Olu,

I think you may have misread what na.rm is supposed to do. I think you are 
getting the correct value. If you want the vectors that contain NA values to be 
evaluated to NA then you will need to set na.rm to FALSE (which is the default 
for prod()).

prod(c(7, 2, NA), na.rm = TRUE)
[1] 14
prod(c(7, 2, NA), na.rm = FALSE)
[1] NA

Mark
R. Mark Sharp, Ph.D.
Director of Data Science Core
Southwest National Primate Research Center
Texas Biomedical Research Institute
P.O. Box 760549
San Antonio, TX 78245-0549
Telephone: (210)258-9476
e-mail: msh...@txbiomed.org









> On Nov 11, 2016, at 2:45 PM, Olu Ola via R-help <r-help@r-project.org> wrote:
>
> Hello,I have a dataset that is similar to the one as follows:
>
>> Df.1 <- data.frame(A = c(5,4,7,6,8,4),B = 
>> (c(1,5,2,4,9,1)),C=(c(2,3,NA,5,NA,9)))
>> Df.1
>  A B  C
> 1 5 1  2
> 2 4 5  3
> 3 7 2 NA
> 4 6 4  5
> 5 8 9 NA
> 6 4 1  9
>> Df.1$D = apply(Df.1, 1, prod, na.rm=T)
>> Df.1$D[1]  10  60  14 120  72  36
>> Df.1
>  A B  C   D
> 1 5 1  2  10
> 2 4 5  3  60
> 3 7 2 NA  14
> 4 6 4  5 120
> 5 8 9 NA  72
> 6 4 1  9  36I intend to obtain a column D that takes into account na.rm=T but 
> 'apply' does not work in R 3.2.2
>
> A way forward will be greatly appreciated.
>
> [[alternative HTML version deleted]]
>
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