a[ii,6]=k
return(a)
}
#
# try / try catch ###
#
metatrialstry<-function(mydata){
a<-matrix(data=NA, nrow=dim(mydata)[3], ncol=5)
#colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se", "counter")
colnames(a)=c("sens", "spec", "corr", "sens_se", "spec_se")
#a[,6]=rep(0, length(a[,6]))
for(ii in 1:dim(mydata)[3]){
tmp<-mydata[,,ii]
tmp1<-as.data.frame(tmp)
names(tmp1)=c("persons", "d1", "tp", "fn", "fp", "fn", "detect", "d0",
"outcome")
lm1<-tryCatch(lmer(outcome~0+d1+d0+(0+d1+d0 | persons),
family=binomial, data=tmp1, nAGQ=3), error=function(e) e)
a[ii,1]=lm1@fixef[1]
a[ii,2]=lm1@fixef[2]
a[ii,3]=vcov(lm1)[1,2]/prod(sqrt(diag(vcov(lm1
a[ii,4:5]=sqrt(diag(vcov(lm1)))
}
return(a)
}
Any guidance would be greatly appreciated...
thanks!
Lisa
--
Lisa Wang
email: wang.li...@gmail.com
cell: +49 -0176-87786557
Tübingen, Germany, 72070
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Tomorrow: The Evolution of Regression from Classical Linear Regression to Modern Ensembles (hands-on)
Tomorrow, Friday March 15 Maybe you missed Part 1 of "The Evolution of Regression Modeling from Classical Linear Regression to Modern Ensembles " webinar series, but you can still join for Parts 2, 3, & 4 > Register Now for Parts 2, 3, 4: > https://www1.gotomeeting.com/register/500959705 > > Course Outline: Overcoming Linear Regression Limitations > > Regression is one of the most popular modeling methods, but the classical > approach has significant problems. This webinar series addresses these > problems. Are you working with larger datasets? Is your data challenging? > Does your data include missing values, nonlinear relationships, local > patterns and interactions? This webinar series is for you! We will cover > improvements to conventional and logistic regression, and will include a > discussion of classical, regularized, and nonlinear regression, as well as > modern ensemble and data mining approaches. This series will be of value to > any classically trained statistician or modeler. > > Part 2 (Hands-on): March 15, 10-11am PST - Hands-on demonstration of concepts > discussed in Part 1 (Classical Regression, Logistic Regression, Regularized > Regression: GPS Generalized Path Seeker, Nonlinear Regression: MARS > Regression Splines) > > Step-by-step demonstration > Datasets and software available for download > Instructions for reproducing demo at your leisure > For the dedicated student: apply these methods to your own data (optional) > · Part 1 recording: > http://www.salford-systems.com/videos/tutorials/805-the-evolution-of-regression-modeling-part-1 > > Part 3: March 29, 10-11am PST - Regression methods discussed > *Part 1 is a recommended pre-requisite > > Nonlinear Ensemble Approaches: TreeNet Gradient Boosting; Random Forests; > Gradient Boosting incorporating RF > Ensemble Post-Processing: ISLE; RuleLearner > Part 4: April 12, 10-11am PST - Hands-on demonstration of concepts discussed > in Part 3 > > Step-by-step demonstration > Datasets and software available for download > Instructions for reproducing demo at your leisure > For the dedicated student: apply these methods to your own data (optional) > > > > > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Hands-on Webinar Series (no charge) The Evolution of Regression from Classical Linear Regression to Modern Ensembles
Maybe you missed Part 1 of "The Evolution of Regression Modeling from Classical Linear Regression to Modern Ensembles " webinar series, but you can still join for Parts 2, 3, & 4 Register Now for Parts 2, 3, 4: https://www1.gotomeeting.com/register/500959705 Download (optional) a free evaluation of the SPM software suite v7.0 (used in the hands-on components of the webinar). As a webinar participant you will qualify for a 60-Day Evaluation of the software at no charge: http://2.salford-systems.com/the-salford-predictive-modeler-download/ Course Outline: Overcoming Linear Regression Limitations Regression is one of the most popular modeling methods, but the classical approach has significant problems. This webinar series addresses these problems. Are you working with larger datasets? Is your data challenging? Does your data include missing values, nonlinear relationships, local patterns and interactions? This webinar series is for you! We will cover improvements to conventional and logistic regression, and will include a discussion of classical, regularized, and nonlinear regression, as well as modern ensemble and data mining approaches. This series will be of value to any classically trained statistician or modeler. Part 2 (Hands-on): March 15, 10-11am PST - Hands-on demonstration of concepts discussed in Part 1 (Classical Regression, Logistic Regression, Regularized Regression: GPS Generalized Path Seeker, Nonlinear Regression: MARS Regression Splines) * Step-by-step demonstration * Datasets and software available for download * Instructions for reproducing demo at your leisure * For the dedicated student: apply these methods to your own data (optional) * Part 1 recording: http://www.salford-systems.com/videos/tutorials/805-the-evolution-of-regression-modeling-part-1 Part 3: March 29, 10-11am PST - Regression methods discussed *Part 1 is a recommended pre-requisite * Nonlinear Ensemble Approaches: TreeNet Gradient Boosting; Random Forests; Gradient Boosting incorporating RF * Ensemble Post-Processing: ISLE; RuleLearner Part 4: April 12, 10-11am PST - Hands-on demonstration of concepts discussed in Part 3 * Step-by-step demonstration * Datasets and software available for download * Instructions for reproducing demo at your leisure * For the dedicated student: apply these methods to your own data (optional) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Results from clogit out of range?
Thanks for the illustrative example. In my project actually my supervisor
wanted to estimate the probabilities using a "conditional MLE" approach,
which happens to be the case that *uses clogit() while trying to achieve
aim b in your words*.
I learned that clogit() is based on the sufficient statistic which is
usually the sum over all positive responses in each stratum. However since
we are supposed to not knowing the responses when trying to predict a new
sample, it can be impossible to do this "out-of-sample" prediction right?
Now what I suggested is to use clogit() to estimate beta (ppl say this beta
is better than betas from unconditional MLE, why??) and derive the linear
predictor for any new sample by multiplying new predictors with this beta,
then
1) if there is only one obs each strata, use the traditional unconditional
formula
phat = exp(xbeta)/(1+exp(xbeta))
to get the so-called predicted probability;
2) if there is a lot of obs each strata, use
phat = exp(xbeta)/sum(exp(xbeta))
to get the so-called predicted probabilities.
My case has only 1 obs per stratum so I used method 1. Though I am not sure
if it is reasonable. Would like to hear opinions from all of you guy. ;P
On Mon, Mar 4, 2013 at 10:04 PM, Terry Therneau wrote:
> I'm late to this discussion, but let me try to put it in another context.
> Assume that I wanted to know whether kids who live west of their school
> or east of their shool are more likely to be early (some hypothesis about
> walking slower if the sun is in their eyes). So I create a 0/1 variable
> east/west and get samples of 10 student arrival times at each of 100
> different schools. Fit the model
>
>lm(arrive ~ factor(school) + east.west)
>
> where "arrive" is in some common scale like "minutes since midnight".
> Since different schools could have different starting times for their
> first class we need an intercept per school.
>
> Two questions:
> 1. Incremental effect: the coefficient of east/west measures the
> incredmental effect across all schools. With n of 1000 it is likely
> estimated with high precision.
> 2. Absolute: predict the average arrival time (on the clock) for
> students.
>
> Conditional logistic is very like this. We have a large number of strata
> ("schools") with a small number of observations in each (often only 2 per
> strata). One can ask incremental questions about variables common to all
> strata, but absolute prediction is pretty worthless. a. You can only do it
> for schools (strata) that have already been seen and b. there are so few
> subjects in each of them that the estimates are very noisy.
> The default prediction from clogit is focused on questions of type 1.
> The documentation doesn't even bother to mention predictions of type 2,
> which would be probabilities of events. I can think of a way to extract
> such output from the routine (being the author gives some insight), but why
> would I want to?
>
> Terry Therneau
>
>
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Results from clogit out of range?
Sorry I was asking a question inspired by the original question, which relates to the out of sample prediction. He was asking why it's not of the probability range, and I also found that clogit only gives out linear predictors in both estimation and prediction, but never the probability. For in sample ones, we can derive it as the last thread says; that's what inspired me of thinking of out of sample cases. And I am guessing but not sure if the probability of that case can also be deduced. I'd like to hear of your opinions. Thanks anyway for your response and sorry for the misunderstandings. Thanks, Lisa On 1 Mar, 2013, at 10:10 PM, David Winsemius wrote: >>>> I still don't think the exp(lp)/(1+exp(lp)) gonna work. Since this is >>>> conditional logit model, while this formula is only used in unconditional >>>> ones. By using this, one neglects the information based on stratum. Though >>>> I don't know how to solve it to. I am also working on a project on this >>>> and I do hope there's someone explaining this problem. Will that be a >>>> possibility that the phat can never be estimated as we never know the >>>> individual intercept? > >>> This appears to be addressed to a thread that appeared almost three years >>> ago. I suspect you have not read all the way to the end of the thread: >>> >>> https://stat.ethz.ch/pipermail/r-help/2010-April/235956.html > On Feb 28, 2013, at 5:45 PM, lisa wrote: > >> I do appreciate this answer. I heard that in SAS, conditional logistic model >> do predictions in the same way. However, this formula can only deal with >> in-sample predictions. How about the out-of-sample one? Is it like one of >> the former responses by Thomas, say, it's impossible to do the out-of-sample >> prediction?? > > I do not understand how an "out-of-sample" conditional estimate makes any > logical sense. The questioner was asking why he was getting values outside > the range of [0,1] which is not the same as asking for estimates for strata > outside the range of the stratum values. Charles Berry gave another sensible > answer, but I did not interpret his suggestion as solving what you seem to be > requesting. > > -- > > David Winsemius > Alameda, CA, USA > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Results from clogit out of range?
