Try this:
my.array <- replicate(10, data.frame(matrix(sample(9), 3)), simplify =
FALSE)
#2
lapply(my.array, replace, list = 3, values = newThirdColumn)
The solution was for matrices, the above works with data.frames.
On Thu, Sep 3, 2009 at 5:03 PM, Carlos Hernandez wrote:
>
> On Thu, Sep 3, 2
On Thu, Sep 3, 2009 at 4:34 PM, Henrique Dallazuanna wrote:
> Try this:
>
> #1
> lapply(my.array, '[', , 3)
>
>
this works! thank you a lot!
> #2
> newThirdColumn <- sample(3)
> lapply(my.array, replace, list = 7:9, values = newThirdColumn)
>
>
i did not understand this last line, so far i could
Try this:
#1
lapply(my.array, '[', , 3)
#2
newThirdColumn <- sample(3)
lapply(my.array, replace, list = 7:9, values = newThirdColumn)
On Thu, Sep 3, 2009 at 11:16 AM, Carlos Hernandez wrote:
> Dear All,
> I created a list (of length Z) in the following way:
>
> my.array <- vector("list", Z)
>
Dear All,
I created a list (of length Z) in the following way:
my.array <- vector("list", Z)
then i assigned a matrix (of T rows by N columns) in each of the elements of
the list my.array in the following way:
my.array[[i]] <- matrix.data ##( matrix.data has dimensions TxN, and i
repeated this
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