Displaying 6 results from an estimated 6 matches for "lst4".
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2013 Jun 04
0
choose the lines2
...ot;\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) {
??? ??? ??? ??? ??? ??? ??? ??? ??? ??? x1<-c(x-2,x-1,x)
??? ??? ??? ??? ??? ??? ??? ??? ??? ??? x2<-x1[!any(x1==0)]
??? ?...
2013 Jun 08
0
data
...t1New[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$dimension*percent))),]??? ???
??? ??? ??? ??? })???
??? ??? ??? ??? })
??? lst4<- lapply(lst3,function(x){
??? ??? ??? ???? ?? lst<- lapply(x,function(y){
??? ??? ??? ??? ??? ??? ??? y[!all(y$dummy==0),]
??? ??? ??? ??? ??? ??? ??? ???? })
??? ??? ??? ??? ?? lstNew<- lst[lapply(lst,nrow)>1]
??? ??? ??? ??? ?? lstNew1<- unique(lstNew)
??? ??? ??? ??? ?? })
??????...
2013 Jun 04
0
choose the lines2
...ot;\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) {
??????????????????????????????????????? x1<-c(x-2,x-1,x)
??????????????????????????????????????? x2<-x1[!any(x1==0)]
?????...
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
...t;\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$evnmt_brutal[line-1]==0){
??? ??? ??? ??? ? vect.brutal=c(vect.brutal,line)
??? ??? ??? ??? ??? }
??? ??? ??? ??? ??? ?? }
??? ??? ??? ????? vect.brutal1<- sort(c(vect.brut...
2013 Jun 30
0
Help: argument is not numeric or logical: returning NA
...gt;300 & V6<3000,select="V6"))
#[1]]
#[1] V6
#<0 rows> (or 0-length row.names)
#
#[[2]]
#[1] V6
#<0 rows> (or 0-length row.names)
#
#[[3]]
#[1] V6
#<0 rows> (or 0-length row.names)
---------------------------------------------
#changed the V1 column
?set.seed(24)
lst4<- lapply(lst3,function(x) {x$V1<- sample(c(0,3,6),nrow(x),replace=TRUE);x})
#I am skipping the first 2, ie. 'id' and 'iat.date' as the filenames were not provided.
res<-t(sapply(lst4,function(x)
{block4<-x[with(x,V1==3 & V6>300 &
V6<3000),"V6"]...