I do appreciate this answer. I heard that in SAS, conditional logistic model do predictions in the same way. However, this formula can only deal with in-sample predictions. How about the out-of-sample one? Is it like one of the former responses by Thomas, say, it's impossible to do the out-of-sample prediction?? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] xyplot and barplot in the same page
Dear Rhelp,
I am trying to get multiple xyplots and barplots in the same page. I couldn't
get it work.
Example code:
library(lattice)
library(grid)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
scales=list(x=list(at=seq(1,8,1),labels=c('A','','C','','E','','G',''))),
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = c(simpleTheme(lty=1:8,
col="black",lwd=2),list(axis.components=list(bottom=list(tck=c(0,1))),layout.heights
= list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0,
axis.bottom=0.8)) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.50, 0.33, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.50, 0.66, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.50, 0.99, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.45), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.45), more = TRUE)
j=j+1
counts <- table(mtcars$gear)
barplot(counts, main="Car Distribution", #this part is not correct
xlab="Number of Gears")
Thanks in advance,
Lisa
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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] Alternate tick labels in xyplot
Dear David and Duncan,
Thank you for the responses.
David:
I was able to fix it with grid.text().
Duncan:
I would certainly look into ?scale.components in latticeExtra package
(xscale.components.default)
The code for the figure is below.
library(lattice)
library(grid)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
scales=list(x=list(at=seq(1,8,1),labels=c('A','','C','','E','','G',''))),
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = c(simpleTheme(lty=1:8,
col="black",lwd=2),list(axis.components=list(bottom=list(tck=c(0,1))),layout.heights
= list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0,
axis.bottom=0.8)) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.50, 0.33, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.50, 0.66, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.50, 0.99, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.45), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.45), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.45), more = FALSE)
grid.text("Upper row", x=0.5, y=0.97, gp=gpar(fontsize=12, col="black"))
grid.text("Lower row", x=0.5, y=0.47, gp=gpar(fontsize=12, col="black"))
Thanks,
Lisa
- Original Message -
From: David Winsemius
To: Lisa Daniel
Cc: "r-help@r-project.org"
Sent: Thursday, February 14, 2013 4:31 PM
Subject: Re: [R] Alternate tick labels in xyplot
On Feb 14, 2013, at 10:52 AM, Lisa Daniel wrote:
> Dear Rhelp,
> I was able to solve the tick labels part. If somebody can help me in
> creating title for each row of figures above the individual headings, it will
> be great.
I think you need to be looking at grid.text() in package grid.
> library(grid)
> library(lattice)
> xy <-xyplot(decrease ~ treatment, OrchardSprays,
> main= "Some plot",
> groups = rowpos, type = "a",
> scales=list(x=list(at=seq(1,8,1),labels=c('A','','C','','E','','G',''))),
> page = function(n){
> grid.text(LETTERS[j],
> y = 0.95,
> x = 0.15,
> default.units = "npc",
> just = c("left"),
> gp = gpar(fontsize = 12,fontface="bold") )
> },
> par.settings = c(simpleTheme(lty=1:8,
>col="black",lwd=2),list(axis.components=list(bottom=list(tck=c(0,1))),layout.heights
> = list(main = 1.2,
> sub = 0,
> axis.top = 0.2,
> top.padding = 0.1,
> bottom.padding = 0)) ),
> auto.key = list(x = 0.2, y = 0.9,
> cex = 0.75, points = FALSE, lines = TRUE))
> j=1
> print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
> j=j+1
> print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
> j=j+1
> print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
> j=j+1
> print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
> j=j+1
> print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
> j=j+1
> print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
>
> Many thanks,
> Lisa
>
>
>
>
> - Original Message -
> From: Lisa Daniel
> To: "r-help@r-project.org"
> Cc:
> Sent: Thursday, February 14, 2013 12:14 PM
> Subject: Alternate tick labels in xyplot
>
> Dear Rhelp,
>
> I would like to get alternate tick labels for the xyplot:
> library(lattice)
> library(grid)
>
> xy <-xyplot(decrease ~ treatment, OrchardSprays,
> main= "Some plot",
> groups = rowpos, type = "a",
> page = function(n){
> grid.text(LETTERS[j],
> y = 0.95,
> x = 0.15,
> default.units = "npc",
> just = c("left"),
> gp = gpar(fontsize =
Re: [R] Alternate tick labels in xyplot
Dear Rhelp,
I was able to solve the tick labels part. If somebody can help me in creating
title for each row of figures above the individual headings, it will be great.
library(grid)
library(lattice)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
scales=list(x=list(at=seq(1,8,1),labels=c('A','','C','','E','','G',''))),
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = c(simpleTheme(lty=1:8,
col="black",lwd=2),list(axis.components=list(bottom=list(tck=c(0,1))),layout.heights
= list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0)) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Many thanks,
Lisa
- Original Message -
From: Lisa Daniel
To: "r-help@r-project.org"
Cc:
Sent: Thursday, February 14, 2013 12:14 PM
Subject: Alternate tick labels in xyplot
Dear Rhelp,
I would like to get alternate tick labels for the xyplot:
library(lattice)
library(grid)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = c(simpleTheme(lty=1:8,
col="black",lwd=2),list(layout.heights = list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0)) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
similar to the x-axis in the code below, but somehow it is not working when I
pasted 'list(axis.components=...) in the above code:
xyplot(rnorm(12) ~ 1:12 , type="l",
scales=list(x=list(at=seq(2,12,2),labels=c(1, ' ', 3 , ' ' , 5 , ' ' ))),
par.settings=list(axis.components=list(bottom=list(tck=c(0,1)
scales=list(x=list(at=seq(1,8,1),labels=c('A', ' ', 'C', ' ' ,'E' , ' ','G','
'))),
If possible, I would like a heading for each rows (justified to center) above
the individual heading.
Please help.
Many thanks,
Lisa.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Alternate tick labels in xyplot
Dear Rhelp,
I would like to get alternate tick labels for the xyplot:
library(lattice)
library(grid)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = c(simpleTheme(lty=1:8,
col="black",lwd=2),list(layout.heights = list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0)) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
similar to the x-axis in the code below, but somehow it is not working when I
pasted 'list(axis.components=...) in the above code:
xyplot(rnorm(12) ~ 1:12 , type="l",
scales=list(x=list(at=seq(2,12,2),labels=c(1, ' ', 3 , ' ' , 5 , ' ' ))),
par.settings=list(axis.components=list(bottom=list(tck=c(0,1)
scales=list(x=list(at=seq(1,8,1),labels=c('A', ' ', 'C', ' ' ,'E' , ' ','G','
'))),
If possible, I would like a heading for each rows (justified to center) above
the individual heading.
Please help.
Many thanks,
Lisa.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting multiple xyplots in same page
Dear Duncan and Pascal,
Thank you for the response.
Duncan:
I was able to fix the space between title and figure. I would also like to
have separate main titles for each row of figures in addition to the individual
titles.
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = c(simpleTheme(lty=1:8,
col="black",lwd=2),list(layout.heights = list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0)) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Pascal:
I would prefer not to use panels as this is for publication purpose.
Many thanks,
Lisa
- Original Message -
From: Duncan Mackay
To: Lisa Daniel ; r-help-r-project.org
Cc:
Sent: Wednesday, February 13, 2013 7:53 PM
Subject: Re: [R] Plotting multiple xyplots in same page
Hi Lisa
have a look at par.settings in ?xyplot
trellis.par.get() shows the settings
trellis.par.get()[[27]]
$left
$left$tck
[1] 1
$left$pad1
[1] 1
$left$pad2
[1] 1
$top
$top$tck
[1] 1
$top$pad1
[1] 1
$top$pad2
[1] 1
$right
$right$tck
[1] 1
$right$pad1
[1] 1
$right$pad2
[1] 1
$bottom
$bottom$tck
[1] 1
$bottom$pad1
[1] 1
$bottom$pad2
[1] 1
as an example:
par.settings = list(
layout.heights = list(main = 0,
sub = 0,
axis.top = 0.2,
top.padding = 0,
bottom.padding = 0) )
HTH
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
At 09:10 14/02/2013, you wrote:
>Dear Rhelp,
>
>I was able to plot 6 figures in one page. I do need help in
>reducing the space between the title and the plot and assigning
>A,B,C,D,E,F to each figure
>library(lattice)
>trellis.par.set(theme = col.whitebg())
>xy <- xyplot(decrease ~ treatment, OrchardSprays,
> main="Some plot",
> groups = rowpos, type = "a",
> auto.key = list(x = 0.2, y = 0.9,
> cex = 0.52, points = FALSE, lines = TRUE))
>
>
>print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
>print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
>print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
>print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
>print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
>print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
>
>Thanks in advance,
>Lisa
>
>
>
>- Original Message -
>From: Lisa Daniel
>To: "r-help@r-project.org"
>Cc:
>Sent: Wednesday, February 13, 2013 5:28 PM
>Subject: Plotting multiple xyplots in same page
>
>Dear Rhelp,
>
>I would like to have 6 xyplots in the same page. Similar to
>(par(mfrow=c(3,3))
>Found this example on the R archives.
