CODEWITHSUNDEEP
rif-statementsapply
Nakx
Nakx

Reputation: 1600

sapply() returns 'logical(0)' when x value is 0 in R

I am trying to make a small script to get rid of extreme values in a large dataset, but when my value is "0" my function returns "logical(0)" instead of NA.

#Getting rid of extreme values
test=NULL
test$value<-runif(200,13.90,14.10)
test$date<-seq(1,200,1)
test$value[125]<-15
test$value[175]<-0

plot(test$date, test$value)

averagei=NULL
averagetot=NULL
for (i in 1:length(test$value)) {
  averagei<-mean(test$value[i-15:i+15])
  averagetot=c(averagetot,averagei)

}

t<-sapply(test$value, function(x) ifelse(x - averagetot[x]>0.5, "NA", x))
t[175]
[[1]]
logical(0)

Why does this happen? I can of course get rid of the 0 before but it drives me crazy that I cannot understand why this happens or fix it.

EDIT: Thanks to Marius comment I fixed it using a for loop, might not be the most efficient but it does the job. Someone has a solution in a sapply loop but I could not make it work in my case: Row/column counter in 'apply' functions

test=NULL
test$value<-runif(200,13.90,14.10)
test$date<-seq(1,200,1)
test$value[125]<-15
test$value[175]<-0

plot(test$date, test$value)

a=NULL
atot=NULL
for (i in 1:length(test$value)){ 
  a<-ifelse(abs(test$value[i] - averagetot[i])>0.5, "NA", test$value[i])
  atot=c(atot,as.numeric(a))
    }

atot[175]
length(atot)

Upvotes: 1

Views: 603

Answers (1)

akrun
akrun

Reputation: 887148

The problem seems to be in the 

averagei<-mean(test$value[i-15:i+15])

It should be

averagei<-mean(test$value[(i-15):(i+15)])

Upvotes: 1

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