On Wed, Feb 08, 2012 at 05:01:01PM -0500, Sam Steingold wrote:
> loop is too slow.
> it appears that sparseMatrix does what I want:
>
> ll <- lapply(l,length)
> i <- rep(1:4, ll)
> vv <- unlist(l)
> j1 <- as.factor(substring(vv,1,1))
> t <- table(j1)
> j <- position of elements of j1 in names(t)
>
loop is too slow.
it appears that sparseMatrix does what I want:
ll <- lapply(l,length)
i <- rep(1:4, ll)
vv <- unlist(l)
j1 <- as.factor(substring(vv,1,1))
t <- table(j1)
j <- position of elements of j1 in names(t)
sparseMatrix(i,j,x=as.numeric(substring(vv,2,2)), dimnames = names(t))
so, the qu
rhead for such
small input vectors).
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 Sam Steingold
> Sent: Wednesday, February 08, 2012 12:56 PM
> To: r-hel
On Wed, Feb 08, 2012 at 03:56:12PM -0500, Sam Steingold wrote:
> To be clear, I can do that with nested for loops:
>
> v <- c("A1B2","A3C4","B5","C6A7B8")
> l <- strsplit(gsub("(.{2})","\\1,",v),",")
> d <- data.frame(A=vector(length=4,mode="integer"),
> B=vector(length=4,mode="int
Sorry, it's in package compiler, now part of the standard distro. My bad.
-- Bert
On Wed, Feb 8, 2012 at 1:23 PM, Sam Steingold wrote:
>> * Bert Gunter [2012-02-08 13:02:00 -0800]:
>>
>> A few extra minutes of computing time to save many more minutes of
>> programming time seems a reasonable tr
> * Bert Gunter [2012-02-08 13:02:00 -0800]:
>
> A few extra minutes of computing time to save many more minutes of
> programming time seems a reasonable tradeoff.
in this case, many hours of computing time.
> see ?compile to compile your
Error: object 'compile' not found
--
Sam Steingold (
Sam:
On Wed, Feb 8, 2012 at 12:56 PM, Sam Steingold wrote:
> To be clear, I can do that with nested for loops:
>
> v <- c("A1B2","A3C4","B5","C6A7B8")
> l <- strsplit(gsub("(.{2})","\\1,",v),",")
> d <- data.frame(A=vector(length=4,mode="integer"),
> B=vector(length=4,mode="integer
I suspect there are cleverer ways to do it, especially using packages
like stringr and gsubfn, but using base tools, you can hack it without
too much effort:
?gregexpr
is the key. To get started (x is your example vector of character strings):
> gregexpr("[[:alpha:]]+[[:digit:]]+",x)
[[1]]
[1] 1
To be clear, I can do that with nested for loops:
v <- c("A1B2","A3C4","B5","C6A7B8")
l <- strsplit(gsub("(.{2})","\\1,",v),",")
d <- data.frame(A=vector(length=4,mode="integer"),
B=vector(length=4,mode="integer"),
C=vector(length=4,mode="integer"))
for (i in 1:len
Suppose I have a vector of strings:
c("A1B2","A3C4","B5","C6A7B8")
[1] "A1B2" "A3C4" "B5" "C6A7B8"
where each string is a sequence of pairs
(fixed width, in this example both value and name are 1 character, in
reality the column name is 6 chars and value is 2 digits).
I need to convert it
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