[R] summarize a vector

[David Winsemius]

On Aug 10, 2012, at 3:42 PM, Michael Weylandt wrote:

> I wouldn't be surprised if one couldn't get an *apply-free solution  
> by using diff(), cumsum() and selective indexing as well.

What about colSums on a matrix extended with the right number of zeros.

 > colSums(matrix (c(v, rep(0, 3- length(v)%%3) ) , nrow=3) )
[1]  6 15 24 10

(My experience is that tapply is generally fairly fast anyway, much  
faster than apply.data.frame. So I do not lump all *apply solutions in  
the same efficiency category.)

-- 
David.
>
> Cheers,
> Michael
>
> On Aug 10, 2012, at 5:07 PM, David Winsemius  
> <dwinsemius at comcast.net> wrote:
>
>>
>> On Aug 10, 2012, at 12:57 PM, Bert Gunter wrote:
>>
>>> ... or perhaps even simpler:
>>>
>>>> sz <- function(x,k)tapply(x,(seq_along(x)-1)%/%k, sum)
>>>> sz(1:10,3)
>>> 0  1  2  3
>>> 6 15 24 10
>>>
>>> Note that this works for k>n, where the previous solution does not.
>>>> sz(1:10,15)
>>> 0
>>> 55
>>
>> I agree that it is more elegant, but I do not get an error or an  
>> unexpected result with my method.
>>
>>> N=10
>>> k=15
>>> w <- tapply( v ,rep(1:(N/k +1), each=k, len=N ) , sum)
>>> w
>> 1
>> 55
>>
>> A different label but the same result. I'm protected from the  
>> typical 1:0 problem that seq_along solves by including +1 in the  
>> second argument to ":"/seq(). Unless, of course, you set N to a  
>> negative number, but that wouldn't make much sense would it, and  
>> you get an error from rep() anyway.
>>
>> Best;
>> David.
>>
>>>
>>> -- Bert
>>>
>>> On Fri, Aug 10, 2012 at 12:37 PM, David Winsemius
>>> <dwinsemius at comcast.net> wrote:
>>>>
>>>> On Aug 10, 2012, at 12:20 PM, Sam Steingold wrote:
>>>>
>>>>> I have a long numeric vector v (length N) and I want create a  
>>>>> shorter
>>>>> vector of length N/k consisting of sums of k-subsequences of v:
>>>>>
>>>>> v <- c(1,2,3,4,5,6,7,8,9,10)
>>>>>
>>>>> N=10, k=3
>>>>> ===> [6,15,24,10]
>>>>>
>>>>> I can, of course, iterate:
>>>>>
>>>>>> w <- vector(mode="numeric",length=ceiling(N/k))
>>>>>> for (i in 1:length(w)) w[i] <- sum(v(i*k:(i+1)*k))
>>>>>
>>>>>
>>>>> (modulo boundary conditions)
>>>>> but I wonder if there is a better way.
>>>>
>>>>
>>>> Well, using v with parentheses instead of square-brackets might  
>>>> not be the
>>>> right way, since v is not a function.
>>>>
>>>> Consider this alternate (no need to pre-allocate 'w'):
>>>>
>>>>> w <- tapply( v ,rep(1:(N/k +1), each=k, len=N ) , sum)
>>>>> w
>>>> 1  2  3  4
>>>> 6 15 24 10
>>>>
>>>> --
>>>>
>>>> David Winsemius, MD
>>>> Alameda, CA, USA
>>>>
>>>> ______________________________________________
>>>> R-help at 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.
>>>
>>>
>>>
>>> -- 
>>>
>>> Bert Gunter
>>> Genentech Nonclinical Biostatistics
>>>
>>> Internal Contact Info:
>>> Phone: 467-7374
>>> Website:
>>> http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
>>
>> David Winsemius, MD
>> Alameda, CA, USA
>>
>> ______________________________________________
>> R-help at 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.

David Winsemius, MD
Alameda, CA, USA