Hi,
If you have a vector of values to compare:
thresh1<- c(30,4,12,65,5)
indx<-findInterval(thresh1-1,cumsum(X))
indx2<-ave(rep(indx,indx),rep(indx,indx),FUN=seq)
X[indx2]
# [1] 1 3 4 5 8 1 1 3 4 1 3 4 5 8 15 1 3
#you can split this into a list
split(X[indx2],cumsum(c(TRUE,diff(indx2)<=0)))
#$`1`
#[1] 1 3 4 5 8
#
#$`2`
#[1] 1
#
#$`3`
#[1] 1 3 4
#
#$`4`
#[1] 1 3 4 5 8 15
#
#$`5`
#[1] 1 3
A.K.
----- Original Message -----
From: "gildororo...@mail-on.us" <gildororo...@mail-on.us>
To: r-help@r-project.org
Cc:
Sent: Saturday, September 14, 2013 10:36 PM
Subject: [R] accumulate() function in R?
I came from Python, newly learning R. is there something like
accumulate() in R?
Example:
accumulate([1,2,3,4,5]) --> 1 3 6 10 15
Or perhaps I should show the problem. The problem I am trying to
solve, is to select elements from X until it accumulate to 30. My
solution is:
> X = c(1,3,4,5,8,15,35,62,78,99)
> X[sapply(seq_len(length(X)), function(x) { sum(X[1:x])}) < 30]
[1] 1 3 4 5 8
Is this already the shortest/canonical way to do it in R?
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and provide commented, minimal, self-contained, reproducible code.
