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 <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])
}
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David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
David Winsemius, MD
West Hartford, CT
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
