[R] asking for help

Anthoni, Peter (IMK) peter.anthoni at kit.edu
Tue Apr 25 14:38:07 CEST 2017


Hi,

the cut function might be helpful.

vec=1: 163863
fcut=cut(vec,seq(1, 163863+1,by= 6069),include.lowest = T,right=F)
aggregate(vec,by=list(fcut),min)
aggregate(vec,by=list(fcut),max)

cheers
Peter



On 25. Apr 2017, at 14:33, PIKAL Petr <petr.pikal at precheza.cz<mailto:petr.pikal at precheza.cz>> wrote:

Hi

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Thomas
Mailund
Sent: Tuesday, April 25, 2017 11:20 AM
To: r-help at r-project.org<mailto:r-help at r-project.org>; Saifuddin, Miah Mohammad
<miah.mohammadsaifudd at mavs.uta.edu<mailto:miah.mohammadsaifudd at mavs.uta.edu>>
Subject: Re: [R] asking for help

If you write something like

indices <- rep(1:(163863/6069), each = 6069)

You can get similar result by

indices2 <- 0:163862%/%6069

but starting with zero.

or the same with
indices2 <- (0:163862%/%6069)+1

Cheers
Petr


you can get the i’th block of rows with

table[indices == i,]

It looks like a little more work than a loop would be since you have to run
through all rows for each block, but the implicit loop in this approach is likely
to be faster than an explicit for-loop.

Cheers
 Thomas

On 25 Apr 2017, 07.01 +0200, Saifuddin, Miah Mohammad
<miah.mohammadsaifudd at mavs.uta.edu<mailto:miah.mohammadsaifudd at mavs.uta.edu>>, wrote:
I have a data frame having 163863 values. I want to subset it so that each set
has 6069 values in it. for example 1:6069 as first, 6070: 6070+6068 as second.
how can I do that, preferably in a loop.


TIA

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