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Sam Steingold

2012-Feb-08 20:34 UTC

[R] "unsparse" a vector

Suppose I have a vector of strings:
c("A1B2","A3C4","B5","C6A7B8")
[1] "A1B2"   "A3C4"   "B5"     "C6A7B8"
where each string is a sequence of <column><value> 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 to a data frame:
data.frame(A=c(1,3,0,7),B=c(2,0,5,8),C=c(0,4,0,6))
  A B C
1 1 2 0
2 3 0 4
3 0 5 0
4 7 8 6

how do I do that?
thanks.

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000
http://mideasttruth.com http://jihadwatch.org http://pmw.org.il
http://openvotingconsortium.org http://iris.org.il http://memri.org
What's the difference between Apathy & Ignorance? -I don't know and
don't care!

Sam Steingold

2012-Feb-08 20:56 UTC

head link

[R] "unsparse" a vector

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:length(l)) {
  l1 <- l[[i]]
  for (j in 1:length(l1)) {
    d[[substring(l1[j],1,1)]][i] <- as.numeric(substring(l1[j],2,2))
  }
}


but I am afraid that handling 1,000,000 (=length(unlist(l))) strings in
a loop will kill me.

> * Sam Steingold <fqf at tah.bet> [2012-02-08 15:34:38 -0500]:
>
> Suppose I have a vector of strings:
> c("A1B2","A3C4","B5","C6A7B8")
> [1] "A1B2"   "A3C4"   "B5"    
"C6A7B8"
> where each string is a sequence of <column><value> 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 to a data frame:
> data.frame(A=c(1,3,0,7),B=c(2,0,5,8),C=c(0,4,0,6))
>   A B C
> 1 1 2 0
> 2 3 0 4
> 3 0 5 0
> 4 7 8 6
>
> how do I do that?
> thanks.
-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000
http://palestinefacts.org http://iris.org.il http://camera.org
http://ffii.org http://www.PetitionOnline.com/tap12009/
An elephant is a mouse with an operating system.

Bert Gunter

2012-Feb-08 20:58 UTC

head link

[R] "unsparse" a vector

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 3
attr(,"match.length")
[1] 2 2
attr(,"useBytes")
[1] TRUE

[[2]]
[1] 1 3
attr(,"match.length")
[1] 2 2
attr(,"useBytes")
[1] TRUE

[[3]]
[1] 1
attr(,"match.length")
[1] 2
attr(,"useBytes")
[1] TRUE

[[4]]
[1] 1 3 5
attr(,"match.length")
[1] 2 2 2
attr(,"useBytes")
[1] TRUE

The components of the result give you indices of the start and stop
values for each "entry" in your final matrix/data frame. You can thus
lapply() on this list to get the column name-value pairs substrings
and decode them.

Alternatively, if all your names are really 6 characters and all your
values are really two digits,
?nchar and ?substring will get you the name-value substrings directly.

I leave the niggling details to you (or to other helpeRs -- especially
those who can suggest a more elegant approach).

-- Bert





On Wed, Feb 8, 2012 at 12:34 PM, Sam Steingold <sds at gnu.org>
wrote:> Suppose I have a vector of strings:
> c("A1B2","A3C4","B5","C6A7B8")
> [1] "A1B2" ? "A3C4" ? "B5" ? ?
"C6A7B8"
> where each string is a sequence of <column><value> 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 to a data frame:
> data.frame(A=c(1,3,0,7),B=c(2,0,5,8),C=c(0,4,0,6))
> ?A B C
> 1 1 2 0
> 2 3 0 4
> 3 0 5 0
> 4 7 8 6
>
> how do I do that?
> thanks.
>
> --
> Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X
11.0.11004000
> http://mideasttruth.com http://jihadwatch.org http://pmw.org.il
> http://openvotingconsortium.org http://iris.org.il http://memri.org
> What's the difference between Apathy & Ignorance? -I don't know
and don't care!
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Petr Savicky

2012-Feb-08 21:32 UTC

head link

[R] "unsparse" a vector

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="integer"),
>                 C=vector(length=4,mode="integer"))
> 
> for (i in 1:length(l)) {
>   l1 <- l[[i]]
>   for (j in 1:length(l1)) {
>     d[[substring(l1[j],1,1)]][i] <- as.numeric(substring(l1[j],2,2))
>   }
> }
Hi.

