Hello,
The issue comes that 'apply' tries to coerce its argument to a matrix. This
means that all your columns will become character class, and the result
will not be what you wanted. I would suggest something more like:
sapply(d, function(x) all(x[!is.na(x)] <= 3))
or
vapply(d, function(x) all(x[!is.na(x)] <= 3), NA)
Also, here is a different method that might look cleaner:
sapply(d, function(x) all(x <= 3, na.rm = TRUE))
vapply(d, function(x) all(x <= 3, na.rm = TRUE), NA)
It's up to you which you choose. I hope this helps!
On Fri, Oct 8, 2021 at 1:50 PM Derickson, Ryan, VHA NCOD via R-help <
r-help@r-project.org> wrote:
> Hello,
>
> I'm seeing unexpected behavior when using apply() compared to a for loop
> when a character vector is part of the data subjected to the apply
> statement. Below, I check whether all non-missing values are <= 3. If I
> include a character column, apply incorrectly returns TRUE for d3. If I
> only pass the numeric columns to apply, it is correct for d3. If I use a
> for loop, it is correct.
>
> > d<-data.frame(d1 = letters[1:3],
> + d2 = c(1,2,3),
> + d3 = c(NA,NA,6))
> >
> > d
> d1 d2 d3
> 1 a 1 NA
> 2 b 2 NA
> 3 c 3 6
> >
> > # results are incorrect
> > apply(d, 2, FUN=function(x)all(x[!is.na(x)] <= 3))
> d1 d2 d3
> FALSE TRUE TRUE
> >
> > # results are correct
> > apply(d[,2:3], 2, FUN=function(x)all(x[!is.na(x)] <= 3))
> d2 d3
> TRUE FALSE
> >
> > # results are correct
> > for(i in names(d)){
> + print(all(d[!is.na(d[,i]),i] <= 3))
> + }
> [1] FALSE
> [1] TRUE
> [1] FALSE
>
>
> Finally, if I remove the NA values from d3 and include the character
> column in apply, it is correct.
>
> > d<-data.frame(d1 = letters[1:3],
> + d2 = c(1,2,3),
> + d3 = c(4,5,6))
> >
> > d
> d1 d2 d3
> 1 a 1 4
> 2 b 2 5
> 3 c 3 6
> >
> > # results are correct
> > apply(d, 2, FUN=function(x)all(x[!is.na(x)] <= 3))
> d1 d2 d3
> FALSE TRUE FALSE
>
>
> Can someone help me understand what's happening?
>
> ______________________________________________
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