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))
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!
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]]
>>>
>>> ______________________________**________________
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help>
>>> PLEASE do read the posting guide http://www.R-project.org/**
>>> posting-guide.html <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
>
>
[[alternative HTML version deleted]]
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