similar to: Loop for converting character columns to Numeric

HI, You can do this in many ways: dat1<-read.table(text=" med1,med2,med3???? ?1,0,1?????? 0,1,1??? 2,0,0 ",sep=",",header=TRUE)?? #1st method library(reshape) dat2<-melt(dat1) dat3<-aggregate(dat2$value,by=list(dat2$variable),sum) ?colnames(dat3)<-c("name","sum(n11)") ?dat3 #? name sum(n11) #1 med1??????? 3 #2 med2??????? 1 #3 med3??????? 2

Hello, I have a data as follow: ID    Visit xa1 xa2 yb1 yc23 yb33   I want to look at frequency of visit for ID and create a new column as response .  For example my response would be 2 for x and 3 for y. I think I need to write a loop, but I don't know how. I really appreciate your help. Thanks a lot. Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

Hi, I have a data set as follow: X         Z x1        102 x2        102 x2        102 x2        77 x3        23   I need to pivot this data as follows and assign the values based on frequency of column Z: X       Z.102   Z.77 Z.23 x1          1        0        0 x2          21  0 x3         00  1 Thanks. Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

Hi, Try: dat1<- read.table(text=" ID Date x2 x1????????? x3 56 25-Jun-01 10 2? 126 56 29-Oct-01 10 2? 140 56 18-Mar-02 10 2? 445 56 6-Jun-03 10 2??? 224 56 16-Jan-04? 10 2??? NA 58 10-Jan-02 10.8 1 715 58 26-Dec-03 10.8 1??? NA ",sep="",header=TRUE,stringsAsFactors=FALSE) ?unlist(with(dat1,by(as.Date(Date,format="%d-%b-%y"),ID,FUN=function(x)

HI, I am not sure about the ?update() method. You could try this: set.seed(232) mat1<-matrix(sample(1:100,80,replace=TRUE),ncol=8) #with 8 columns dat1<-data.frame(mat1) ?names(dat1)[1]<-"Y" ?fit<-lm(Y~.,data=dat1) res<-coef(summary(fit)) ?res #???????????????? Estimate Std. Error???? t value?? Pr(>|t|) #(Intercept) 143.763081344 21.9902865? 6.53757200 0.02260698

Hi Farnoosh, Use ?as.Date() dat1<- read.table(text="V1 2012-01-12 08:23:00 2012-01-19 15:29:00",sep=",",header=TRUE,stringsAsFactors=FALSE) dat1$V1<- as.Date(dat1$V1) ?dat1 #????????? V1 #1 2012-01-12 #2 2012-01-19 A.K. ________________________________ From: farnoosh sheikhi <farnoosh_81 at yahoo.com> To: "smartpink111 at yahoo.com"

Hi, I have a R output that looks as follow: Rad:0 Rad1:2 Rad3:3 I want to make a new matrix that looks like : sample size is 2400 Variable    n11  n12 Rad            0     2400-0=2400 Rad1          2       2400-2 Rad3  3      2400-3   Thanks a lot for your time and help:) Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

Hello, I have a data which looks like below: Some of the patients have multiple diagnosis. ID(200 patients)   Diag (100 unique Diag-200 in general)   Proc (50 uniqe Proc)  DOB (200)   Gender (200)    a                           daig1 b                           diag2 c                            diag1 I want to reformat this data to : ID   diag1 diag 2 diag 3..  diagx   proc1   proc2  

Hi Farnoosh, set.seed(25) Data.All.Var<- data.frame(ID= sample(LETTERS[1:10],25,replace=TRUE), value=rnorm(25),stringsAsFactors=FALSE) set.seed(28) Y.Out<- data.frame(ID=sample(LETTERS[1:5],25,replace=TRUE),value=rnorm(25),stringsAsFactors=FALSE) unique(Y.Out$ID) #[1] "A" "C" "E" "D" "B" ?unique(Data.All.Var$ID) # [1] "E"

Hi, I have a column of DOB in mm/dd/yyyy. I want to convert this format to age. Thanks a lot.   Best,Farnoosh Sheikhi [[alternative HTML version deleted]]

Hi, ?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.? DataA<- read.table(text=" ID,Status,Date1,Date2 ??? ??? ?????? 1,A,3-Feb-01,15-May-01 ??? ??? 1,B,15-May-01,16-May-01 ??? ??? 1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ??? 1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ??? 1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ???

Hi Arun, I have a question about choosing a subset of data. I have a matrix of 5000 in 3500. I want to choose specific variables such as proc1 to proc1000 and Lab1 to Lab1600 and put it into a new matrix to run some correlation analysis. Since I have 3500 variables, I don't know from what column my lab or procedures start. I really appreciate your help. Thanks.   Best,Farnoosh Sheikhi

Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0<-lm(Y~1, data= mydata) fit.final<- lm(Y~X1+X2+X3+.....+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction="forward")

Thanks Jeff. This is a nice way of solving this problem. What about the cases with 0015-02-21?Many thanks.?Best,Farnoosh On Wednesday, June 28, 2017 10:49 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: I doubt your actual file looks like the mess that made it to my email software (below) because you posted HTML-format email. Read the Posting Guide, and in particular

I doubt your actual file looks like the mess that made it to my email software (below) because you posted HTML-format email. Read the Posting Guide, and in particular figure out how to send plain text email. You might try the "anytime" contributed package, though I suspect it too will choke on your mess. Otherwise, that will pretty much leave only a brute-force series of regular

Hello, I have a data with following format:      Predictors          n11     n12     n21     n22     Odds.Ratio    log.ratio    se.log.odds. 1 ProcOR respirato     2 ProcVaric vein      3 DiagCardiac anom   4 DiagAllergy        5  DiagOth skin dx    6    DiagGastritis          I want to plot odds ratio by command: forestplot in rmeta library, but I get the following error constantly.  Error in

Hey, Are all the dates connected? So no comma or space btw? Regards, Christoph > On 29 Jun 2017, at 2:02 pm, Farnoosh Sheikhi via R-help <r-help at r-project.org> wrote: > > Hi, > I have a data set with various date formats in one column and not sure how to unify it.Here is a few formats: > 02091702/22/170221201703/17/160015-08-239/2/1500170806May-2-201522-March-2014

Left as an exercise for the student. -- Sent from my phone. Please excuse my brevity. On June 29, 2017 7:25:36 PM EDT, Farnoosh Sheikhi <farnoosh_81 at yahoo.com> wrote: >Thanks Jeff. This is a nice way of solving this problem. What about the >cases with 0015-02-21?Many thanks.?Best,Farnoosh > > > >On Wednesday, June 28, 2017 10:49 PM, Jeff Newmiller ><jdnewmil at

Hi Farnoosh, Perhaps this will help: drop_dollar<-function(x) return(as.numeric(as.character(gsub("\\$","",x)))) sapply(My.Data,drop_dollar) Jim On Thu, Apr 19, 2018 at 7:23 AM, Farnoosh Sheikhi via R-help <r-help at r-project.org> wrote: > Hello, > I have a data frame with 400 columns and wanted to filter character columns with "$" in it.For

Your code is full of syntactic errors. What do you think 1.71(se) means? After you clean up your code, something like this might be what you want: out <- lapply(seq(40,500,by = 25), f) To get plots, just stick in a plot statement after you define m and d. Have you gone through any R tutorials? You seem to be confused about basics that tutorials could help you with. This list is not meant