On Sat, 2011-03-19 at 15:58 +0100, Bodnar Laszlo EB_HU wrote:
> Hi,
I'll top-post as the original Q is very lengthy:
tabs <-lapply(df[,2:6],
function(x, id){ t(table(addNA(x), id, useNA = "ifany")) }, df$id)
is one way of doing what you want. More details are here:
http://stackoverflow.com/questions/5362702/persuading-tabulate-function-to-count-nas-in-a-data-frame-in-r
where you also posted your Q.
HTH
G
> I'd like to ask you a question again. It is basically about data frames, NAs
> and tabulate function.
>
> I have this data frame. I already used this in one of the previous questions
> of mine. It intentionally looks this simple, my real 'df' dataframe is much
> bigger actually and again, I am not willing to annoy anyone with huge
> databases... So, my database:
>
> id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
> a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
> b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
> c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
> d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
> e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,4)
> df <-data.frame(id,a,b,c,d,e)
> df
>
> I have managed to calculate the distributions of the numbers occurring in
> columns 'b' to 'e' but considering the fact at the very same time that these
> distributions should be 'groupped by' the id numbers in column 'id'. It works
> fine, check it ->
>
> matrix(matrix(unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,2]))))
> [[1]])),ncol=3,nrow=3,byrow=TRUE)
> matrix(matrix(unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,3]))))
> [[2]])),ncol=3,nrow=3,byrow=TRUE)
> matrix(matrix(unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,4]))))
> [[3]])),ncol=3,nrow=3,byrow=TRUE)
> matrix(matrix(unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,5]))))
> [[4]])),ncol=3,nrow=3,byrow=TRUE)
> matrix(matrix(unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,6]))))
> [[5]])),ncol=4,nrow=3,byrow=TRUE)
>
> Now my problem is: what if my data frame contains NA values here and there
> and what if I want my in-built tabulate function to collect these NAs as
> well? So what if I want it to count how many occurrences I have from these
> NAs?
>
> Here's my modified data frame with the NAs:
> id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
> a <-c(NA,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
> b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
> c <-c(1,3,2,3,2,1,2,3,3,2,2,3,NA,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
> d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
> e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,NA,1,4)
> df <-data.frame(id,a,b,c,d,e)
> df
>
> At first I tried something like this (you see, the only thing I did was that
> I tried to apply this "exclude=NULL" thing).
> unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,2],exclude=NULL)))) [[1]])
>
> At least my code realizes the fact that I have 4 different levels in column
> 'a' (1,2,3,NA) and not only three (1,2,3). Check it here:
> nlevels(factor(df[,2],exclude=NULL))
>
> But you see in the result that somehow it could not calculate the NAs. It says
> 3 0 6 0(!) 4 3 3 0 4 1 5 0
>
> Instead of the correct:
> 3 0 6 1(!) 4 3 3 0 4 1 5 0
>
> Or in case of:
> unlist(lapply(df[,(-(1))],function(x)
> tapply(x,df$id,tabulate,nbins=nlevels(factor(df[,4],exclude=NULL)))) [[3]])
>
> It says
> 2 4 4 0 2 3 4 0(!) 1 5 4 0
>
> Instead of the correct
> 2 4 4 0 2 3 4 1(!) 1 5 4 0
> etc.
>
> Does someone have any ideas how to "persuade" the function tabulate to count
> NAs? Is it possible at all?
> Thanks very much and have a pleasant weekend,
> Laszlo
>
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