thats beautiful
> apply(m[, 3:14], 2,
+ function(x) tapply(x, m[,2], function(x) sum(!is.na(x))))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
1800 3 3 2 3 3 2 3 3 3 3 3 3
1801 3 3 2 3 3 2 3 3 3 3 3 3
1802 3 3 3 3 3 3 2 3 3 3 3 3
1803 3 3 3 3 3 2 3 3 3 3 3 3
1804 3 3 3 3 3 2 3 3 3 3 3 3
i was thinking of doing 'by' inside an apply, but this is perfect. thx
On Wed, Jun 9, 2010 at 6:16 PM, Erik Iverson <er...@ccbr.umn.edu> wrote:
> Hello,
>
>
> steven mosher wrote:
>
>> # create a matrix with some random NAs in it
>>
>>> m<-matrix(NA,nrow=15,ncol=14)
>>> m[,3:14]<-52
>>> m[13,9]<-NA
>>> m[4:7,8]<-NA
>>> m[1:2,5]<-NA
>>> m[,2]<-rep(1800:1804, by=3)
>>> y<-order(m[,2])
>>> m<-m[y,]
>>> m[,1]<-rep(1:3,by=5)
>>>
>>
>
>> # what we want is a result that looks like this
>> 1800 3 3 2 3 3 2 3 3 3 3 3 3
>> 1801 3 3 2 3 3 2 3 3 3 3 3 3
>> 1802 3 3 3 3 3 3 2 3 3 3 3 3
>> 1803 3 3 3 3 3 2 3 3 3 3 3 3
>> 1804 3 3 3 3 3 2 3 3 3 3 3 3
>>
>>
> This should work:
>
> apply(m[, 3:14], 2,
> function(x) tapply(x, m[,2], function(x) sum(!is.na(x))))
>
> It uses tapply inside of apply to break up the groups by m[, 2].
>
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