[R] using mapply()

[dxc13]

Yes, that's it.  Thank you again!



Dimitris Rizopoulos wrote:
> 
> do you mean the following?:
> 
> out <- rowMeans(matrix(unlist(res), ncol = length(res)))
> dim(out) <- dim(res[[1]])
> out
> 
> 
> Best,
> Dimitris
> 
> ----
> Dimitris Rizopoulos
> Ph.D. Student
> Biostatistical Centre
> School of Public Health
> Catholic University of Leuven
> 
> Address: Kapucijnenvoer 35, Leuven, Belgium
> Tel: +32/(0)16/336899
> Fax: +32/(0)16/337015
> Web: http://med.kuleuven.be/biostat/
>       http://www.student.kuleuven.be/~m0390867/dimitris.htm
> 
> 
> Quoting dxc13 <dxc13 at health.state.ny.us>:
> 
>>
>> useR's,
>>
>> This is a follow up question to one I previously asked.  Consider the
>> 3-element list below
>>
>>> res
>> [[1]]
>>        [,1] [,2]  [,3]  [,4]  [,5] [,6]  [,7] [,8]  [,9] [,10] [,11]
>> [,12]
>> [,13] [,14]
>> [1,]   NA   NA   NA 1.25 0.25 0.75   NA   NA   NA    NA    NA    NA  1.25
>> 0.25
>> [2,] 1.25 0.25 0.75   NA   NA   NA   NA   NA   NA  1.25  0.25  0.75    NA
>> NA
>> [3,]   NA   NA   NA   NA   NA   NA 1.25 0.25 0.75    NA    NA    NA    NA
>> NA
>> [4,]   NA   NA 1.25 0.25 0.75   NA   NA   NA   NA    NA    NA  1.25  0.25
>> 0.75
>> [5,]   NA   NA   NA   NA   NA 1.25 0.25 0.75   NA    NA    NA    NA    NA
>> NA
>>
>> [[2]]
>>        [,1] [,2]  [,3] [,4]  [,5] [,6]  [,7]  [,8] [,9] [,10] [,11] [,12]
>> [,13] [,14]
>> [1,]   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA
>> 1.25
>> [2,] 1.25   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA
>> NA
>> [3,] 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25  0.25  0.25  0.25  0.25
>> 0.25
>> [4,]   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA
>> NA
>> [5,]   NA   NA   NA   NA   NA  1.25   NA   NA   NA    NA    NA    NA   
>> NA
>> NA
>>
>> [[3]]
>>        [,1] [,2]  [,3]  [,4] [,5] [,6]  [,7]  [,8] [,9] [,10] [,11] [,12]
>> [,13]  [,14]
>> [1,]   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA
>> 1.25
>> [2,] 0.25   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA
>> NA
>> [3,] 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25  0.25  0.25  0.25  0.25
>> 0.25
>> [4,]   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA    NA    NA    NA
>> NA
>> [5,]   NA   NA   NA   NA   NA  0.25   NA   NA   NA    NA    NA    NA   
>> NA
>> NA
>>
>> What I want to do is find the values that are not NA simultaneously for
>> each
>> value in each element of the list, and then average these numbers and
>> store
>> them in a 14x1 vector.  For example, in position (2,1) in each of the 3
>> elements of the res list, there is a value, 1.25 in [[1]] and [[2]], 0.25
>> in
>> [[3]].  Since there is a value in this position in all elements of the
>> list
>> I want to average these numbers: (1.25+1.25+0.25)/3 and store this value
>> in
>> a numeric vector.   So, if there are not 3 values to average, this means
>> that a given position (i,j) does not have a value in each element of the
>> list and thus I want to store NA in the 14x1 vector.  In the end, the
>> 14x1
>> vector will be mixed with values and NA's or NaN's.
>>
>> Does anyone know a efficient way to do this?  Maybe using mapply()?  I
>> hope
>> it is clear.  Thank you for any input.
>>
>> Derek
>> --
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>>
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>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
> 
> 
> 
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