Yes, I meant to say drop=FALSE
Also, I made a mistake in my "desired answer" structure example, sorry for
that confusion.
The apply results when margin=1 are very unintuitive. That transposition
issue has caused me numerous headaches. I think it's a design error, but
that changing it would be a disaster.
On Thu, Jul 28, 2011 at 2:45 PM, David Winsemius <dwinsem...@comcast.net>wrote:
>
> On Jul 28, 2011, at 3:13 PM, Gene Leynes wrote:
>
> Very clever (as usual)
It works, but since I wanted to switch the rows
>> and columns, which would require this:
>> answer.slightly.clumsy =
>> lapply(exampBad, function(x) matrix(apply(x ,1, cumsum),
>> ncol=nrow(x)))
>>
>> However, with a slight modification of your code I can use a wrapper
>> function for apply. This gives me the functionality, with clean syntax. I
>> will probably add to my standard library.
>> apply1 = function(mat, fun){
>> matrix(apply(mat ,1, fun), nrow=nrow(mat), byrow=T)
>> }
>> res = lapply(exampBad, function(x) apply1(x , cumsum))
>>
>
> Hmmm. I originally wrote nrow=nrow(mat) but changed it after it did not
> give your specified output. But `apply` typically transposes its results so
> you are using this matrix property: t(t(.)) == (.)
>
>
>
>> As I mentioned in my email to Dennis, I am typically dealing with highly
>> nested lists of matrices that have one thing in common: 1000 rows. This
>> fact makes fact the lapply, sapply, apply, and do.call family extremely
>> effective, and makes this apply problem something that is worth solving.
>>
>> PS: I still wish that there were a drop=TRUE option in apply!
>>
>
> ITYM ..., drop=FALSE, since drop = TRUE is the default behavior that causes
> loss of matrix dimensions in "[".
>
> --
> david.
>
>
>
>> Thanks again,
>>
>> Gene
>>
>> On Thu, Jul 28, 2011 at 12:05 PM, David Winsemius <dwinsem...@comcast.net>
>> wrote:
>>
>> 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 youre
>> doing row margins, the as.matrix doesnt help because its 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 <dwinsem...@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 <dwinsem...@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]]
>>
>
> David Winsemius, MD
> West Hartford, CT
>
>
[[alternative HTML version deleted]]
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