mapply(function(x1,x2)x1[[x2]],all.predicted.values,max.growth,SIMPLIFY=FALSE)
gives a list of factors.
On 10/18/2010 8:40 PM, Gregory Ryslik wrote:
> Hi Everyone,
>
> This is closer to what I need but this returns me a matrix where each
> element is a factor. Instead I would want a list of lis
Does this do what you want:
> x <- lapply(seq_along(MaxGrowth), function(.num){
+ AllPredictedValues[[.num]][[MaxGrowth[[.num
+ })
> x
[[1]]
[1] 2 1 2 2 2 2 0 2 2 0 0 2 2 0 0 2 2 1 2 0 1 1 0 0 0 2 0 0 0 2 2 0
0 1 0 0 2 0 1 0 2 0 0 2 1 0 0 0 2 1 0 2 2
[54] 2 2 0 2 0 1 0 2 0 1 0 0 0 1 0 0
Hi,
It seems that the files did not make it through the mailer. Perhaps it didn't
like my extensions. I have now attached the files as .txt's as well as copied
in the contents of each file:
list(list(structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
Hi Everyone,
This is closer to what I need but this returns me a matrix where each element
is a factor. Instead I would want a list of lists. The first entry of the list
should equal the first column of the matrix that mapply makes, the second entry
to the second column etc...
I've attached th
You probably need mapply since you have 2 list of arguments which you want to
use "in sync"
mapply(function(x1,x2)x1[[x2]],all.predicted.values,max.growth)
might be what you want.
On Oct 18, 2010, at 5:17 PM, Gregory Ryslik wrote:
> Unfortunately, that gives me null everywhere. Here's the da
Try posting your data using 'dput' so it is easily read for testing.
On Mon, Oct 18, 2010 at 11:17 AM, Gregory Ryslik wrote:
> Unfortunately, that gives me null everywhere. Here's the data I have for
> all.predicted.values and max.growth. Perhaps this will help. Thus I want
> all.predicted.valu
Unfortunately, that gives me null everywhere. Here's the data I have for
all.predicted.values and max.growth. Perhaps this will help. Thus I want
all.predicted.values[[1]][[4]] then all.predicted.values[[2]][3]] and then
all.predicted.values[[3]][[4]].
I've attached what your statement outputs
Try this:
diag(sapply(all.predicted.values, '[[', 'max.growth'))
On Mon, Oct 18, 2010 at 12:59 PM, Gregory Ryslik wrote:
> Hi,
>
> I have a list of n items and the ith element has m_i elements within it.
>
> I want to do something like:
>
> predicted.values<- lapply(all.predicted.values,'[[',m
Hi,
I have a list of n items and the ith element has m_i elements within it.
I want to do something like:
predicted.values<- lapply(all.predicted.values,'[[',max.growth[[i]])
Where max.growth[[i]] is the element I want to extract from each of the ith
predicted elements. Thus, for example, I wa
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