The inner loop may be vectorized.

  d <- as.matrix(d)
  for (i in 1:length(l)) {
    l1 <- l[[i]]
    d[i, substring(l1,1,1)] <- as.numeric(substring(l1,2,2))
  }

    A B C
  1 1 2 0
  2 3 0 4
  3 0 5 0
  4 7 8 6

If the number of rows of d is large, a matrix is probably
more efficient.

Hope this helps.

Petr Savicky.

William Dunlap

2012-Feb-08 21:53 UTC

head link

[R] "unsparse" a vector

When compute time is important it often helps
to loop over columns instead of over rows (assuming
there are fewer columns than rows, the usual
case).  E.g., putting your code into a function f0
and the column-looping version into f1:

f0 <- function(v) {
  n <- length(v)
  d <- data.frame(A=vector(length=n,mode="integer"),
                  B=vector(length=n,mode="integer"),
                  C=vector(length=n,mode="integer"))
  l <- strsplit(gsub("(.{2})","\\1,",v),",")
  for (i in seq_along(l)) {
    l1 <- l[[i]]
    for (j in seq_along(l1)) {
      d[[substring(l1[j],1,1)]][i] <- as.integer(substring(l1[j],2,2))
    }
  }
  d
}

f1 <- function(v) {
  n <- length(v)
  letters <- c("A", "B", "C")
  names(letters) <- letters
  data.frame(lapply(letters,
              function(letter) {
                  retval <- integer(n)
                  hasLetter <- grepl(letter, v)
                  retval[hasLetter] <- as.integer(
                     gsub(sprintf("^.*%s([[:digit:]]+).*$", letter), 
                          "\\1", 
                          v[hasLetter]))
                  retval
              }))
}

I get the following times for a 10,000 long v like yours
(and the results are the same):
> vv <- rep(v, len=10000)
> system.time(r1 <- f1(vv))   user  system elapsed 
   0.13    0.00    0.14 > system.time(r0 <- f0(vv))   user  system elapsed 
  10.75    0.19   10.99 > all.equal(r0, r1)[1] TRUE

If I double the length of v, your code takes 53 seconds (5x slower,
quadratic behavior?) while mine takes 0.17 (less than linear, suggesting
that its time is still dominated by the function call overhead for such
small input vectors).

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at
r-project.org] On Behalf Of Sam Steingold
> Sent: Wednesday, February 08, 2012 12:56 PM
> To: r-help at r-project.org
> Subject: Re: [R] "unsparse" a vector
> 
> 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:length(l)) {
>   l1 <- l[[i]]
>   for (j in 1:length(l1)) {
>     d[[substring(l1[j],1,1)]][i] <- as.numeric(substring(l1[j],2,2))
>   }
> }
> 
> 
> but I am afraid that handling 1,000,000 (=length(unlist(l))) strings in
> a loop will kill me.
> 
> 
> > * Sam Steingold <fqf at tah.bet> [2012-02-08 15:34:38 -0500]:
> >
> > Suppose I have a vector of strings:
> > c("A1B2","A3C4","B5","C6A7B8")
> > [1] "A1B2"   "A3C4"   "B5"    
"C6A7B8"
> > where each string is a sequence of <column><value> 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 to a data frame:
> > data.frame(A=c(1,3,0,7),B=c(2,0,5,8),C=c(0,4,0,6))
> >   A B C
> > 1 1 2 0
> > 2 3 0 4
> > 3 0 5 0
> > 4 7 8 6
> >
> > how do I do that?
> > thanks.
> 
> --
> Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X
11.0.11004000
> http://palestinefacts.org http://iris.org.il http://camera.org
> http://ffii.org http://www.PetitionOnline.com/tap12009/
> An elephant is a mouse with an operating system.
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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