Hi:
Here's another option:
rep(b, rle(a)$lengths)
> identical(a1, rep(b, rle(a)$lengths))
[1] TRUE
rle(a)$lengths computes a table of the number of consecutive repeats
of a number (or run lengths). It will have the same length as b in
this case. Using rep() with the table of lengths as repetiti
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 Lisa
> Sent: Wednesday, June 01, 2011 12:28 PM
> To: r-help@r-project.org
> Subject: Re: [R] Recode
Thank you, Duncan,
Here “a” has the length of 24, and “b” has the length of 20 with numbers
from 1 to 20 uniquely. I just want encode “a” from 1 to 20 based on “a”
current order using “b”. So, a1[1] = b[1] = 1
a1[2] = b[2] = 5
a1[3] = a1[4] = b[3] = 8 (since third and fourth numbers are the same i
(The attributions are a little messed up here:)
I have two sets of numbers that look like
a<- c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2, 3,
4)
b<- c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
19)
I just want to use “b” to encode “a” so that “a”
Thank you for your help, Pete. I tried b[a], but it is not a1.
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Dear all,
I have two sets of numbers that look like
a <- c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2, 3,
4)
b <- c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
19)
I just want to use “b” to encode “a” so that “a” looks like
a1<- c(1, 5, 8, 8, 9, 9,
Lisa wrote:
>
> Dear all,
>
> I have two sets of numbers that look like
>
> a <- c(1, 2, 3, 3, 4, 4, 5, 6, 1, 2, 2, 3, 1, 2, 1, 2, 3, 3, 4, 5, 1, 2,
> 3, 4)
>
> b <- c(1, 5, 8, 9, 14, 20, 3, 10, 12, 6, 16, 7, 11, 13, 17, 18, 2, 4, 15,
> 19)
>
> I just want to use “b” to encode “a” so that “a”
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