[R] "unsparse" a vector

[Bert Gunter]

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
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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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