[R] Extending each element in a list, or rbind()-ing arrays of different length without recycling

[Gabor Grothendieck]

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
<ggrothendieck at 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.23 at 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.
>>
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>>
>