>library(lattice)
>trellis.par.set(theme = col.whitebg())
>xy <- xyplot(decrease ~ treatment, OrchardSprays,
> groups = rowpos, type = "a",
> auto.key = list(x = 0.2, y = 0.9,
> cex = 0.75, points = FALSE, lines = TRUE))
>
>print(xy, pos = c(0.0, 0.0, 0.5, 0.5), more = TRUE)
>print(xy, pos = c(0.0, 0.5, 0.5, 1.0), more = TRUE)
>print(xy, pos = c(0.5, 0.0, 1.0, 0.5), more = TRUE)
>print(xy, pos = c(0.5, 0.5, 1.0, 1.0), more = FALSE)
>
>
>
>
>When I changed the print() to:
>
> print(xy, pos=c(0.0,0.0,0.33,0.33),more = TRUE),
>the figure was not looking good.
>
>Also, is it possible to assign A, B, C,D, etc for each figure.
>
>
>
>a<- rnorm(1000,0,1)
>b<- rnorm(1000,0,2)
>c<- rnorm(1000,0,3)
>d<- rnorm(1000,0,4)
>par(mfrow=c(2,2))
>hist(a,main="")
>title(main="A",adj=0)
>hist(b,main="")
>title(main="B",adj=0)
>hist(c,main="")
>title(main="C",adj=0)
>hist(d,main="")
>title(main="D",adj=0)
>
>
>I
Re: [R] Plotting multiple xyplots in same page
Dear Duncan,
Thank you. It helped a lot.
I was able to get it correct though I couldn't get it working under
simpleTheme() (for publication purpose).
#this gave me the correct format.
library(grid)
library(lattice)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = list(
layout.heights = list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
#pdf("trial.pdf",width=12)
#jpeg("trial.jpeg",quality=100,width=680,height=480)
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
#dev.off()
# I tried the same with simpleTheme(), but I am not getting it correct, (the
lines are colored):
pset <- simpleTheme(lty = 1:8, col="black",lwd=2)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = list(pset,
layout.heights = list(main = 1.2,
sub = 0,
axis.top = 0.2,
top.padding = 0.1,
bottom.padding = 0) ),
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
#the output from the below code is B&W, but space between title and figure not
corrected
pset <- simpleTheme(lty = 1:8, col="black",lwd=2)
xy <-xyplot(decrease ~ treatment, OrchardSprays,
main= "Some plot",
groups = rowpos, type = "a",
page = function(n){
grid.text(LETTERS[j],
y = 0.95,
x = 0.15,
default.units = "npc",
just = c("left"),
gp = gpar(fontsize = 12,fontface="bold") )
},
par.settings = pset,
auto.key = list(x = 0.2, y = 0.9,
cex = 0.75, points = FALSE, lines = TRUE))
j=1
print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Also, I would like to create a main title for the first row and second row of
figures.
Many thanks in advance,
Lisa
- Original Message -
From: Duncan Mackay
To: Lisa Daniel ; r-help-r-project.org
Cc:
Sent: Wednesday, February 13, 2013 7:53 PM
Subject: Re: [R] Plotting multiple xyplots in same page
Hi Lisa
have a look at par.settings in ?xyplot
trellis.par.get() shows the settings
trellis.par.get()[[27]]
$left
$left$tck
[1] 1
$left$pad1
[1] 1
$left$pad2
[1] 1
$top
$top$tck
[1] 1
$top$pad1
[1] 1
$top$pad2
[1] 1
$right
$right$tck
[1] 1
$right$pad1
[1] 1
$right$pad2
[1] 1
$bottom
$bottom$tck
[1] 1
$bottom$pad1
[1] 1
$bottom$pad2
[1] 1
as an example:
par.settings = list(
layout.heights = list(main = 0,
sub = 0,
axis.top = 0.2,
top.padding = 0,
bottom.padding = 0) )
HTH
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New
Re: [R] Plotting multiple xyplots in same page
Dear Rhelp, I was able to plot 6 figures in one page. I do need help in reducing the space between the title and the plot and assigning A,B,C,D,E,F to each figure library(lattice) trellis.par.set(theme = col.whitebg()) xy <- xyplot(decrease ~ treatment, OrchardSprays, main="Some plot", groups = rowpos, type = "a", auto.key = list(x = 0.2, y = 0.9, cex = 0.52, points = FALSE, lines = TRUE)) print(xy, pos = c(0.0, 0.5, 0.33, 1.0), more = TRUE) print(xy, pos = c(0.33, 0.5, 0.66, 1.0), more = TRUE) print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE) print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE) print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE) print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE) Thanks in advance, Lisa - Original Message - From: Lisa Daniel To: "r-help@r-project.org" Cc: Sent: Wednesday, February 13, 2013 5:28 PM Subject: Plotting multiple xyplots in same page Dear Rhelp, I would like to have 6 xyplots in the same page. Similar to (par(mfrow=c(3,3)) Found this example on the R archives. library(lattice) trellis.par.set(theme = col.whitebg()) xy <- xyplot(decrease ~ treatment, OrchardSprays, groups = rowpos, type = "a", auto.key = list(x = 0.2, y = 0.9, cex = 0.75, points = FALSE, lines = TRUE)) print(xy, pos = c(0.0, 0.0, 0.5, 0.5), more = TRUE) print(xy, pos = c(0.0, 0.5, 0.5, 1.0), more = TRUE) print(xy, pos = c(0.5, 0.0, 1.0, 0.5), more = TRUE) print(xy, pos = c(0.5, 0.5, 1.0, 1.0), more = FALSE) When I changed the print() to: print(xy, pos=c(0.0,0.0,0.33,0.33),more = TRUE), the figure was not looking good. Also, is it possible to assign A, B, C,D, etc for each figure. a<- rnorm(1000,0,1) b<- rnorm(1000,0,2) c<- rnorm(1000,0,3) d<- rnorm(1000,0,4) par(mfrow=c(2,2)) hist(a,main="") title(main="A",adj=0) hist(b,main="") title(main="B",adj=0) hist(c,main="") title(main="C",adj=0) hist(d,main="") title(main="D",adj=0) I tried this to xyplot, but it was not working. Please help me. Many thanks in advance, Lisa. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting multiple xyplots in same page
Dear Rhelp, I would like to have 6 xyplots in the same page. Similar to (par(mfrow=c(3,3)) Found this example on the R archives. library(lattice) trellis.par.set(theme = col.whitebg()) xy <- xyplot(decrease ~ treatment, OrchardSprays, groups = rowpos, type = "a", auto.key = list(x = 0.2, y = 0.9, cex = 0.75, points = FALSE, lines = TRUE)) print(xy, pos = c(0.0, 0.0, 0.5, 0.5), more = TRUE) print(xy, pos = c(0.0, 0.5, 0.5, 1.0), more = TRUE) print(xy, pos = c(0.5, 0.0, 1.0, 0.5), more = TRUE) print(xy, pos = c(0.5, 0.5, 1.0, 1.0), more = FALSE) When I changed the print() to: print(xy, pos=c(0.0,0.0,0.33,0.33),more = TRUE), the figure was not looking good. Also, is it possible to assign A, B, C,D, etc for each figure. a<- rnorm(1000,0,1) b<- rnorm(1000,0,2) c<- rnorm(1000,0,3) d<- rnorm(1000,0,4) par(mfrow=c(2,2)) hist(a,main="") title(main="A",adj=0) hist(b,main="") title(main="B",adj=0) hist(c,main="") title(main="C",adj=0) hist(d,main="") title(main="D",adj=0) I tried this to xyplot, but it was not working. Please help me. Many thanks in advance, Lisa. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Webinar: Advances in Gradient Boosting: the Power of Post-Processing. TOMORROW, 10-11 a.m., PST
Webinar: Advances in Gradient Boosting: the Power of Post-Processing TOMORROW: December 14, 10-11 a.m., PST Webinar Registration: http://2.salford-systems.com/gradientboosting-and-post-processing/ Course Outline: I. Gradient Boosting and Post-Processing: o What is missing from Gradient Boosting? o Why post-processing techniques are used? II. Applications Benefiting from Post-Processing: Examples from a variety of industries. o Financial Services o Biomedical o Environmental o Manufacturing o Adserving III. Typical Post-Processing Steps IV. Techniques: o Generalized Path Seeker (GPS): modern high-speed LASSO-style regularized regression. o Importance Sampled Learning Ensembles (ISLE): identify and reweight the most influential trees. o Rulefit: ISLE on "steroids." Identify the most influential nodes and rules. V. Case Study Example: o Output/Results without Post-Processing o Output/Results with Post-Processing o Demo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Webinar signup: Advances in Gradient Boosting: the Power of Post-Processing. December 14, 10-11 a.m., PST
Webinar signup: Advances in Gradient Boosting: the Power of Post-Processing December 14, 10-11 a.m., PST Webinar Registration: http://2.salford-systems.com/gradientboosting-and-post-processing/ Course Outline: *Gradient Boosting and Post-Processing: o What is missing from Gradient Boosting? o Why post-processing techniques are used? *Applications Benefiting from Post-Processing: Examples from a variety of industries. o Financial Services o Biomedical o Environmental o Manufacturing o Adserving *Typical Post-Processing Steps *Techniques: o Generalized Path Seeker (GPS): modern high-speed LASSO-style regularized regression. o Importance Sampled Learning Ensembles (ISLE): identify and reweight the most influential trees. o Rulefit: ISLE on "steroids." Identify the most influential nodes and rules. *Case Study Example: o Output/Results without Post-Processing o Output/Results with Post-Processing o Demo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Webinar signup: Gradient Boosting and Classification Trees: A Winning Combination. November 9, 10-11 a.m., PST
Webinar signup: Tree Ensembles and Classification Trees: A Winning Combination November 9, 10-11 a.m., PST. Webinar Registration: http://2.salford-systems.com/gradientboosting/ Understand major shortcomings of using only decision trees and how tree ensembles can help overcome these challenges and improve your model building. Combine all of the advantages of using classification and regression trees with the power-house interpretability of stochastic gradient boosting. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] choosing the correct number of clusters usinf NbClust
I am attempting to use NbClust to examine different indices to determine the number of clusters. The code appears to be working and I am getting results. However, I have noticed that each time I run it I may get different answers. For example looking at the DB index one run recommended 6 clusters while the next 5 - the code is exactly the same. I assume there is some randomization that I don't see. Is there a way to get a consistent answer. Thanks. -Lisa [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] simple matching with R
Hello, I would like to use the simple matching coefficient in R to compare genotypic data from two years (there are some inconsistencies in the genotype notations in the 2 years and I want to find out the error). Is there a R code I could use? How does my matrix needs to look like? I have 50 genotypes and each genotype is genotyped with 780 markers for two different years. So I need to compare the markers for genotype 1 form year 1 versus year 2, the markers from genotype 2 from year 1 to year 2 and so on, so I can calculate the matching coefficients. Is there a R code for that already? Thank you very much in advance, Lisa-Marie [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] resistanceDistance representation
Dear all,
i'm using gdistance to model animal movement across landscape.
