[R] apply is making me crazy...

[David Winsemius]

On Jul 28, 2011, at 12:31 PM, Gene Leynes wrote:

> (As I mentioned in my other reply to Dennis, I think I'll stick with  
> for loops, but I wanted to respond.)
>
> By "almost does it" I meant that using as.matrix helps because it  
> puts the vector into a column, that "almost does it” because half  
> the problem is that the output is a non dimensional vector when  
> apply is passed a matrix with one column.
>
> However, since the output of the apply function is transposed when  
> you’re doing row margins, the as.matrix doesn’t help because it’s  
> putting your result into a column, while the apply function is  
> putting everything else into rows. I tried several combination of  
> using t() before, after, and during (changing margin=1 to margin=2)  
> the function; but none did the trick.
>
> I was not as diligent about using your margin=1:1 suggestion in all  
> my trials, that didn't seem to be different from using margin=1.
>
> The problem is a bit hard to describe using a natural language, and  
> I think more apparent from the code.  Of course, that could my  
> shortcoming.
>
> I still think that the structure of the proposed solution, which I  
> think makes the problem apparent.
> > str(answerProposed)
> List of 3
>  $ : num [1:1000, 1] 0.5658 0.1759 1.2444 -0.0456 0.0236 ...
>  $ : num [1:2, 1:1000] 0.0392 0.7047 0.1834 -0.6644 -0.6952 ...
>  $ : num [1:3, 1:1000] -0.835 -0.0461 -0.1725 0.8365 0.7835 ...
> >

Sometimes I need to be hit over the head a few times for things to  
sink in. I hadn't noticed the reversal of dimensions in the "1" row  
case:

  answer.not.Bad  = lapply(exampBad, function(x) matrix(apply(x , 
1,cumsum),  ncol=nrow(x)))

 > str(answer.not.Bad)
List of 3
  $ : num [1, 1:1000] -0.159 -0.035 -0.386 -1.81 1.123 ...
  $ : num [1:2, 1:1000] -0.7801 0.6004 -0.0869 -0.1611 -0.3594 ...
  $ : num [1:3, 1:1000] -1.14 -2.81 -3.45 3.16 2.54 ...

The 1:1 dodge was useless, anyway. And just to be sure, you did want  
the row and col dimensions reversed? And you did want the first  
element to just be a (transposed) copy of its argument?

Are we good now?

-- 
david.

