Thanks, Peter. Why not cbind your idea for the first column with my idea
for the second column and get it done in one line?:
v <- c(1,2,5,6,7,8,25,30,31,32,33)
M <- cbind( v[ c(1, which( diff(v) !=1 ) + 1 ) ] , rle( v - 1:length(v)
)$lengths )
M
[,1] [,2]
[1,] 1 2
[2,] 5 4
[3,] 25 1
[4,] 30 4
I find that pretty appealing and I'll probably stick with it. It seems
quite fast. Here's an example:
# make fairly long vector
v <- sort(unique(round(100000*runif(100000))))
length(v)
[1] 63274
# time the procedure:
ptm <- proc.time() ; M <- cbind( v[ c(1, which( diff(v) !=1 ) + 1 ) ] , rle( v
- 1:length(v) )$lengths ) ; proc.time() - ptm
user system elapsed
0.03 0.00 0.03
dim(M)
[1] 23212 2
I probably won't be using vectors any longer than that, and this isn't the
kind of thing that I do over and over again, so that speed is excellent.
Mike
On Mon, 5 Jan 2015, Peter Alspach wrote:
Tena koe Mike
An alternative, which is slightly fast:
diffv <- diff(v)
starts <- c(1, which(diffv!=1)+1)
cbind(v[starts], c(diff(starts), length(v)-starts[length(starts)]+1))
Peter Alspach
-----Original Message-----
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Mike Miller
Sent: Monday, 5 January 2015 1:03 p.m.
To: R-Help List
Subject: [R] counting sets of consecutive integers in a vector
I have a vector of sorted positive integer values (e.g., postive integers after
applying sort() and unique()). For example, this:
c(1,2,5,6,7,8,25,30,31,32,33)
I want to make a matrix from that vector that has two columns: (1) the first
value in every run of consecutive integer values, and (2) the corresponding
number of consecutive values. For example:
c(1:20) would become this...
1 20
...because there are 20 consecutive integers beginning with 1 and
c(1,2,5,6,7,8,25,30,31,32,33) would become
1 2
5 4
25 1
30 4
What would be the best way to accomplish this? Here is my first effort:
v <- c(1,2,5,6,7,8,25,30,31,32,33)
L <- rle( v - 1:length(v) )$lengths
n <- length( L )
matrix( c( v[ c( 1, cumsum(L)+1 ) ][1:n], L), nrow=n)
[,1] [,2]
[1,] 1 2
[2,] 5 4
[3,] 25 1
[4,] 30 4
I suppose that works well enough, but there may be a better way, and besides, I
wouldn't want to deny anyone here the opportunity to solve a fun puzzle. ;-)
The use for this is that I will be doing repeated seeks of a binary file to
extract data. seek() gives the starting point and readBin(n=X) gives the
number of bytes to read. So when there are many consecutive variables to be
read, I can multiply the X in n=X by that number instead of doing many
different seek() calls. (The data are in a transposed format where I read in
every record for some variable as sequential elements.) I'm probably not the
first person to deal with this.
Best,
Mike
--
Michael B. Miller, Ph.D.
University of Minnesota
http://scholar.google.com/citations?user=EV_phq4AAAAJ
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