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### Re^2: string match using with an N in any position

by biobee07 (Novice)
 on Nov 18, 2011 at 04:31 UTC ( #938739=note: print w/replies, xml ) Need Help??

Hi BrowserUk, You code has worked perfectly for my problem!!! I agree that my problem was one that needed more than just regular expression. However I was struggling to get an elegant solution in perl. Could I please ask you to explain me these two lines:
```my \$matched = ( \$q ^ substr( \$t, 0, length( \$q ) ) ) =~ tr[\0]
+[\0];
if(  \$matched >= ( length( \$q ) -1 ) )

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Re^3: string match using with an N in any position
by BrowserUk (Pope) on Nov 18, 2011 at 05:03 UTC

Sure.

During the first iteration, the first of those lines will come down to:

```
my \$matched = ( 'GCGAT' ^ 'GNGAT' ) =~ tr[\0][\0];

In words, the first of the query strings is exclusive-Or'd (^) with the first 5 characters of the first target string.

Where the aligned characters in those two string are the same, the result of xoring them will be zero. Where the two characters differ, the result will be non-zero:

```\$xord = 'GCGAT' ^ 'GNGAT';;

print unpack 'C*', \$xord;;
0 13 0 0 0

In order to know how many characters matched, all we need to do is count the zeros, which is efficiently done using the transliteration operator:

```\$matched = ( 'GCGAT' ^ 'GNGAT' ) =~ tr[\0][\0];;

print \$matched;;
4

The result of the first line is that \$matched is assigned the number of characters that matched between the two strings.

As you want all characters, except one in any position, to match, the second line tests the number that matched against the length of the query string minus 1:

```if(  \$matched >= ( length( \$q ) -1 ) ) {
## we have a winner

If you know all your query strings are always five characters, then you could hard code the value 4 instead of ( length( \$q ) -1 ). Ditto for the usage of length in the substr, but it never hurts to be flexible if there is little cost involved in doing so.

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