> I want it to do this:
> > str(answerDesired)
> List of 3
>  $ : num [1, 1:1000,] 0.5658 0.1759 1.2444 -0.0456 0.0236 ...
>  $ : num [1:2, 1:1000] 0.0392 0.7047 0.1834 -0.6644 -0.6952 ...
>  $ : num [1:3, 1:1000] -0.835 -0.0461 -0.1725 0.8365 0.7835 ...
> >
>
> There are a lot of reasons why I would want the apply function to  
> work this way, or at least have an option to work this way.  One  
> reason is so that you could perform do.call(rbind, mylist) at the  
> later
>
> I guess this behavior is described in the apply documentation:
> “If each call to FUN returns a vector of length n, then apply  
> returns an array of dimension c(n, dim(X)[MARGIN]) if n > 1. If  
> nequals 1, apply returns a vector if MARGIN has length 1 and an  
> array of dimension dim(X)[MARGIN] otherwise. If n is 0, the result  
> has length 0 but not necessarily the ‘correct’ dimension.”
>
> I just wish that it had an option to do return an array of dimension  
> c(n, dim(X)[MARGIN]) if n >= 1
>
> On Wed, Jul 27, 2011 at 8:25 PM, David Winsemius <dwinsemius at comcast.net 
> > wrote:
>
> On Jul 27, 2011, at 7:44 PM, Gene Leynes wrote:
>
>> David,
>>
>> Thanks for the suggestion, but I think your answer only works  
>> because I was printing the wrong thing (because apply with margin=1  
>> transposes the results,
>
> And if you want to change that,  then the t() function is readily at  
> hand.
>
>> something I always forget).
>>
>> Check this to see what I mean:
>>     str(answerGood)
>>     str(answerBad)
>>
>> Adding "as.matrix" is interesting and almost does it,
>
> "It" ... What is "it"? In a natural language,  ...  English  
> preferably.
>
> -- 
> david.
>
>> however the results are still transposed.
>>
>> Sorry to be confusing with the initial example.
>>
>> Here's an updated example (adding as.matrix doesn't make a  
>> difference)
>>
>>
>> ## Make three example matricies
>> exampGood = lapply(2:4, function(x)matrix(rnorm(1000*x),ncol=x))
>> exampBad  = lapply(1:3, function(x)matrix(rnorm(1000*x),ncol=x))
>> ## Two ways to see what was created:
>> for(k in 1:length(exampGood)) print(dim(exampGood[[k]]))
>> for(k in 1:length(exampBad)) print(dim(exampBad[[k]]))
>>
>> ##  Take the cumsum of each row of each matrix
>> answerGood =      lapply(exampGood, function(x) apply(x ,1,cumsum))
>> answerBad  =      lapply(exampBad, function(x) apply(x ,1,cumsum))
>> answerProposed  = lapply(exampBad, function(x) as.matrix(apply(x , 
>> 1:1,cumsum)))
>> str(answerGood)
>> str(answerBad)
>> str(answerProposed)
>>
>> ##  Take the first element of the final column of each answer
>> for(mat in answerGood){
>>     mat = t(mat)  ## To get back to 1000 rows
>>     LastColumn = ncol(mat)
>>     print(mat[2,LastColumn])
>> }
>> for(mat in answerBad){
>>     mat = t(mat)  ## To get back to 1000 rows
>>     LastColumn = ncol(mat)
>>     print(mat[2,LastColumn])
>> }
>> for(mat in answerProposed){
>>     mat = t(mat)  ## To get back to 1000 rows
>>     LastColumn = ncol(mat)
>>     print(mat[2,LastColumn])
>> }
>>
>>
>>
>> On Wed, Jul 27, 2011 at 5:45 PM, David Winsemius <dwinsemius at comcast.net 
>> > wrote:
>>
>> On Jul 27, 2011, at 6:22 PM, Gene Leynes wrote:
>>
>> I have tried a lot of ways around this, but I can't find a way to  
>> make apply
>> work in a generalized way because it causes a failure whenever  
>> reduces the
>> dimensions of its output.
>> The following example is easier to understand than the question.
>>
>> I wish it had a "drop=TRUE/FALSE" option like the "["  (and I wish  
>> I had
>> found the drop option a year ago, and I wish that I had 1e6  
>> dollars... Oops,
>> I mean euros).
>>
>>
>>   ## Make three example matricies
>>   exampGood = lapply(2:4, function(x)matrix(rnorm(1000*x),ncol=x))
>>   exampBad  = lapply(1:3, function(x)matrix(rnorm(1000*x),ncol=x))
>>   ## Two ways to see what was created:
>>   for(k in 1:length(exampGood)) print(dim(exampGood[[k]]))
>>   for(k in 1:length(exampBad)) print(dim(exampBad[[k]]))
>>
>>   ##  Take the cumsum of each row of each matrix
>>   answerGood = lapply(exampGood, function(x) apply(x ,1,cumsum))
>>   answerBad  = lapply(exampBad, function(x) apply(x ,1,cumsum))
>>
>> Try instead:
>>
>> answerBad  = lapply(exampBad, function(x) as.matrix(apply(x , 
>> 1:1,cumsum)))
>>
>>
>> I also find wrapping as.matrix() around vector results inside a  
>> print() call often makes my console output much more to my liking.
>>
>>
>>   str(answerGood)
>>   str(answerBad)
>>
>>   ##  Take the first element of the final column of each answer
>>   for(mat in answerGood){
>>       LastColumn = ncol(mat)
>>       print(mat[1,LastColumn])
>>   }
>>   for(mat in answerBad){
>>       LastColumn = ncol(mat)
>>       print(mat[1,LastColumn])
>>   }
>>
>>        [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>> David Winsemius, MD
>> West Hartford, CT
>>
>>
>
> David Winsemius, MD
> West Hartford, CT
>
>
>

David Winsemius, MD
West Hartford, CT