I have imported 11 rasters with roads, freeways, slope, use-of-land, lakes
(...) after recoding them with GRASS with a HSI value ranging from 1 to 4.
I've assigned zero to the NAs and then transformed all the rasters in
TransitionLayers (function=mean,directions=4) and later summed all of them
into a new transition:
> MY_transitionclass : TransitionLayer
dimensions : 2181, 1648, 3594288 (nrow, ncol, ncell)
resolution : 100.0049, 99.98945 (x, y)
extent : 1460708, 1625516, 4947383, 5165460 (xmin, xmax, ymin, ymax)
coord. ref. : +proj=tmerc +lat_0=0 +lon_0=9 +k=0.9996 +x_0=150
+y_0=0 +ellps=intl +towgs84=-225,-65,9,0,0,0,0 +units=m +no_defs
values : conductance
matrix class: dsCMatrix
I applied two different geocorrections to this same transition:
> MY_correction<-geoCorrection(MY_transition,"c",F,scl=T)
> MY_Rcorrection<-geoCorrection(MY_transition,"r",F,scl=T)
My coords are 44 points where samples were taken (some further genetic
analyses are programmed), but i've also tried with only two points:
> coords<-read.csv("coords_md.csv",header=F)
> mycoords<-as.matrix(coords)
> my_spatialpoints<-SpatialPoints(mycoords)
Finally I calculated three geographical distance matrices:
> geodist <- pointDistance(my_spatialpoints,longlat=FALSE)
> summary(geodist) V1 V2 V3 V4
> V5
Min. : 0 Min. : 0 Min. : 0 Min. : 0
Min. : 0
1st Qu.:133788 1st Qu.:133788 1st Qu.:133788 1st Qu.:133788
1st Qu.:133788
Median :133788 Median :133788 Median :133788 Median :133788
Median :135168
Mean :114704 Mean :117371 Mean :120166 Mean :123097
Mean :126174
3rd Qu.:137971 3rd Qu.:137971 3rd Qu.:137971 3rd Qu.:137971
3rd Qu.:137971
Max. :137971 Max. :137971 Max. :137971 Max. :137971
Max. :137971
NA's : 1 NA's : 2 NA's : 3
NA's : 4
()
> costdist <- costDistance(MY_correction, my_spatialpoints)
> summary(costdist) Min. 1st Qu. MedianMean 3rd Qu.Max.
0.000.00 21.92 68.53 210.90 244.70
> resdist <- resistanceDistance(MY_Rcorrection, my_spatialpoints)
> summary(resdist) Min. 1st Qu. Median Mean 3rd Qu. Max. 0 0 1022
1707 4286 4830
Now, I'd like to represent this measures but I couldn't find any examples
of how to do it; plotting or imaging them doesn't return a meaningful
graph, I guess I should transform them somehow but I don't know how.
ANY help would be really appreciated, along with comments about the rest of
the work done.
Thanks in advance,
Lisa
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with SEM package: Error message
Dear John, Thank you for your reply. My data is actually simulated under the model X = Lambda*F + E. Since my post, I've simplified the simulation of my data and I still get the error. This is what I've done since my last post. I constructed Lambda apriori (so I know exactly which observed variables load onto which factors), E follows a Gaussian with mean 0 and var-cov matrix given by the Identity matrix. For my particular model, I sample the factor scores F_i (for sample i) from a multivariate normal F_i ~ N(mu_i, Phi). mu_i is fixed to Phi*z_i, where z_i is a 5x1 vector. Thinking I could have an ill-conditioned var-cov matrix, I looked at the condition number of Phi (the factor var-cov matrix). I recently adjusted Phi to ensure that the condition number was indeed small (it is now about 2). I then sample Y_i ~ N(Lambda*F_i, Psi). If the data I'm simulating is ill conditioned, I'm not even sure how to fix it because the simulation itself is pretty straightforward. Even with a well conditioned factor var-cov matrix Phi that I used to sample my factor scores, I still get that same problem. In any case, I am so grateful for your help- I've been working on this all day and I can't seem to figure out where I go wrong. I made Lambda pretty sparse and with 150 samples, I certainly don't have too many parameters... besides identifiability, I'm not sure what to check for if its not a problem with my coding. Your post has already helped me to think about this problem a little differently. Sincerely, Lisa On Tue, Nov 8, 2011 at 9:32 PM, John Fox wrote: > Dear Lisa, > > There doesn't seem to be anything logically wrong with your model. > > I don't have much time today to look into it, but trying different > optimizers in version 2.0-0 of sem, using the correlation matrix in place > of the covariance matrix, and setting the par.size parameter, I was unable > to obtain an admissible solution. I also was unable using factanal() to fit > an exploratory factor analysis for five factors to your data. I expect that > the problem is ill-conditioned. > > Best, > John > > > John Fox > Sen. William McMaster Prof. of Social Statistics > Department of Sociology > McMaster University > Hamilton, Ontario, Canada > http://socserv.mcmaster.ca/jfox/ > On Tue, 8 Nov 2011 08:18:28 -0800 (PST) > lisamp85 wrote: > > Hello. > > > > I started using the sem package in R and after a lot of searching and > trying > > things I am still having difficulty. I get the following error message > when > > I use the sem() function: > > > > Warning message: > > In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = > > vars, : > > Could not compute QR decomposition of Hessian. > > Optimization probably did not converge. > > > > I started with a simple example using the specify.model() function, but > it > > is really straight forward. I uploaded my specify.model script and my > data > > covariance matrix here too so I wouldn't clutter this email with the > entire > > model (20 observed variables, 5 factors). Could this error message be > from > > the data itself and not from my path model? > > > > I have my observed variables X and my unobserved variables F. I have > ONLY > > exogenous latent variables (i.e. they never appear on the right side of > the > > single head arrow ->). I include all possible factor covariances FjFk, > and > > the only constraints I've made was to restrict the Factor variances to 1. > > My model follows in this basic format (as you can see from my uploaded > > file): > > > > # Factors (where I specify which observed variables load on to which > > factors) > > # I have only exogenous latent variables > > F.i -> X.j, lamj.i, NA > > . > > . > > . > > # Observed variable variances > > X.j <-> X.j, ej, NA > > . > > . > > . > > # Factor variances (I fixed all factor variances to 1) > > F.i <-> F.i, NA, 1 > > . > > . > > . > > # Factor covariances (I represent all factor covariances, i.e. the upper > or > > lower triangle of a covariance matrix) > > F.i <-> F.k, FiFk, NA > > . > > . > > . > > > > Did I do something wrong here? > > Here are my uploaded files: > > CFA script: http://r.789695.n4.nabble.com/file/n4016569/CFA_script.txt > > CFA_script.txt > > Covariance matrix: > > http://r.789695.n4.nabble.com/file/n4016569/covariance_matrix.RData > > covariance_matrix.RData > > > > > > Thank you s
[R] (no subject)
can i be taken off of this mailing list please? is there another way that you can access this without having to get all the emails?? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Text annotation of a graph
Thank you. I will try this. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Text-annotation-of-a-graph-tp3719775p3722163.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Text annotation of a graph
Thanks you for your reply. I consider only the performance of “vars” (v1 to v10), so I just want to indicate then on the y-axis at the point they should be located. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Text-annotation-of-a-graph-tp3719775p3721799.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Text annotation of a graph
Dear All,
I am trying to add some text annotation to a graph using matplot() as
follows:
vars <- c("v1", "v2", "v3", "v4", "v5", "v6", "v7", "v8", "v8", "v10")
id <- seq(5.000, 0.001, by = -0.001)
sid <- c(4.997, 3.901, 2.339, 0.176, 0.151, 0.101, 0.076, 0.051, 0.026,
0.001)
rn <- sample(seq(0, 0.6, by = 0.001), 5000, replace = T)
matplot(rbind(rn, rep(0, length(rn))), rbind(id, id), xlim = c(0, 1),
type = "l", lty = 1, lwd = 1, col = 1, xlab = "",
ylab = "", axes = F)
abline(v = 0, lty = 2)
axis(1)
mtext(side = 2, text = c(vars), at = sid, las = 2, line = 0.8)
axis(3)
But the text annotation can not be displayed correctly, i.e., some of them
stick together.
