perl

[semna]

Hi,
I have two list like this:
a1 = [2,5,7,9,10]
a2 = [1,4,7,8,10]
first I tried to find the first common element by perl script:

#!/usr/bin/perl -w
use strict;
use warnings;
my $index1 = length(@a1)-1;
my $index2 = length(@a2)-1;
my $commonElement = "";
while ($a1[$index1] eq $a2[$index2]){
$commonElement = $a1[$index1];
$index1--;
$index2--;
}
return "$commonElement\n";

if I want to find get distance simultaneously, how can do that? for instance in this exapmle for both list should be 3.
I wrote this but it doesn't work:
$index1 = 1;
$index2 =1;
if ($a1[$index1] ne $a2[$index2]){
$index1++;
$index2++;

}elsif ($a1[$index1] eq $a2[$index2]){
return $index1, $index2;
exit;
}
Any response would be appreciated.

[rglover]

Hi Semna,

There are a number of things that are making your code not work, but first a few pointers:

1. Enclosing your code in

   code tags

   is a really good idea if you want people to help you out - it's much easier to read.
2. Look up http://www.perlmonks.com - it's an excellent resource for perl questions, especially as yours isn't a bioinformatics problem, but a perl one.

I'm in a good mood today so I've put together some code that will do what I think you want. You wanted to find the first elements that matched and then find the position of those elements and their value - so where your two arrays match (with values of 7) and find the position of those elements, rather than the elements that match with values of 10?

Firstly, you're working backwards through your arrays, which will not give you the first matching values, but the last. I don't think the code you've written there would ever work... I don't understand your logic. It says "while the length of my two arrays are the same, the common element is the length of the array. Reduce the size of the array and start again".

Secondly, you're trying to use "length" to get the size of your array. The way to get the number of elements present in an array is to use "scalar" (see the code below). Using "length" would kill your code straight away because both of your index values would be 0 all of the time.

Thirdly, you say you want to find the "distance" simultaneously. I've understood that to be that you want the element position within the array. In that case, the element position for your example arrays will be 2, not 3, because array positions are counted from 0, not 1.

Hopefully the code below will show you enough when you run it - try it with different values in your array and see if you understand how it is working and why. I think it would be a good learning exercise if you're not familiar with Perl.

use strict;
use warnings;

my @a1 = (2,5,7,9,10);
my @a2 = (1,4,7,8,10);
my $size = scalar(@a1);

my $common_element = undef;
my $common_element_array_position = undef;

for (my $i=0; $i<$size; $i++) {
  print "a1:".$a1[$i]."\n";
  print "a2:".$a2[$i]."";
  if ($a1[$i] == $a2[$i]) {
    print " They match so we'll skip the rest of the array\n";
    $common_element = $a1[$i];
    $common_element_array_position = $i;
    last;
  }
  else {
    print " They don't match\n";
  }
}

if (defined $common_element && defined $common_element_array_position) {
  print "CE: $common_element\nPosition: $common_element_array_position\n";
}
else {
  print "Could not find any common elements\n";
}

[westerman]

Finding the first common element of two lists, eh? Sounds like a 1st year programming assignment. I suspect that there is some algorithmic function you could look up

Of course in Perl TIMTOWTDI but I would

1) Convert the first list into a hash via the map command
2) Use a foreach loop to go through the second list to find the first element in the second list that has a definition in the hash.
3) Then, since you also want the index in the first list, do a foreach to find the index of the element.

I can see problems with the above with large lists or sparse hit lists. I.e., it isn't the most efficient code in the world.

Or you could use List::Compare.

[westerman]

Ah, rglover's solution gives the first position where both elements are common. My suggestion just gives the first common element. It was nice of her to give a solution -- Rachel must be having a better day than I. :-)

[rglover]

report writing distraction technique I'm not normally someone who answers "my code doesn't work" questions...

[damiankao]

To find first common element of two lists in python...

a = [1,2,3,4,5]
b = [6,7,8,9,1,10,12,2,3,4]

intersection = [overlap for overlap in a if overlap in b]

print intersection[0]

[simonandrews]

Aargh. Shouldn't have started playing with this.

#!/usr/bin/perl
use warnings;
use strict;
my @ar1 = (2,5,7,9,10,12,16);
my @ar2 = (1,4,5,7,8,10);
my %c;

for (0..$#ar1){$c{$ar1[$_]}->[0] = $_ unless (defined $c{$ar1[$_]}->[0])};
for (0..$#ar2){$c{$ar2[$_]}->[1] = $_ unless (defined $c{$ar2[$_]}->[1])};

my @common;
for (keys %c){push @common,$c{$_}->[0] if (defined $c{$_}->[0] and defined $c{$_}->[1])};
@common = sort {$a<=>$b} @common;

print "First common value is ",$ar1[$common[0]]," at indices ",$common[0]," and ",$c{$ar1[$common[0]]}->[1],"\n";

Finds the first common value and only needs to do a single pass through each list. Pretty trivially modified to get all overlaps or just those with the same index in both sets.

Figuring out how it works is left as an exercise to the reader, but looking back at it even makes my eyes hurt!

Off to do something more productive....