Re: [R] partial cumsum

2011-05-30 Thread mayouf.k
thank u very much Wiliam Dunlap, your function is very helpful, i been trying to find the right algo,,,and i saw yours, it works perfectly. i even checked with system.time function, and it's clear the cum.reset is the best. system.time(ave(J, rev(cumsum(rev(is.na(J, FUN=cumsum)) utilisateur

[R] partial cumsum

2009-11-11 Thread smu
Hello, I am searching for a function to calculate partial cumsums. For example it should calculate the cumulative sums until a NA appears, and restart the cumsum calculation after the NA. this: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) should become this: 1 3 6 NA 5 11 18 26 35 45 any

Re: [R] partial cumsum

2009-11-11 Thread William Dunlap
Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of smu Sent: Wednesday, November 11, 2009 7:58 AM To: r-help@r-project.org Subject: [R] partial cumsum Hello, I am

Re: [R] partial cumsum

2009-11-11 Thread Karl Ove Hufthammer
On Wed, 11 Nov 2009 08:53:50 -0800 William Dunlap wdun...@tibco.com wrote: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) should become this: 1 3 6 NA 5 11 18 26 35 45 Perhaps ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum) [1] 1 3 6 NA 5 11 18 26 35 45 Clever way of

Re: [R] partial cumsum

2009-11-11 Thread Karl Ove Hufthammer
On Wed, 11 Nov 2009 18:11:51 +0100 Karl Ove Hufthammer k...@huftis.org wrote: I wish cumsum took a 'na.rm' argument, though. It would make stuff like this much easier. Or, more specifically, a 'na.value', which could perhaps default to 'NA' to get the current behaviour, but which one could

Re: [R] partial cumsum

2009-11-11 Thread smu
On Wed, Nov 11, 2009 at 08:53:50AM -0800, William Dunlap wrote: Perhaps ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum) [1] 1 3 6 NA 5 11 18 26 35 45 it takes some time to understand how it works, but it's perfect. thank you, stefan

Re: [R] partial cumsum

2009-11-11 Thread William Dunlap
-Original Message- From: smu [mailto:m...@z107.de] Sent: Wednesday, November 11, 2009 9:26 AM To: William Dunlap Cc: r-help@r-project.org Subject: Re: [R] partial cumsum On Wed, Nov 11, 2009 at 08:53:50AM -0800, William Dunlap wrote: Perhaps ave(x, rev(cumsum(rev(is.na

Re: [R] partial cumsum

2009-11-11 Thread Tony Plate
William Dunlap wrote: Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of smu Sent: Wednesday, November 11, 2009 7:58 AM To: r-help@r-project.org Subject: [R] partial cumsum

Re: [R] partial cumsum

2009-11-11 Thread Carl Witthoft
quote: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) rev(cumsum(rev(is.na(x [1] 1 1 1 1 0 0 0 0 0 0 A more natural way to do this is cumsum(is.na(c(NA,x[-length(x)]))) [1] 1 1 1 1 2 2 2 2 2 2 endquote Both of which suggest the original problem could also be dealt with by using

Re: [R] partial cumsum

2009-11-11 Thread Marc Schwartz
On Nov 11, 2009, at 4:45 PM, Carl Witthoft wrote: By Bill Dunlap: quote: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) rev(cumsum(rev(is.na(x [1] 1 1 1 1 0 0 0 0 0 0 A more natural way to do this is cumsum(is.na(c(NA,x[-length(x)]))) [1] 1 1 1 1 2 2 2 2 2 2 endquote Both of which