Can anybody help me with this particular problem? Thanks in advance.
Lisa
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Re: [R] R package: pbatR
Thanks. I have installed PBAT on my computer. -- View this message in context: http://r.789695.n4.nabble.com/R-package-pbatR-tp3667844p3667907.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R package: pbatR
Dear All,
Does anybody have experience with R package pbatR
(http://cran.r-project.org/web/packages/pbatR/index.html)? I am trying to
use it to analyze the family-based case-control data, but the package
totally doesn’t work on my computer. I contacted the authors of the package,
but I haven’t heard anything from them.
Following the package manual, I tried the simple example as below:
library(pbatR)
library(tcltk)
pbat.set("C:/pbat")
x <- data.frame(pid = c(1,1,1,2,2,2,2,3,3,3), # three families
id = c(1,2,3,1,2,3,4,1,2,3),
idfath = c(0,0,1,0,0,1,1,0,0,1),
idmoth = c(0,0,2,0,0,2,2,0,0,2),
sex = c(1,2,1,1,2,2,2,1,2,1),
AffectionStatus = c(0,0,1,0,0,1,0,0,0,1), # 1 for case, 0 for control
m1.1 = c(1,1,2,2,1,1,2,2,2,1), # two SNPs with two columns
for each SNP
m1.2 = c(1,2,1,2,1,2,1,1,2,2),
m2.1 = c(1,1,2,2,2,1,1,1,2,1),
m2.2 = c(2,1,2,1,2,2,2,1,1,1))
x1 <- as.ped(x)
y <- data.frame(pid = c(1,1,1,2,2,2,2,3,3,3),
id = c(1,2,3,1,2,3,4,1,2,3),
age = c(55,50,22,38,37,15,11,42,41,17),
weight = c(185,170,130,165,170,90,60,170,160,120))
y1 <- as.phe(y)
1. I first consider a model with the disease as a phenotype, and two SNPs as
predictors (on covariates) as bellow:
pbat.m(AffectionStatus ~ NONE, y1, x1, fbat="gee",
distribution='categorical', offset='none')
But some error messages were returned:
Error in writeCommandStrMatch("distribution", distribution, c("default", :
'distribution' can only take on the following values: 'default', 'jiang',
'murphy', 'naive', 'observed'. You passed the invalid value 'categorical'.
Then I removed last two arguments
pbat.m(AffectionStatus ~ NONE, y1, x1, fbat="gee")
This time, a box appeared on the console:
R for Windows GUI front-end has encountered a problem and needs to close.
2. I consider a model with the disease as a phenotype, and two covariates
(age, weigth) and two SNPs as predictors as bellow:
pbat.m(AffectionStatus ~ age + weight, y1, x1, fbat="gee")
The function had been running for a very long time and no output was
returned until I had to stop it.
Any help would be greatly appreciated.
Thanks.
Lisa
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[R] Colors in R
HI everyone,
I'm trying to assign colors to multiple lines in a graph. Problem is I don't
want to type in as many colors as there are linesis there a way around
this? In brief, I'm plotting the logratio for up to 60 samples and want a
different color for each sample. Here is the code I'm using now..
Any help is greatly appreciated..
Best
LT
data <- read.table("data.csv", header=T, sep=",", stringsAsFactors= F)
data$Pt <-as.numeric(data$Pt)
npts <- unique(data$Pt)
xrange = c(min(data$Coordinate), max(data$Coordinate))
yrange = c(min(data$Log2Ratio), max(data$Log2Ratio))
colors
<-c("red","blue","green","darkmagenta","darkgreen","darkorange","darkred","gold","midnightblue","seagreen1","tomato","slateblue","violet","purple4","palegreen","darkviolet","forestgreen","firebrick"...up
to 60 colors)
labels <- c('a','b', 'c', 'd', 'e', up to 60 labels)
#add lines
for (i in 1:length(npts)) {
color=colors[i]
pt = data[data$Pt==npts[i],]; plot(pt$Coordinate, pt$Log2Ratio, type="l",
lty=3,col=color, xlim=xrange, ylim=yrange); par(new=T)}
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Re: [R] Multiple line graphs
Thank you!
LT
On 7/7/11 3:46 PM, "Joshua Wiley" wrote:
Hi lt2,
I would use the ggplot2 or lattice package. It strikes me as more
effort to do in traditional graphics. Anyway, here are some examples.
Lattice is a very nice package, but I am not quite as familiar with
it, so my examples for it are not representative of its full power.
Cheers,
Josh
# Your data
dat <- read.table(textConnection("
Pt Coordinate Log2Ratio
1 34046382-0.023639869
1 341414370.016456776
1 34232498-0.027852121
1 34261900-0.167065347
2 34364856-0.182864057
2 34461485-0.228341688
2 349526941.90875531
3 341747051.312169128
3 342478331.381729262
3 343202731.218914649
3 343905450.992551194"), header = TRUE)
closeAllConnections()
require(ggplot2)
require(lattice)
Solution using the ggplot2 package
## Using colours by patient
ggplot(dat, aes(x = Coordinate, y = Log2Ratio, colour = factor(Pt))) +
geom_line()
## Mapping Pt to type of line
ggplot(dat, aes(x = Coordinate, y = Log2Ratio)) +
geom_line(aes(linetype = factor(Pt)))
## Nothing, but creating separate plots
ggplot(dat, aes(x = Coordinate, y = Log2Ratio)) +
geom_line() +
facet_grid(Pt ~ .)
Solution using the lattice package
## Colouring lines
xyplot(Log2Ratio ~ Coordinate, data = dat, group = Pt,
type = "l", auto.key = list(points = FALSE, lines = TRUE))
## Condition on lines
xyplot(Log2Ratio ~ Coordinate | factor(Pt), data = dat, type = "l")
On Thu, Jul 7, 2011 at 9:38 AM, lt2 wrote:
> HI everyone, I'm just starting to get into graphing with R and I need to
> generate one graph that illustrates the pattern of gene expression for
> various patients. My data is in .csv format and is as follows and i'm
> showing below a portion of the data.
>
> Pt Coordinate Log2Ratio
> 1 34046382-0.023639869
> 1 341414370.016456776
> 1 34232498-0.027852121
> 1 34261900-0.167065347
> 2 34364856-0.182864057
> 2 34461485-0.228341688
> 2 349526941.90875531
> 3 341747051.312169128
> 3 342478331.381729262
> 3 343202731.218914649
> 3 343905450.992551194
>
> I can generate the multiple line graphs but i can't get the different colors
> per patient, nor do i know how to change the pattern of the line.
>
> Any ideas and help is greatly appreciated.
>
>
>
>
> --
> View this message in context:
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> Sent from the R help mailing list archive at Nabble.com.
