Very nice!
On Sat, Feb 20, 2010 at 10:37 AM, Henrique Dallazuanna wrote:
> Try this:
>
> do.call(rbind, lapply(unstack(x, V2 ~ V1), '[', 1:max(with(x,
> tapply(V2, V1, length)
>
> On Sat, Feb 20, 2010 at 9:26 AM, Newbie19_02 wrote:
>>
>> Hi there,
>>
>> I think I'm struggling with a fairly s
David,
Because the recycling, is needed index each element from 1 to maximum
of length of the elements. ( '[' ).
On Sat, Feb 20, 2010 at 1:31 PM, David Winsemius wrote:
> Henrique;
>
> I wonder if you can comment on why that works as it does. I have tried to
> pull it apart and reassemble it pie
Dear Henrique,
THanks this works in the way that I need it to. THanks for your help.
Natalie
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Henrique;
I wonder if you can comment on why that works as it does. I have tried
to pull it apart and reassemble it piece by piece, but always end up
with argument recycling.
> do.call(rbind, c(unstack(x, V2 ~ V1)[1], unstack(x, V2 ~ V1)[3]))
[,1] [,2] [,3] [,4]
Try this:
do.call(rbind, lapply(unstack(x, V2 ~ V1), '[', 1:max(with(x,
tapply(V2, V1, length)
On Sat, Feb 20, 2010 at 9:26 AM, Newbie19_02 wrote:
>
> Hi there,
>
> I think I'm struggling with a fairly simple problem but can't seem to solve
> it. I have multiple observations for one unique
Hi Jim,
Your method works really well except that I lose the format of my dates and
I'm not sure how to fix this?
THanks,
Natalie
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Is this what you want:
> x <- read.table(textConnection("CAO0337134 05/09/95
+ CAO0337134 27/09/05
+ CAO0347741 10/10/04
+ CAO0347741 12/10/04
+ CAO0367128 11/07/05
+ CAO0367128 12/07/05
+ CAO0367128 14/07/05
+ CAO0367128 19/09/97
+ CAO0367128 20/09/97
Hi there,
I think I'm struggling with a fairly simple problem but can't seem to solve
it. I have multiple observations for one unique identifier. Ultimately I
want to end up with one line per identifier with multiple observations in
rows. I'm really stuck any help would be really appreciated.
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