Quoting "Zhang Weiwu" :
Jim Holtman asked me to elaborate the problem:
It is a common problem in reading sparse variable-lenght record data
file. Records are stored in file one next to another. The length of
each record is known in advance, but a lot of them records are invalid,
Quoting "PIKAL Petr" :
diff(c(0,which(1:10 %in% c(3, 5, 6, 10
[1] 3 2 1 4
That solves the problem! Thanks.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org
I can do this:
1:10 %in% c(3, 5, 6, 10)
[1] FALSE FALSE TRUE FALSE TRUE TRUE FALSE FALSE FALSE TRUE
but what I wish to get is:
[1] 3 2 1 4
let me explain:
3 # [1:3] ends with TRUE, i.e. FALSE FALSE TRUE
2 # [4:5] ends with TRUE, i.e. FALSE TRUE
1 # [6:6] ends with TRUE, i.e. TRUE
4 #
Quoting "Gabor Grothendieck" :
Try this:
library(sqldf)
Thank you for introducing me to SQLDF package! I wasn't aware I could
use SQL in R!
Right now Arun sent me (forgetting to cc the list) a solution using
apply/sapply, which inspired me of my own version:
x <- which(sapply(seq_len
Quoting "Berend Hasselman" :
Your trim-point in B is not unique (at least for the data you provided).
Indeed. It's quit a surprise to me. I couldn't figure out why my
evaluation expression results in multiple trimp-points
In a general way, I am looking for x and y, where:
A[x, condition]
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
This is better explained by example:
A <- data.frame(force = sort(runif(10, 0, 1)), condition =
sort(sample(0:100, 10)))
B <- data.frame(counterforce = sort(runif(15, 0, 1), decreasing=T),
condition = sort(sample(0:100, 15)))
So we have:
A
force condition
1 0.03515542 1
7 matches
Mail list logo