>
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>
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
https://joshuawiley.com/
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Re: [R] Derivative of a function
Yes. I need to do implicit differentiation. After rearrangement, I got (x2 – x1) * b = log(1 / y - 1) Take derivative of both sides with respect to y, I have (x2 – x1) * b’[y] = - 1/y(1-y) Since both (x2 – x1) and b’[y] are vectors, I cannot move (x2 – x1) to RHS. This is why I posted my question here to see if there is some R functions or some idea that can help me solve this problem. Thanks. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3633947.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Derivative of a function
This is not a homework. I just want to see if there are some R functions or some ideas I can borrow to solve my problem. -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3633071.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Derivative of a function
Dear all, I just want to get the derivative of a function that looks like: y = exp(x1*b) / (exp(x1*b) + exp(x2*b)) where y is a scalar, x1, x2, and b are vectors. I am going to take the derivative of b with respect to y, but I cannot derive an expression in which b is function of y. I know there is another way to get the similar result, i.e., first take the derivative of y with respect to each element of b, and then take its reciprocal. But it is not what I want. Could someone please tell me how to solve this problem? Thank you in advance. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Derivative-of-a-function-tp3631814p3631814.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RES: Recode numbers
Thank you very much, Bill. Your script works very well. -- View this message in context: http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566847.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RES: Recode numbers
Thank you so much, Filipe. Your R script is what I am looking for. -- View this message in context: http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566818.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Recode numbers
Thank you, Duncan, Here “a” has the length of 24, and “b” has the length of 20 with numbers from 1 to 20 uniquely. I just want encode “a” from 1 to 20 based on “a” current order using “b”. So, a1[1] = b[1] = 1 a1[2] = b[2] = 5 a1[3] = a1[4] = b[3] = 8 (since third and fourth numbers are the same in “a”) a1[5] = a1[6] = b[4] = 9 a1[7] = b[5] = 14 a1[8] = b[6] = 20 a1[9] = b[7] = 3 a1[10] = a1[11] = b[8] = 10 ... a1[23] = b[19] = 15 a1[24] = b[20] = 19 -- View this message in context: http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566681.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Recode numbers
Thank you for your help, Pete. I tried b[a], but it is not a1. -- View this message in context: http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566534.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Recode numbers
Dear all, I have two sets of numbers that look like a <- c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2, 3, 4) b <- c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15, 19) I just want to use “b” to encode “a” so that “a” looks like a1<- c(1, 5, 8, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13, 17, 18, 2, 4, 15, 19) Does anyone have a suggestion how to deal with this? Thank you in advance. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Recode-numbers-tp3566395p3566395.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Arrange a multi-level list to a one-level list
Hi, Phil, Yes. That's what I am looking for. Thank you so much. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Arrange-a-multi-level-list-to-a-one-level-list-tp3556500p3556601.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Arrange a multi-level list to a one-level list
Dear all, I just want to arrange a multi-level list to a one-level list. For example: x <- list(list(matrix(sample(c(0,1), 4, replace = T), 2), matrix(sample(c(2,3), 4, replace = T), 2)), list(matrix(sample(c(0,1), 6, replace = T), 2), matrix(sample(c(2,3), 6, replace = T), 2))) > x [[1]] [[1]][[1]] [,1] [,2] [1,]11 [2,]00 [[1]][[2]] [,1] [,2] [1,]32 [2,]33 [[2]] [[2]][[1]] [,1] [,2] [,3] [1,]100 [2,]010 [[2]][[2]] [,1] [,2] [,3] [1,]233 [2,]333 If I do this work one time, I will arrange it like this: c(x[[1]], x[[2]]) # Only consider the first level of the list. [[1]] [,1] [,2] [1,]11 [2,]00 [[2]] [,1] [,2] [1,]32 [2,]33 [[3]] [,1] [,2] [,3] [1,]100 [2,]010 [[4]] [,1] [,2] [,3] [1,]233 [2,]333 But when I repeat this work many times and the levels will change each time in a loop, how can I pass the arguments to c()? For example, in the first iteration, the first level of a list is 3, in the second iteration, the first level of a list is 5? …. Any help would be greatly appreciated. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Arrange-a-multi-level-list-to-a-one-level-list-tp3556500p3556500.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Suppress intermediate results on console
Thanks for your comments and suggestion. I didn’t show all my own function here because it has many lines. “x” is the results of another function. I am calling summary because I want to extract some values from the results. -- View this message in context: http://r.789695.n4.nabble.com/Suppress-intermediate-results-on-console-tp3553276p3553405.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Suppress intermediate results on console
Dear all,
I have a question about how to suppress intermediate results in a function
on console. For example, I will use summary() in my own function that looks
like:
myfunction <- function(…)
{
…
Summary(x)
…
}
Then myfunction() will print “x” on console that is intermediate result and
doesn’t need showing.
Does someone have any idea or any suggestion? Thank you in advance.
Lisa
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Re: [R] Order a matrix
Hi, Bill, Thank you for your help. Your R script works very well. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Order-a-matrix-tp3547923p3548017.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Order a matrix
Dear all, I just want to order a matrix using several columns in a matrix. For example: x <- matrix(sample(c(1:5), 60, replace = T), 10, 6). If I order the matrix by the first two columns, I will do it like this: x[order(x[, 1], x[, 2]), ]. But when I repeat this work many times and the columns will change each time in a simulation study, how can I assign the arguments of order()? For example, in the first iteration, the columns needing to order is the first two columns, in the second iteration, the columns needing to order is the first, second, and fourth column? …. I would appreciate any help on this question. Thanks a lot. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Order-a-matrix-tp3547923p3547923.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Line feed for a long character string
Hi, Duncan,
On your and William Dunlap’s suggestion, I tried this:
cat(sep="\n", strwrap("my long character string")),
and it works very well. Thanks so much for your help.
Lisa
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[R] Line feed for a long character string
Dear all,
Does anyone know how to make a line feed automatically based on the width of
console window? For example, when you cat() a long character string just
like this:
cat("Seminar series is an opportunity for students to learn about ongoing
researches in the field of mathematics, computer science, physics,
chemistry, and some other related programs. Students must complete a seminar
attendance form and return to their mentors.\n")
In my computer, about half of the string is displayed with a symbol “$”at
the end that looks like this:
> cat("Seminar series is an opportunity for students to learn about ongoing
> researches in the field of mathematics, computer science, physics,
> chemistry, a$
Any help will be appreciated!
Lisa
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Re: [R] Line numbers of scan(0
I am trying to write a function to check unusual values in my datasets and correct them. As some R users suggested, I try to use readline() and scan() in my function. Suppose there are several unusual values in a dataset. I want to change the line numbers in scan() to something like: unusual value 1, unusual value 2 and so on. I think this can make the correction much easier. -- View this message in context: http://r.789695.n4.nabble.com/Line-numbers-of-scan-0-tp3482314p3482802.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Line numbers of scan(0
Dear all, Does everyone know how to change the line numbers of scan()? For example, > scan(n = 3) 1: 4 2: 6 3: 9 Read 3 items [1] 4 6 9 I just want to change the line numbers 1, 2, and 3 to, say a, b, and c that look like this: > scan(n = 3) a: 4 b: 6 c: 9 Read 3 items [1] 4 6 9 Any help will be appreciated. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Line-numbers-of-scan-0-tp3482314p3482314.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Pause the execution of a function
Hi, Rolf, Thank you for your help. I am trying to use readline() and I have a question about the maximum length of prompt string. The R help on readline() says: “The prompt string will be truncated to a maximum allowed length, normally 256 chars (but can be changed in the source code).” I don’t know how to change the source code. Could you please help how to get this done? Thanks. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Pause-the-execution-of-a-function-tp3478535p3482192.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Pause the execution of a function
Thanks. I will try them. -- View this message in context: http://r.789695.n4.nabble.com/Pause-the-execution-of-a-function-tp3478535p3479240.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Pause the execution of a function
Dear all, I am trying to write a script to pause the execution of a function and provide some additional commands to the function and then continue execution of the function. For example, when my function detects a wrong number in a dataset, the function pauses automatically and returns information on the screen: “There is a wrong number in the dataset. Would you like to correct it? ” If I input “yes” from the keyboard and hit the enter key, the function asks again: “What is the correct number?” After inputting a correct number from the keyboard and hitting the enter key, the function continues to execute. Can anybody please help how to get this done? Thanks a lot in advance Lisa -- View this message in context: http://r.789695.n4.nabble.com/Pause-the-execution-of-a-function-tp3478535p3478535.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tell the difference between characters
Yes. That is what I want. Thank you very much. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Tell-the-difference-between-characters-tp3476130p3476288.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tell the difference between characters
Thanks a lot. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Tell-the-difference-between-characters-tp3476130p3476352.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Tell the difference between characters
Dear all,
I just want to determine if the characters in a character string are the
same or not. For example,
temp <- c("aa", "aA", "ab")
How do I determine the first one have the two same “a”, and the second and
third have the different characters? Thanks in advance.
Lisa
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Re: [R] Factor function
Thank you so much! Lisa -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-tp3473984p3475549.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Factor function
Thank you for your help. Your R code works well. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-tp3473984p3474196.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Factor function
Thank you for your reply again. I really know that NA is not "NA". I just want to figure out how to remove "NA" from the levels. Thanks again. -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-tp3473984p3474127.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Factor function
Did you see the data frame "d"? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-tp3473984p3474065.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Factor function
Dear All,
I just want to remove “NA” from the levels of a factor. For example:
d<-data.frame(matrix(c("ww","ww","xx","yy","ww","yy","xx","yy","NA"),
ncol=3, byrow=TRUE))
> factor(d[, 3], exclude=NA)
[1] xx yy NA
Levels: NA xx yy
But “NA” is still listed in the levels. How can I solve this problem? Thanks
in advance.
