mattgarrett has asked for the wisdom of the Perl Monks concerning the following question:

I'm trying to parse a string that looks like this (Fortran namelist input):

WB(1,2)=    0.000, 1.23, TB(1,2)=     0.0,  253.0, TMB(1,2)=  0.0,  1.0,  SL(1,2)= 0.00, 1.00

I'd like to read the second number after WB(1,2)= (1.23 in the above example) and change it.

Is there a way to do this with regular expressions?

Thanks,
Matt

Replies are listed 'Best First'.
Re: parsing a line of text with inconsistent delimiters
by moritz (Cardinal) on Sep 12, 2007 at 14:59 UTC
[reply]

      Does Fortran::Format do namelist i/o? I see no references to it in its documentation. Fortran::Namelist or Fortran::F90Namelist may be more appropriate. I've not used either, even though, as a long-time Fortran programmer, I find namelist i/o extremely handy.

      emc

      Information about American English usage here and here.

      Any New York City or Connecticut area jobs? I'm currently unemployed.

[reply]
        Thanks Swampy. Fortran::Namelist looks like it will do the trick.
[reply]
Re: parsing a line of text with inconsistent delimiters
by erroneousBollock (Curate) on Sep 12, 2007 at 15:05 UTC
    A regexp something like the following should match each such "second number" in the line.
      / 
    [A-Z]+               # WB
    \(\d+,\d+\)          # (1,2)
    \s*=\s*              # =
    (?:                  # start non-capturing group
       \d+\.\d+,         # a float followed by comma
       \s*(\d+\.\d+),?   # capture a float (followed by comma?)
    )                    # end-group
  /gx

    [download]

    To change it use the substitution operator (perhaps with /e flag so that you can pass the matched number to a function to get back the value you want it to be changed to).

    -David

    Update: fixed a typo.
    Update2: and another, thanks naikonta.

[reply]
[d/l]
[select]
      Oops, 
      Unmatched ) in regex; marked by <-- HERE in m/
    [A-Z]+               # WB
    \(\d+,\d+) <-- HERE            # (1,2)
    \s*=\s*              # =
    (?:                  # start non-capturing group
       \d+\.\d+,         # a float followed by comma
       \s*(\d+\.\d+),?   # capture a float (followed by comma?)
    )                    # end-group
  / at /tmp/wb line 10.

      [download]
      Matt, you can do it also with split but you have to join the parts back together after changing the part you want. 
      $_ = 'WB(1,2)=    0.000, 1.23, TB(1,2)=     0.0,  253.0, TMB(1,2)=  0.
+0,  1.0,  SL(1,2)= 0.00, 1.00';
my @parts = split /, /;
$parts[1] = 3.21; # change 1.23 with 3.21
$_ = join ', ', @parts;
print $_, "\n";

# output:
# WB(1,2)=    0.000, 3.21, TB(1,2)=     0.0,  253.0, TMB(1,2)=  0.0,  
+1.0,  SL(1,2)= 0.00, 1.00

      [download]

      Open source softwares? Share and enjoy. Make profit from them if you can. Yet, share and enjoy!

[reply]
[d/l]
[select]
Re: parsing a line of text with inconsistent delimiters
by mayaTheCat (Scribe) on Sep 12, 2007 at 15:11 UTC
    You can both extract the number and replace with the new ones, as given in the following snippet; 
    
# assuming; ...
my $str = "WB(1,2)=    0.000, 1.23, TB(1,2)=     0.0,  253.0, TMB(1,2)
+=  0.0, 1.0,  SL(1,2)= 0.00, 1.00";

sub replace {
    my ($extracted_number) = @_;

    # calculate or query the new number. For example;
    my $new_number = $extracted_number + 1;

    return $new_number;
}

$str =~ s/
    (?<=,)            # Precondition: There should be a comma before,
    \s*
    (\d+\.\d+)        # ... and it should be a floating number!
/replace($1)/xges;

print "$str\n";

    [download]

    ---------------------------------
    life is ... $mutation = sub { $_[0] =~ s/(.)/rand()<0.1?1-$1:$1/ge };

[reply]
[d/l]
[select]
      Thank you to everyone who responded.
[reply]

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