http://qs321.pair.com?node_id=950374

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

Hi everyone, I am a beginner to programming and perl. I have a hash (a=>2, b=>1, c=>5, d=>3, e=>4); all i have to do is take a string and slide through the string and calculate the i+(i+3) value and print it. Forgive me I make it so complex so i coudn't paste the entire code. Plz help me. Here is the pseudo code

%hash = (a=>2, b=>1, c=>5, d=>3, e=>4); \$input = "abcdaeec"; \$calculate = \$i+(\$i+n) of \$input; #where n = 2 or 3 (sliding window:- #like a+c b+d c+a d+e a+e e+c 2+5 1+3 5+2 3+4 2+4 4+5) prints= 7 4 7 7 6 9

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Re: Sliding window
by Eliya (Vicar) on Jan 27, 2012 at 14:59 UTC

As I understand your task (which seems to be different from BrowserUk), this should do it:

#!/usr/bin/perl -w use strict; my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4); my \$input = "abcdaeec"; for (2, 3) { print "n=\$_: "; my \$n = \$_ - 1; my @sums; while ( \$input =~ /(.)(?=.{\$n}(.))/g ) { push @sums, \$hash{\$1} + \$hash{\$2}; } print "@sums\n"; } __END__ n=2: 7 4 7 7 6 9 n=3: 5 3 9 7 7

Update: let YAPE::Regex::Explain explain what the regex does:

use YAPE::Regex::Explain; print YAPE::Regex::Explain->new( qr/(.)(?=.{1}(.))/ )->explain; __END__ The regular expression: (?-imsx:(.)(?=.{1}(.))) matches as follows: NODE EXPLANATION ---------------------------------------------------------------------- (?-imsx: group, but do not capture (case-sensitive) (with ^ and \$ matching normally) (with . not matching \n) (matching whitespace and # normally): ---------------------------------------------------------------------- ( group and capture to \1: ---------------------------------------------------------------------- . any character except \n ---------------------------------------------------------------------- ) end of \1 ---------------------------------------------------------------------- (?= look ahead to see if there is: ---------------------------------------------------------------------- .{1} any character except \n (1 times) ---------------------------------------------------------------------- ( group and capture to \2: ---------------------------------------------------------------------- . any character except \n ---------------------------------------------------------------------- ) end of \2 ---------------------------------------------------------------------- ) end of look-ahead ---------------------------------------------------------------------- ) end of grouping ----------------------------------------------------------------------

And see perlop for what m//g does in scalar context (like the while (...) here).

Re: Sliding window
by choroba (Archbishop) on Jan 27, 2012 at 14:22 UTC
\$s = "abcdef"; print "\$1,\$2\n" while \$s =~ /(.)(?=.(.))/g;
Update: Using while instead of for to simplify getting the pairs.
Re: Sliding window
by BrowserUk (Patriarch) on Jan 27, 2012 at 14:47 UTC

"Sliding window" is a decptive title, but the problem is interesting.

A not very perlish solution:

#! perl -slw use strict; my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4); my \$input = "abcdaeec"; our \$N //= 2; \$,= ' '; print map{ my \$iplus = ord() - ord('a'); \$iplus = ( \$iplus + \$N ) % keys %hash; \$iplus = chr( ord('a') + \$iplus ); \$hash{ \$_ } + \$hash{ \$iplus }; } split '', \$input; __END__ C:\test>junk41 -N=2 7 4 9 5 7 5 5 9 C:\test>junk41 -N=3 5 5 7 4 5 9 9 7 C:\test>junk41 -N=4 6 3 6 8 6 7 7 6

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The start of some sanity?

Re: Sliding window
by JavaFan (Canon) on Jan 27, 2012 at 15:21 UTC
my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4); my \$input = "abcdaeec"; my @input = split //, \$input; for my \$w (2, 3) { for (my \$i = \$w; \$i < @input; \$i++) { printf "%s+%s = %d+%d = %d\n", \$input[\$i-\$w], \$input[\$i], \$hash{\$input[\$i-\$w]}, \$hash{\$input[\$i]}, \$hash{\$input[\$i-\$w]} + \$hash{\$input[\$i]}; } } __END__ a+c = 2+5 = 7 b+d = 1+3 = 4 c+a = 5+2 = 7 d+e = 3+4 = 7 a+e = 2+4 = 6 e+c = 4+5 = 9 a+d = 2+3 = 5 b+a = 1+2 = 3 c+e = 5+4 = 9 d+e = 3+4 = 7 a+c = 2+5 = 7
Re: Sliding window
by repellent (Priest) on Jan 29, 2012 at 00:28 UTC
my %hash = (a=>2, b=>1, c=>5, d=>3, e=>4); my \$input = "abcdaeec"; my @v = map \$hash{\$_}, (split //, \$input); for my \$n (2, 3) { my @w = map \$v[\$_] + \$v[\$_ + \$n], 0 .. \$#v - \$n; print "@w\n"; } __END__ 7 4 7 7 6 9 5 3 9 7 7