Lisa
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Re: [R] Select subset of data
Thank you for your kind help. Your R code works very well. -- View this message in context: http://r.789695.n4.nabble.com/Select-subset-of-data-tp3416012p3416307.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Select subset of data
Dear All, I have a dataset that looks like this: group subject result v4 v5 1 1 1 0 1 0 2 1 2 1 0 0 3 1 3 0 0 0 4 1 4 1 0 0 5 2 1 0 1 1 6 2 2 0 0 1 7 2 3 0 1 1 8 3 1 0 1 0 9 3 2 0 0 1 10 3 3 1 0 0 11 3 4 0 1 0 12 4 1 1 0 0 13 4 2 1 1 0 14 4 3 0 0 1 15 4 4 0 0 0 16 4 5 1 0 1 …… I only show 4 groups here. There are several subjects within each group. I want to select some groups in which for the firs two subjects, the results are equal to 0, and for the other subjects, only one has the result being equal to 1. So, for the data above, only the group 3 satisfies these conditions. Therefore, the new dataset is: group subject result v4 v5 8 3 1 01 0 9 3 2 00 1 10 3 3 10 0 11 3 4 01 0 Can anybody please help how to get this done? Your help would be greatly appreciated. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Select-subset-of-data-tp3416012p3416012.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Randomly generating data
Hi, everybody, I have a problem and need your help. There are two columns that look like this: [1,] "t" "f" [2,] "f" "t" [3,] "t" "f" [4,] "t" "t" [5,] "f" "f" I just want to generate the third column based on these two columns. First, I randomly choose one of the two columns, say, the first column. So, the first character of the third column is “t” that looks like this: [1,] "t" "f" "t" [2,] "f" "t" [3,] "t" "f" [4,] "t" "t" [5,] "f" "f" Second, I determine the second character of the third column with an additional Bernoulli trial with a probability, for example, rbinom(1, 1, 0.3). If the random generation in R is 0, we keep "f", the second character in the first column because the first column has been chosen in the first step, while if the random generation in R is 1, we choose "t" instead, the second character in the second column. Third, I determine the third character of the third column with a Bernoulli trial but a different probability, for example, rbinom(1, 1, 0.7). Repeat these processes… How can I do this efficiently if there are more than thousand records (rows)? Can anybody please help how to get this done? Thanks a lot in advance Lisa -- View this message in context: http://r.789695.n4.nabble.com/Randomly-generating-data-tp3394250p3394250.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Arguments of a function
Thank you very much! -- View this message in context: http://r.789695.n4.nabble.com/Arguments-of-a-function-tp3387643p3388615.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Arguments of a function
Hi, everybody, I just want to pass arguments to a function as below: range <- c(0.1, 0.5) runif(1, range) But it doesn’t work. Does anyone have any suggestions to offer? Thanks. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Arguments-of-a-function-tp3387643p3387643.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Add columns of dataset
Your R script works very well. Thank you very much. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Add-columns-of-dataset-tp3071722p3072513.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Add columns of dataset
Dear all, I have a dataset that looks like id var1 var2 var4 var7 var8 10.0 0.10.30.90.0 20.4 0.60.00.00.2 30.0 0.00.00.80.7 Some columns are missed, for example, here the fourth (var3), sixth(var5) and seventh (var6) columns. I want to first determine which columns are missed in a huge dataset and then add the missed columns with the values of zeros that looks like id var1 var2 var3 var4 var5 var6 var7 var8 10.0 0.1 0.00.3 0.00.0 0.90.0 20.4 0.6 0.00.0 0.00.0 0.00.2 30.0 0.0 0.00.0 0.00.0 0.80.7 Can anybody please help how to get this done? Thanks a lot in advance Lisa -- View this message in context: http://r.789695.n4.nabble.com/Add-columns-of-dataset-tp3071722p3071722.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data transformation
Thank you so much. Lisa -- View this message in context: http://n4.nabble.com/Data-transformation-tp1289899p1289915.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data transformation
Dear all,
I have a dataset that looks like this:
x <- read.table(textConnection("col1 col2
3 1
2 2
4 7
8 6
5 10"), header=TRUE)
I want to rewrite it as below:
var1 var2 var3 var4 var5 var6 var7 var8 var9 var10
1 0 1 0 0 0 0 0 0 0
0 2 0 0 0 0 0 0 0 0
0 0 0 1 0 0 1 0 0 0
0 0 0 0 0 1 0 1 0 0
0 0 0 0 1 0 0 0 0 1
Can anybody please help how to get this done? Your help would be greatly
appreciated.
Lisa
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Re: [R] Delete components of a list
It works well. Thank you very much. Lisa -- View this message in context: http://n4.nabble.com/Delete-components-of-a-list-tp1289732p1289745.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Delete components of a list
Dear all, I have a question about deleting components of a list. For example, I have a list that looks like this: [[1]] var1 var2 var3 134 221 [[2]] var1 var2 var3 251 217 [[3]] var1 var2 var3 167 384 How to delete, say, the first columns of the list or the second rows? Thank in advance. Lisa -- View this message in context: http://n4.nabble.com/Delete-components-of-a-list-tp1289732p1289732.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Arguments of a function
That's what I want. Thank you so much.
Lisa
Henrique Dallazuanna wrote:
>
> Try this:
>
> my.f <- function(a, b) {
> x1 <- 2 * 3
> x2 <- 3 / 6
> x3 <- 4 * 4 / 5 - sqrt(2)
> y <- get(deparse(substitute(a))) + get(deparse(substitute(b)))
> return(y)
> }
>
> my.f(x1, x2)
>
> On Fri, Jan 8, 2010 at 4:15 PM, Lisa wrote:
>>
>> Dear all,
>>
>> I have a question about how to set arguments in my own function. For
>> example, I have a function that looks like this:
>>
>> my.f <- function(a = x1, b = x2)
>> {
>> x1 = equation 1
>> x2 = equation 2
>> x3 = equation 3
>> y = a + b
>> }
>>
>> x1, x2, and x3 are temporary variables (intermediate results) calculated
>> from other variables within the funciton. I want to use two of these
>> three
>> variables to calculate y, and write R script as below:
>>
>> my.f(a = x1, b = x2)
>>
>> or
>>
>> my.f(a = x2, b = x3)
>>
>> The error information shows that: “objects 'x1', 'x2', or 'x3' not
>> found”.
>>
>> Can anybody help me solve this problem? Thanks in advance.
>>
>> Lisa
>>
>> --
>> View this message in context:
>> http://n4.nabble.com/Arguments-of-a-function-tp1009883p1009883.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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[R] Arguments of a function
Dear all,
I have a question about how to set arguments in my own function. For
example, I have a function that looks like this:
my.f <- function(a = x1, b = x2)
{
x1 = equation 1
x2 = equation 2
x3 = equation 3
y = a + b
}
x1, x2, and x3 are temporary variables (intermediate results) calculated
from other variables within the funciton. I want to use two of these three
variables to calculate y, and write R script as below:
my.f(a = x1, b = x2)
or
my.f(a = x2, b = x3)
The error information shows that: “objects 'x1', 'x2', or 'x3' not found”.
Can anybody help me solve this problem? Thanks in advance.
Lisa
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Re: [R] Data replacement
Thank you for your kind help. Your R script works well. Lisa Dieter Menne wrote: > > > > Lisa wrote: >> >> I have a dataset that looks like this: >> >>> data >> idcode1code2 >> 1 114 >> 2 123 >> 3 244 >> .. >> >> I want to change some numbers in the columns of “code1” and “code2” based >> on “indx” as below >> >>> indx >> [[1]] >> code >> 1 1 >> 2 3 >> 3 4 >> For example, for the first ten records (rows) of my dataset, I want to >> change 2 to 3, 3 to 4, 4 to 6, and 5 to 8 in both “code1” and “code2”, >> while for the last ten records, I want to change 3 to 4 and 4 to 6. >> >> > > You might check for "recode", for example in package car, or for > "transform". You could also do it the quick and dirty way, good to learn > indexing. Be careful if you have NA in your data, or data out of the > recode range. > > Dieter > > > data = data.frame(code1=sample(1:5,10,TRUE),code2=sample(1:5,10,TRUE)) > data = > rbind(data,data.frame(code1=sample(1:4,10,TRUE),code2=sample(1:4,10,TRUE))) > > # The recode table as in your example > #indx = list(data.frame(code=c(1,3,4,6,8)),data.frame(code=c(1,2,4,6))) > > #easier to read > recode1 = c(1,3,4,6,8) > recode2 = c(1,2,4,6) > > data$code1T[1:10] = recode1[data$code1[1:10]] > data$code2T[1:10] = recode1[data$code2[1:10]] > > data$code1T[11:20] = recode2[data$code1[11:20]] > data$code2T[11:20] = recode2[data$code2[11:20]] > > > > > -- View this message in context: http://n4.nabble.com/Data-replacement-tp999060p999342.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data replacement
Dear all, I have a question and need your help. I have a dataset that looks like this: > data idcode1code2 1 114 2 123 3 244 4 315 5 324 6 411 7 434 8 643 9 622 10752 11114 12132 13344 14414 15432 16511 17543 18714 19723 20811 I want to change some numbers in the columns of “code1” and “code2” based on “indx” as below > indx [[1]] code 1 1 2 3 3 4 4 6 5 8 [[2]] code 1 1 2 2 3 4 4 6 For example, for the first ten records (rows) of my dataset, I want to change 2 to 3, 3 to 4, 4 to 6, and 5 to 8 in both “code1” and “code2”, while for the last ten records, I want to change 3 to 4 and 4 to 6. Can anybody please help how to get this done? Thanks a lot in advance Lisa -- View this message in context: http://n4.nabble.com/Data-replacement-tp999060p999060.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove double quotation marks
Your script works well. Thank you very mcuh.
Lisa
Nutter, Benjamin wrote:
>
> It seems from your example that you're assuming all of the vectors have
> the same length. If this is the case, then a data.frame might be your
> friend.
