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in which column is an entry?

5 messages · Chuck Cleland, Christian Schulz, Brian Ripley

Christian Schulz
# 
Hi,

df  is a data.frame with 43 colums and 29877 rows with lot of NA.
I want the column number for all respondendts in one column 
where is the first entry >=0 as columnnumber.

my first step:
 time <- function(df)
+     {        for (i in 1:length(df [,1])) {    
+     which(df[i,1]:df[i,43] >= 0)
+     }
+            }
Error in df[i, 1]:df[i, 43] : NA/NaN argument

Many thanks for help , regards christian
Chuck Cleland
#
Christian Schulz wrote:
I am not sure, but I think you might want something like this:

t1 <- apply(df, 1, function(x){
             ifelse(all(is.na(x)) | all(na.omit(x) < 0),
             NA, which(x >= 0))})

hope this helps,

Chuck Cleland
Christian Schulz
# 
Yes, many thanks i have really to  avoid think in loops  :-)
christian


Am Samstag, 31. Januar 2004 16:57 schrieb Chuck Cleland:
Brian Ripley
#
On Sat, 31 Jan 2004, Christian Schulz wrote:

            

      
      
      
    

        
Unfortunately Chuck's solution is a loop over rows, disguised by the use 
of apply.

Let us assume that the dataframe has all numeric entries and coerce to a 
matrix (as apply will, BTW).

tmp <- as.matrix(df)
tmp[is.na(tmp)] <- -1   # get rid of the NAs
tmp <- tmp >= 0         # a logical matrix
tmp <- cbind(tmp, TRUE) # add a fence column

I am happy to loop over 43 columns, though, so

for(i in 2:44) tmp[, i] <- tmp[, i] | tmp[, i-1]
for(i in 44:2) tmp[, i] <- tmp[, i] & !tmp[, i-1]
rtmp <- t(tmp)
z <- row(rtmp)[rtmp]
z[z==44] <- NA
z

is what you want.  It's a lot faster (about 12x).

            

      
      
      
    

        

      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Christian Schulz
    

    

      
      #
    

  


  

      
Thanks, speed is sometimes reaaly important , but
to understand in a advanced practice guide about  loops.
i'm always not sure when indexing is necessary for the objects in the loop:
Different versions accept as function but didn'T run without error.

many thanks & regards,
christian

special <-  function(const,modeldat,YS) {
                  for(i in 1:10) {
const[i]=const[i]+1   #i'know  only here indexing make no sense!?
t1 <-  apply(YS,1, function(x) {  ifelse(all(is.na(x)) | all(na.omit(x)<0),
                                NA, which( x > const[i]))})
t1[is.na(t1)] <-  13
t2 <-  sapply(t1,function(x) { ifelse(x ==13,0,1)})
modeldat$MONTH <- t1
modeldat$ACTIVE <- t2
modeldats <- na.omit(modeldat)
mod1<-  coxph(Surv(MONTH,ACTIVE)  ~ ALTER+LEISTUNG,data=modeldats)
pdf(file = "~/Survival.pdf",    width = 6, height = 6, onefile = TRUE, family 
= "Helvetica",title = "R Graphics Output")
plot(survfit(mod1),ylim=c(.7,1),xlab='Month',ylab='Proportion not Active') }
                dev.off() }

Error in fitter(X, Y, strats, offset, init, control, weights = weights,  : 
	NA/NaN/Inf in foreign function call (arg 6)
In addition: Warning message: 
Ran out of iterations and did not converge in: fitter(X, Y, strats, offset, 
init, control, weights = weights,  



Am Samstag, 31. Januar 2004 18:54 schrieben Sie: