Re: [R] Differenciate numbers from reference for rows

[David Winsemius]

On Oct 30, 2010, at 8:42 AM, Gabor Grothendieck wrote:
On Fri, Oct 29, 2010 at 6:54 PM, M.Ribeiro  
<mresende...@yahoo.com.br> wrote:

So, I am having a tricky reference file to extract information from.

The format of the file is

x   1 + 4 * 3 + 5 + 6 + 11 * 0.5

So, the elements that are not being multiplied (1, 5 and 6) and the  
elements
before the multiplication sign (4 and 11) means actually the  
reference for
the row in a matrix where I need to extract the element from.

The numbers after the multiplication sign are regular numbers
Ex:

x<-matrix(20:35)

I would like to read the rows 1,4,5,6 and 11 and sum then. However  
the
numbers in the elements row 4 and 11 are multiplied by 3 and 0.5

So it would be
20 + 23 * 3 + 24 + 25 + 30 * 0.5.

And I have this format in different files so I can't do all by hand.
Can anybody help me with a script that can differentiate this?


I assume that every number except for the second number in the pattern
number * number is to be replaced by that row number in x.  Try this.
We define a regular expression which matches the first number ([0-9]+)
of each potential pair and optionally (?) spaces ( *) a star (\\*),
more spaces ( *) and digits [0-9.]+ passing the first and second
backreferences (matches to the parenthesized portions of the regular
expression) to f and inserting the output of f where the matches had
been.

library(gsubfn)
f <- function(a, b) paste(x[as.numeric(a)], b)
s2 <- gsubfn("([0-9]+)( *\\* *[0-9.]+)?", f, s)

If the objective is to then perform the calculation that that
represents then try this:
sapply(s2, function(x) eval(parse(text = x)))

For example,

s <- c("1 + 4 * 3 + 5 + 6 + 11 * 0.5", "1 + 4 * 3 + 5 + 6 + 11 *  
0.5")
x <- matrix(20:35)
f <- function(a, b) paste(x[as.numeric(a)], b)
s2 <- gsubfn("([0-9]+)( *\\* *[0-9.]+)?", f, s)
s2
[1] "20  + 23  * 3 + 24  + 25  + 30  * 0.5" "20  + 23  * 3 + 24  +  
25 + 30  * 0.5"
sapply(s2, function(x) eval(parse(text = x)))
20  + 23  * 3 + 24  + 25  + 30  * 0.5 20  + 23  * 3 + 24  + 25  +  
30  * 0.5
                                  
153                                   153

For more see the gsubfn home page at http://gsubfn.googlecode.com


I am scratching my head regarding the gsubfn workings. It appears that  
as gsubfn moves across the input strings that it will either match  
just "[0-9+]" or it will match "[0-9+] *\\* *[0-9.]+?".

In either case the match will do a lookup in x[] for the first match  
using the "a" index, and if there is a match for the second position  
assigned to "*b" then that x[a] will be followed by "*b"  and is  
therefore destined to be multiplied by "b". I cannot quite figure out  
how the NULL value gets not-matched to the second back-reference and  
then doesn't screw up the f() function by only providing one argument  
to a two argument function. Maybe it's due to this? (So can you  
comment on how optional back-references return values?)

> paste("a", NULL)
[1] "a "

Furthermore, somehow (and this is further functiona magic I am  
missing) these results are concatenated in a string, and then  
evaluated, a step which I do get.

--

David Winsemius, MD
West Hartford, CT

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