>
>> df <- data.frame(
> v1 = c(0, 1, 0),
> v2 = c(1, 1, 0),
> v3 = c(2, 1, 2),
> v4 = c(2, 2, 1),
> v5 = c(0, 1, 1) )
>
>> x <- 5
>
>> get.var <- c("v1", "v2", paste("v", x, sep=""))
>
>> df[, get.var]
>
> Or, if you know your variables in the data.frame will be named
> sequentially
>
>> df[, c(1, 2, x)]
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Lisa
> Sent: Tuesday, December 29, 2009 3:55 PM
> To: r-help@r-project.org
> Subject: Re: [R] Remove double quotation marks
>
>
> Thank you for your help. But here I just want to combine some vectors by
> column based on the numbers determined by other R script.
>
> For example, I have five vectors:
>
> v1 <- c(0, 1, 0)
> v2 <- c(1, 1, 0)
> v3 <- c(2, 1, 2)
> v4 <- c(2, 2, 1)
> v5 <- c(0, 1, 1)
>
> If I am going to combine the first two vectors, and one more other
> vector determined by my other R script, say, vector 5. Then my R script
> is
>
> x <- 5
> cbind(v1, v2, paste("v", x, sep = ""))
>
> The output is
>
> v1 v2
> [1,] "0" "1" "v5"
> [2,] "1" "1" "v5"
> [3,] "0" "0" "v5"
>
> This is not what I want. I want to get this:
>
> v1 v2 v5
> [1,] 0 1 0
> [2,] 1 1 1
> [3,] 0 0 1
>
> Can you give me further suggestions or comments? Thanks a lot.
>
> Lisa
>
>
>
> Barry Rowlingson wrote:
>>
>> On Tue, Dec 29, 2009 at 6:31 PM, Lisa wrote:
>>>
>>> Thank you for your reply. But in the following case, "cat()" or
> "print()"
>>> doesn't work.
>>>
>>> data.frame(cbind(variable 1, variable 2, cat(paste("variable", x),
>>> "\n"))), where x is a random number generated by other R script.
>>>
>>> Lisa
>>
>> Yes, because you are Doing It Wrong. If you have data that is indexed
>
>> by an integer, don't store it in variables called variable1, variable2
>
>> etc, because very soon you will be posting a message to R-help that is
>
>> covered in the R FAQ...
>>
>> Store it in a list:
>>
>> v = list()
>> v[[1]] = c(1,2,3,4,5)
>> v[[2]] = c(4,5,6,7,8,9,9,9)
>>
>> then you can do v[[i]] for integer values of i.
>>
>> If you really really must get values of variable by name, perhaps
>> because someone has given you a data file with variables called
>> variable1 to variable99, then use the paste() construction together
>> with the 'get' function:
>>
>> [ not tested, but should work ]
>>
>> > v1="99"
>> > v2="102"
>> > i=2
>> > get(paste("v",i,sep=""))
>> [1] 102
>>
>> Barry
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
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Re: [R] Remove double quotation marks
Thanks. I also need column name "v5".
Lisa
jholtman wrote:
>
> ?get
>
> something like:
>
>> v1 <- c(0, 1, 0)
>> v2 <- c(1, 1, 0)
>> v3 <- c(2, 1, 2)
>> v4 <- c(2, 2, 1)
>> v5 <- c(0, 1, 1)
>> x <- 5
>> cbind(v1, v2, get(paste('v', x, sep='')))
> v1 v2
> [1,] 0 1 0
> [2,] 1 1 1
> [3,] 0 0 1
>>
>
>
> On Tue, Dec 29, 2009 at 3:54 PM, Lisa wrote:
>
>>
>> Thank you for your help. But here I just want to combine some vectors by
>> column based on the numbers determined by other R script.
>>
>> For example, I have five vectors:
>>
>> v1 <- c(0, 1, 0)
>> v2 <- c(1, 1, 0)
>> v3 <- c(2, 1, 2)
>> v4 <- c(2, 2, 1)
>> v5 <- c(0, 1, 1)
>>
>> If I am going to combine the first two vectors, and one more other vector
>> determined by my other R script, say, vector 5. Then my R script is
>>
>> x <- 5
>> cbind(v1, v2, paste("v", x, sep = ""))
>>
>> The output is
>>
>> v1 v2
>> [1,] "0" "1" "v5"
>> [2,] "1" "1" "v5"
>> [3,] "0" "0" "v5"
>>
>> This is not what I want. I want to get this:
>>
>> v1 v2 v5
>> [1,] 0 1 0
>> [2,] 1 1 1
>> [3,] 0 0 1
>>
>> Can you give me further suggestions or comments? Thanks a lot.
>>
>> Lisa
>>
>>
>>
>> Barry Rowlingson wrote:
>> >
>> > On Tue, Dec 29, 2009 at 6:31 PM, Lisa wrote:
>> >>
>> >> Thank you for your reply. But in the following case, cat() or
>> print()
>> >> doesnt work.
>> >>
>> >> data.frame(cbind(variable 1, variable 2, cat(paste("variable", x),
>> >> "\n"))),
>> >> where x is a random number generated by other R script.
>> >>
>> >> Lisa
>> >
>> > Yes, because you are Doing It Wrong. If you have data that is indexed
>> > by an integer, don't store it in variables called variable1, variable2
>> > etc, because very soon you will be posting a message to R-help that is
>> > covered in the R FAQ...
>> >
>> > Store it in a list:
>> >
>> > v = list()
>> > v[[1]] = c(1,2,3,4,5)
>> > v[[2]] = c(4,5,6,7,8,9,9,9)
>> >
>> > then you can do v[[i]] for integer values of i.
>> >
>> > If you really really must get values of variable by name, perhaps
>> > because someone has given you a data file with variables called
>> > variable1 to variable99, then use the paste() construction together
>> > with the 'get' function:
>> >
>> > [ not tested, but should work ]
>> >
>> > > v1="99"
>> > > v2="102"
>> > > i=2
>> > > get(paste("v",i,sep=""))
>> > [1] 102
>> >
>> > Barry
>> >
>> > __
>> > R-help@r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> >
>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>> >
>>
>> --
>> View this message in context:
>> http://n4.nabble.com/Remove-double-quotation-marks-tp990502p990586.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
> [[alternative HTML version deleted]]
>
>
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>
>
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Re: [R] Remove double quotation marks
Thank you for your help. But here I just want to combine some vectors by
column based on the numbers determined by other R script.
For example, I have five vectors:
v1 <- c(0, 1, 0)
v2 <- c(1, 1, 0)
v3 <- c(2, 1, 2)
v4 <- c(2, 2, 1)
v5 <- c(0, 1, 1)
If I am going to combine the first two vectors, and one more other vector
determined by my other R script, say, vector 5. Then my R script is
x <- 5
cbind(v1, v2, paste("v", x, sep = ""))
The output is
v1 v2
[1,] "0" "1" "v5"
[2,] "1" "1" "v5"
[3,] "0" "0" "v5"
This is not what I want. I want to get this:
v1 v2 v5
[1,] 0 1 0
[2,] 1 1 1
[3,] 0 0 1
Can you give me further suggestions or comments? Thanks a lot.
Lisa
Barry Rowlingson wrote:
>
> On Tue, Dec 29, 2009 at 6:31 PM, Lisa wrote:
>>
>> Thank you for your reply. But in the following case, “cat()” or “print()”
>> doesn’t work.
>>
>> data.frame(cbind(variable 1, variable 2, cat(paste("variable", x),
>> "\n"))),
>> where x is a random number generated by other R script.
>>
>> Lisa
>
> Yes, because you are Doing It Wrong. If you have data that is indexed
> by an integer, don't store it in variables called variable1, variable2
> etc, because very soon you will be posting a message to R-help that is
> covered in the R FAQ...
>
> Store it in a list:
>
> v = list()
> v[[1]] = c(1,2,3,4,5)
> v[[2]] = c(4,5,6,7,8,9,9,9)
>
> then you can do v[[i]] for integer values of i.
>
> If you really really must get values of variable by name, perhaps
> because someone has given you a data file with variables called
> variable1 to variable99, then use the paste() construction together
> with the 'get' function:
>
> [ not tested, but should work ]
>
> > v1="99"
> > v2="102"
> > i=2
> > get(paste("v",i,sep=""))
> [1] 102
>
> Barry
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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Re: [R] Remove double quotation marks
Thank you for your reply. But in the following case, “cat()” or “print()”
doesn’t work.
data.frame(cbind(variable 1, variable 2, cat(paste("variable", x), "\n"))),
where x is a random number generated by other R script.
Lisa
Duncan Murdoch wrote:
>
> On 29/12/2009 1:16 PM, Lisa wrote:
>> Dear All,
>>
>> I am not sure how to remove double quotation marks in a string, e.g.,
>> paste("variable", 1). Can anybody please help me solve it? Thank you in
>> advance.
>
> I think you need to tell us what is wrong with what you get from that.
> When I look at the result:
>
> > cat(paste("variable", 1), "\n")
> variable 1
>
> I see no quotation marks. (If you use print() you'll see some, but they
> aren't part of the string, they are just used in the default display by
> print().)
>
> Duncan Murdoch
>
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>
>
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[R] Remove double quotation marks
Dear All,
I am not sure how to remove double quotation marks in a string, e.g.,
paste("variable", 1). Can anybody please help me solve it? Thank you in
advance.
Lisa
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