in reply to Re: replace a column
in thread replace a column

essentially: 
0.00  X  1.000  
0.530 X  0.753
# output values for residue    1
     0
# input   
0.00   X  1.00  
0.555  X  0.683
# output values for residue    2
     0

would become

0.00  0.99  1.000  
0.530 0.99  0.753
# output values for residue    1
     0
# input   
0.00   0.99  1.00  
0.555  0.99  0.683
# output values for residue    2
     0

[download]

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Re^3: replace a column
by davidj (Priest) on Jul 06, 2004 at 15:03 UTC
    Ok, if you simply want to replace the values of 1 arbitrary column in 1 file with the values of 1 arbitrary column in another file, the modified version of what I previously posted will do it.

    given the following original file:
    1:2:3:4
5:6:7:8
9:10:11:12
13:14:15:16
17:18:19:20

    [download]
    and the following replacement file:
    a:b:c:d:e
f:g:h:i:j
k:l:m:n:o
p:q:r:s:t
u:v:w:x:y

    [download]
    The following example will replace column 2 of the orginal file with column 3 of the replacement file
    #!/usr/bin/perl
use File::Copy;
use strict;
my (@col, @vals, $val, $i);
open(COL, "<replace.txt");

while(<COL>) {
    chomp($_);
    (@vals) = split(":", $_);
    push @col, $vals[1]; #store values for column 2

}
close(COL);

open(OUT, ">out.txt");
open(COL, "<original.txt");
while(<COL>) {
    chomp($_);
    (@vals) = split(":", $_);
    $vals[2] = $col[$i++];    #replace values in column 3
    print OUT join(":", @vals) . "\n";
}
close(COL);
close(OUT);
copy("out.txt", "original.txt");

    [download]
    original file after running the program:
    1:2:b:4
5:6:g:8
9:10:l:12
13:14:q:16
17:18:v:20

    [download]

    davidj
[reply]
[d/l]
[select]
Re^3: replace a column
by Anonymous Monk on Jul 06, 2004 at 14:37 UTC
    i ahve tried 
    while (<FILE>) {

#foreach $line = <FILE> {

for ($count = 0; $count < $end; $count++) {

s/^(\s+\S+\s+\S+\s+\S+\s+\S+\s+\S+\s+\S+\s+\S+)\s+(.*)/$1 $array[$coun
+t]  $2/;
                                           }
print OUT $_;

}

but the entire array is printed at the substitution point?

# input 
  0.00  1.000  0.00  0.00  1.00  0.00  1.00 0.99
  0.99
  0.710  0.000
  0.22  0.780  0.61  0.39  0.17  0.00  0.50 0.99
  0.99
  0.530  0.753
# output values for residue    1
     0
# input 
  0.00  1.000  1.00  1.00  0.00  0.00  0.00 0.99
  0.99
  0.264  0.000
  0.22  0.780  0.56  0.33  0.22  0.00  0.56 0.99
  0.99
  0.555  0.683
# output values for residue    2
     0

    [download]
[reply]
[d/l]
      but will use the code u have provided, thanks!
[reply]

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