Displaying 8 results from an estimated 8 matches for "lst3".
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lst
2013 Jun 04
0
choose the lines2
...is helps:
dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t")
dat.bru<- dat1[!is.na(dat1$evnmt_brutal),]
fun2<- function(dat){??
????? lst1<- split(dat,dat$patient_id)
??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,])
??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])
??? lst4<- lst3[lapply(lst3,nrow)!=0]
??? lst5<- lapply(seq_along(lst4),function(i){
??? ??? ??? ??? ???? do.call(rbind,lapply(which(lst4[[i]]$evnmt_brutal==1),function(x) {
??? ??? ??? ??? ??? ??? ??? ??? ???...
2013 Jun 08
0
data
...to select only a pair of rows (dummy=0 and dummy=1) for each lists.
fun1<- function(dat,percent,number){???
??? lst1<- split(dat,list(dat$year,dat$industry))
??? lst1New<- lapply(lst1,function(x) x[!(all(x$dummy==0)|all(x$dummy==1)),])
??? lst2<- lst1New[lapply(lst1New,nrow)>0]
??? lst3<- lapply(lst2,function(x){
??? ??? ??? ??? lapply(x$dimension,function(y){
??? ??? ??? ??? ? x1<- x[(y < (x$dimension+(x$dimension*percent))) & (y > (x$dimension-(x$dimension*percent))),]??? ???
??? ??? ??? ??? })???
??? ??? ??? ??? })
??? lst4<- lapply(lst3,function(x){
??? ??...
2013 Jun 04
0
choose the lines2
...do this:
dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t")
dat.bru<- dat1[!is.na(dat1$evnmt_brutal),]
fun2<- function(dat){?
????? lst1<- split(dat,dat$patient_id)
??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,])
??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])
??? lst4<- lst3[lapply(lst3,nrow)!=0]
??? lst5<- lapply(seq_along(lst4),function(i){
???????????????????? do.call(rbind,lapply(which(lst4[[i]]$evnmt_brutal==1),function(x) {
????????????????????????????????????...
2013 Jun 07
4
matched samples, dataframe, panel data
I R-helpers
#I have a data panel of thousands of firms, by year and industry and
#one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not
#and another variable the represents the firm dimension (total assets in thousand of euros)
#I need to create two separated samples with the same number os firms where
#one firm in the first have a corresponding
2013 May 27
0
choose the lines
...ry this:
dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t")
dat.bru<- dat1[!is.na(dat1$evnmt_brutal),]
fun1<- function(dat){???
? ??? lst1<- split(dat,dat$patient_id)
??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,])
??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])
??? lst4<-lapply(lst3,function(x) {vect.brutal=c()
??? ??? ??? ??? for(line in which(x$evnmt_brutal==1)){
??? ??? ??? ??? ?? if(x$evnmt_brutal[line-1]==0){
??? ??? ??? ??? ? vect.brutal=c(vect.brutal,line)
??? ???...
2013 May 02
3
R issue with unequal large data frames with multiple columns
I'm a bit of an amateur R programmer. I can do simple R scenarios but my
handle on complex grammatical issues isn't steady.
I have 12 CSV files that I've read into dataframes. Each has 8 columns and
over 2000000 rows. Each dataframe has data associated by time component
and a date component in the format of:
X.DATE and then X.TIME
X.DATE is in the format of MMDDYYYY and X.TIME is
2013 Jun 30
0
Help: argument is not numeric or logical: returning NA
...t one having different dimensions.? I think it was
based on `summarized`.
Then, the second problem I encountered for testing is that the dataset you provided doesn't fulfill the criteria.
lst1<- list(structure(list......) #output of dput
?
lst2<- lst1[1:10] #deleted the last list element
lst3<-lapply(lst2,function(x)
{x$V3<- as.character(x$V3);x})? #here if I apply one of the
conditions, none of the list elements fulfill the criteria
#for example
lapply(lst3,function(x) subset(x,V1==6 & V6>300 & V6<3000,select="V6"))
#[1]]
#[1] V6
#<0 rows> (or 0-le...
2013 Feb 28
11
new question
Hi,
directory<- "/home/arunksa111/data.new"
#first function
filelist<-function(directory,number,list1){
setwd(directory)
filelist1<-dir(directory)
direct<-dir(directory,pattern = paste("MSMS_",number,"PepInfo.txt",sep=""), full.names = FALSE, recursive = TRUE)
list1<-lapply(direct, function(x) read.table(x,header=TRUE, sep =