r-help@r-project.org" ; "s...@gnu.org"
; Bert Gunter
Sent: Friday, August 10, 2012 6:58 PM
Subject: Re: [R] summarize a vector
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
> dif
Certainly ... but this is of course limited to the few C coded
functions available. Back to apply-type stuff for, say, median as a
summary statistic.
-- Bert
On Fri, Aug 10, 2012 at 3:58 PM, David Winsemius wrote:
>
> On Aug 10, 2012, at 3:42 PM, Michael Weylandt wrote:
>
>> I wouldn't be surpri
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(
I wouldn't be surprised if one couldn't get an *apply-free solution by using
diff(), cumsum() and selective indexing as well.
Cheers,
Michael
On Aug 10, 2012, at 5:07 PM, David Winsemius wrote:
>
> On Aug 10, 2012, at 12:57 PM, Bert Gunter wrote:
>
>> ... or perhaps even simpler:
>>
>>> sz
Oh yes, I stand corrected. I didn't look at your code carefully enough.
-- Bert
On Fri, Aug 10, 2012 at 3:07 PM, David Winsemius 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
Oh yes, I stand corrected. I didn't look at your code carefully enough.
-- Bert
On Fri, Aug 10, 2012 at 3:07 PM, David Winsemius 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
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
Thanks David & Bert.
It turned out that what I actually wanted was much simpler.
my vector's elements are 0&1 and the right way to "summarize" it is
hist(which(v==1))
however, your replies were quire educational!
Thanks again,
Sam.
> * Bert Gunter [2012-08-10 12:57:40 -0700]:
>
>> sz <- function(
... 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
-- Bert
On Fri, Aug 10, 2012 at 12:37 PM, David Winsemius
wrote:
>
> On Aug 10, 201
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="num
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