Here is one other solution. Its nearly the same as the last but
uses ts instead of zoo:
> all3 <- do.call(cbind, lapply(apply(peaks, 1, unique), ts))
> colnames(all3) <- make.names(1:ncol(all3))
> all3
Time Series:
Start = 1
End = 5
Frequency = 1
X1 X2 X3
1 2 10 1
2 7 6 9
3 9 8 7
4 NA NA 2
5 NA NA 0
or unclass(all3) if you don't want it as a ts object. (Similar comment
for the zoo solution.)
On Thu, Feb 12, 2009 at 3:16 PM, Gabor Grothendieck
<ggrothendi...@gmail.com> wrote:
> Try this. After the apply from your post we use lapply
> to make each series into a zoo series so that we can later
> use zoo's multiway merge. Finally we actually merge them
> and in the next statement just makes nice column names:
>
>> library(zoo)
>
>> all3 <- do.call(merge, lapply(apply(peaks, 1, unique), zoo))
>> colnames(all3) <- make.names(1:ncol(all3))
>> all3
> X1 X2 X3
> 1 2 10 1
> 2 7 6 9
> 3 9 8 7
> 4 NA NA 2
> 5 NA NA 0
>
>
> On Thu, Feb 12, 2009 at 2:31 PM, Jason Shaw <jason.shaw...@gmail.com> wrote:
>> Hi,
>>
>> I'm trying to take a matrix such as
>>
>> [,1] [,2] [,3] [,4] [,5]
>> [1,] 2 7 2 7 9
>> [2,] 10 10 6 8 6
>> [3,] 1 9 7 2 0
>>
>> and generate a new matrix which contains only the unique values in each row:
>>
>> [,1] [,2] [,3] [,4] [,5]
>> [1,] 2 7 9 NA NA
>> [2,] 10 6 8 NA NA
>> [3,] 1 9 7 2 0
>>
>> My problem is that I can use apply(matrix,MARGIN=1,FUN=unique) to find
>> the unique values, but this leaves me with a list with arrays of
>> different length:
>>
>>> x <- apply(peaks,MARGIN=1,FUN=unique)
>> [[1]]
>> [1] 2 7 9
>>
>> [[2]]
>> [1] 10 6 8
>>
>> [[3]]
>> [1] 1 9 7 2 0
>>
>> and using do.call("rbind",x) recycles the values of the shorter
>> vectors instead of filling them with NA:
>>
>>> do.call("rbind",x)
>> [,1] [,2] [,3] [,4] [,5]
>> [1,] 2 7 9 2 7
>> [2,] 10 6 8 10 6
>> [3,] 1 9 7 2 0
>>
>> So, I'd like to either take every element of the list and extend it
>> with NAs to the length of the longest element, or rbind every element
>> where missing places are filled with NAs instead of recycled values.
>> Is this possible? Of course, the solution is trivial using a loop,
>> but I'm trying to avoid this.
>>
>> Thanks for any suggestions.
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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