[R] Problem with lapply

[arun]

Hi,

No problem.
Regarding your code:

#changed name from 'dataset' to "Testcor"
#Your code

  firm.names=colnames(Testcor)[2:length(colnames(Testcor))]
 firm.names
# [1] "PAXN"   "PED"    "PEDP"   "PM"     "PRFN"   "PRWN"   "ZG"     "ZUBN"  
 #[9] "ZURN"   "ZWM"    "Market" #####includes "Market"

#If you look my code:
firm.names <- colnames(Testcor)[!grepl("DATE|Market",colnames(Testcor))]
 firm.names
# [1] "PAXN" "PED"  "PEDP" "PM"   "PRFN" "PRWN" "ZG"   "ZUBN" "ZURN" "ZWM" 

If you wanted to use the "formula" method:
res1 <- sapply(firm.names,function(x) { cor.results <- cor.test(as.formula(paste("~", paste(x,"Market", sep="+")) ),data=Testcor,na.action=na.exclude); cor.results$estimate})
res1
#   PAXN.cor     PED.cor    PEDP.cor      PM.cor    PRFN.cor    PRWN.cor 
#-0.31113122 -0.09359550 -0.17056943  0.40025112  0.34385888  0.17935045 
 #    ZG.cor    ZUBN.cor    ZURN.cor     ZWM.cor 
 #0.20762797  0.39238270  0.74336028 -0.09166795 
 identical(res,res1)
#[1] TRUE

#or just
res2 <- sapply(firm.names,function(x) { cor.test(~get(x) + Market,data=Testcor,na.action=na.exclude)$estimate })
 identical(res,res2)
#[1] TRUE



A.K.






On Tuesday, October 15, 2013 3:15 PM, rissa <rissa_r at gmx.ch> wrote:
Thank you! Worked perfectly!

By the way: What was wrong with my original code?



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