arn to love lists.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of matteo
> Sent: Monday, June 24, 2013 11:32 AM
> To: Bert Gunter
> Cc: r-help@r-project
First of all, thanks for all the replies!!
What you have written helps, but is not entirely the answer to my problem.
What I'd have is the creation of new data.frames each of one named with
the ID of the original dataframe and with all the columns.
For example, in the original dataframe one co
Inline below...
On Mon, Jun 24, 2013 at 11:31 AM, matteo wrote:
> First of all, thanks for all the replies!!
> What you have written helps, but is not entirely the answer to my problem.
>
> What I'd have is the creation of new data.frames each of one named with the
> ID of the original dataframe
Oh yes, that's even better!
-- Bert
On Mon, Jun 24, 2013 at 10:33 AM, Rui Barradas wrote:
> Hello,
>
> I had forgotten the much simpler solutions. The following should do it.
>
> split(dataset, dataset$ID)
>
>
> Rui Barradas
>
> Em 24-06-2013 18:13, Bert Gunter escreveu:
>>
>> First of all, is y
On Jun 24, 2013, at 10:03 AM, matteo wrote:
> Hi,
>
>> result <- lapply(unique(dataset$ID), function(uid) dataset[dataset$ID ==
>> uid, ])
> Ok, I have the element result as a list
>
>> names(result) <- unique(dataset$ID)
> Nothing happens. I don't have any submatrix...
>
> Matteo
>
> _
Hello,
I had forgotten the much simpler solutions. The following should do it.
split(dataset, dataset$ID)
Rui Barradas
Em 24-06-2013 18:13, Bert Gunter escreveu:
First of all, is your data structure a matrix or a data frame? They
are different!
Assuming the latter, a shorter version of Rui'
Sorry for the blank message. The default behavior of the Mac Mail.app spell
checker has me confused.
On Jun 24, 2013, at 10:03 AM, matteo wrote:
> Hi,
>
>> result <- lapply(unique(dataset$ID), function(uid) dataset[dataset$ID ==
>> uid, ])
> Ok, I have the element result as a list
>
>> names(
Hello,
You don't have a sub-data.frame, what you have is a list, with each
element of that list a df. Try to see, for instance, result[[1]]. This
should be a data.frame corresponding to the first ID.
Rui Barradas
Em 24-06-2013 18:03, matteo escreveu:
Hi,
result <- lapply(unique(dataset$ID
First of all, is your data structure a matrix or a data frame? They
are different!
Assuming the latter, a shorter version of Rui's answer that avoids
unique() and automatically takes care of names is:
result <- by(dataset, dataset$ID,I)
See ?by, ?tapply, and ?split
-- Bert
On Mon, Jun 24, 2013
Hi,
result <- lapply(unique(dataset$ID), function(uid) dataset[dataset$ID
== uid, ])
Ok, I have the element result as a list
names(result) <- unique(dataset$ID)
Nothing happens. I don't have any submatrix...
Matteo
__
R-help@r-project.org mailing
Hello,
Try the following.
result <- lapply(unique(dataset$ID), function(uid) dataset[dataset$ID ==
uid, ])
names(result) <- unique(dataset$ID)
Hope this helps,
Rui Barradas
Em 24-06-2013 15:36, matteo escreveu:
Hi guys,
I'm a newby, so sorry for the easy question.
I have a matrix (459x2
Hi guys,
I'm a newby, so sorry for the easy question.
I have a matrix (459x28) in which a large number of observations are
repeated (same placed sampled in different times).
One of the columns is refers to the ID of the place of sampling.
What I would like is to extract subset matrix for every
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