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The top answer in this post: How can I create a multidimensional array in Perl? suggests building a multi-dimensional array as follows:
my #array = ();
foreach my $i ( 0 .. 10 ) {
foreach my $j ( 0 .. 10 ) {
push #{ $array[$i] }, $j;
}
}
I am wondering if there is a way of building the array more compactly and avoiding the nested loop, e.g. using something like:
my #array = ();
my #other_array = (0 ... 10);
foreach my $i ( 0 .. 10 ) {
$array[$i] = #other_array; # This does not work in Perl
}
}
Does Perl support any syntax like that for building multi-dimensional arrays without nested looping?
Similarly, is there a way to print the multidimensional array without (nested) looping?
There is more than one way to do it:
Generating
push accepts LISTs
my #array;
push #{$array[$_]}, 0 .. 10 for 0 .. 10;
Alternative syntax:
my #array;
push #array, [ 0 .. 10 ] for 0 .. 10;
map eye-candy
my #array = map { [ 0 .. 10 ] } 0 .. 10;
Alternative syntax:
my #array = map [ 0 .. 10 ], 0 .. 10;
Printing
With minimal looping
print "#$_\n" for #array;
On Perl 5.10+
use feature 'say';
say "#$_" for #array;
With more formatting control
print join( ', ', #$_ ), "\n" for #array; # "0, 1, 2, ... 9, 10"
"No loops" (The loop is hidden from you)
use Data::Dump 'dd';
dd #array;
Data::Dumper
use Data::Dumper;
print Dumper \#array;
Have a look at perldoc perllol for more details
You are close, you need a reference to the other array
my #array; # don't need the empty list
my #other_array = (0 ... 10);
foreach my $i ( 0 .. 10 ) {
$array[$i] = \#other_array;
# or without a connection to the original
$array[$i] = [ #other_array ];
# or for a slice
$array[$i] = [ #other_array[1..$#other_array] ];
}
}
You can also make anonymous (unnamed) array reference directly using square braces [] around a list.
my #array;
foreach my $i ( 0 .. 10 ) {
$array[$i] = [0..10];
}
}
Edit: printing is probably easiest using the postfix for
print "#$_\n" for #array;
for numerical multidimensional arrays, you can use PDL. It has several constructors for different use cases. The one analogous to the above would be xvals. Note that PDL objects overload printing, so you can just print them out.
use PDL;
my $pdl = xvals(11, 11);
print $pdl;
I want Perl (5.8.8) to find out what word has the most letters in common with the other words in an array - but only letters that are in the same place. (And preferably without using libs.)
Take this list of words as an example:
BAKER
SALER
BALER
CARER
RUFFR
Her BALER is the word that has the most letters in common with the others. It matches BAxER in BAKER, xALER in SALER, xAxER in CARER, and xxxxR in RUFFR.
I want Perl to find this word for me in an arbitrary list of words with the same length and case. Seems I've hit the wall here, so help is much appreciated!
What I've tried until now
Don't really have much of a script at the moment:
use strict;
use warnings;
my #wordlist = qw(BAKER SALER MALER BARER RUFFR);
foreach my $word (#wordlist) {
my #letters = split(//, $word);
# now trip trough each iteration and work magic...
}
Where the comment is, I've tried several kinds of code, heavy with for-loops and ++ varables. Thus far, none of my attempts have done what I need it to do.
So, to better explain: What I need is to test word for word against the list, for each letterposition, to find the word that has the most letters in common with the others in the list, at that letter's position.
One possible way could be to first check which word(s) has the most in common at letter-position 0, then test letter-position 1, and so on, until you find the word that in sum has the most letters in common with the other words in the list. Then I'd like to print the list like a matrix with scores for each letterposition plus a total score for each word, not unlike what DavidO suggest.
What you'd in effect end up with is a matrix for each words, with the score for each letter position, and the sum total score fore each word in the matrix.
Purpose of the Program
Hehe, I might as well say it: The program is for hacking terminals in the game Fallout 3. :D My thinking is that it's a great way to learn Perl while also having fun gaming.
Here's one of the Fallout 3 terminal hacking tutorials I've used for research: FALLOUT 3: Hacking FAQ v1.2, and I've already made a program to shorten the list of words, like this:
#!/usr/bin/perl
# See if one word has equal letters as the other, and how many of them are equal
use strict;
use warnings;
my $checkword = "APPRECIATION"; # the word to be checked
my $match = 4; # equal to the match you got from testing your checkword
my #checkletters = split(//, $checkword); #/
my #wordlist = qw(
PARTNERSHIPS
REPRIMANDING
CIVILIZATION
APPRECIATION
CONVERSATION
CIRCUMSTANCE
PURIFICATION
SECLUSIONIST
CONSTRUCTION
DISAPPEARING
TRANSMISSION
APPREHENSIVE
ENCOUNTERING
);
print "$checkword has $match letters in common with:\n";
foreach my $word (#wordlist) {
next if $word eq $checkword;
my #letters = split(//, $word);
my $length = #letters; # determine length of array (how many letters to check)
my $eq_letters = 0; # reset to 0 for every new word to be tested
for (my $i = 0; $i < $length; $i++) {
if ($letters[$i] eq $checkletters[$i]) {
$eq_letters++;
}
}
if ($eq_letters == $match) {
print "$word\n";
}
}
# Now to make a script on to find the best word to check in the first place...
This script will yield CONSTRUCTION and TRANSMISSION as its result, just as in the game FAQ. The trick to the original question, though (and the thing I didn't manage to find out on my own), is how to find the best word to try in the first place, i.e. APPRECIATION.
OK, I've now supplied my own solution based on your help, and consider this thread closed. Many, many thanks to all the contributers. You've helped tremendously, and on the way I've also learned a lot. :D
Here's one way. Having re-read your spec a couple of times I think it's what you're looking for.
It's worth mentioning that it's possible there will be more than one word with an equal top score. From your list there's only one winner, but it's possible that in longer lists, there will be several equally winning words. This solution deals with that. Also, as I understand it, you count letter matches only if they occur in the same column per word. If that's the case, here's a working solution:
use 5.012;
use strict;
use warnings;
use List::Util 'max';
my #words = qw/
BAKER
SALER
BALER
CARER
RUFFR
/;
my #scores;
foreach my $word ( #words ) {
my $score;
foreach my $comp_word ( #words ) {
next if $comp_word eq $word;
foreach my $pos ( 0 .. ( length $word ) - 1 ) {
$score++ if substr( $word, $pos, 1 ) eq substr( $comp_word, $pos, 1);
}
}
push #scores, $score;
}
my $max = max( #scores );
my ( #max_ixs ) = grep { $scores[$_] == $max } 0 .. $#scores;
say "Words with most matches:";
say for #words[#max_ixs];
This solution counts how many times per letter column each word's letters match other words. So for example:
Words: Scores: Because:
ABC 1, 2, 1 = 4 A matched once, B matched twice, C matched once.
ABD 1, 2, 1 = 4 A matched once, B matched twice, D matched once.
CBD 0, 2, 1 = 3 C never matched, B matched twice, D matched once.
BAC 0, 0, 1 = 1 B never matched, A never matched, C matched once.
That gives you the winners of ABC and ABD, each with a score of four positional matches. Ie, the cumulative times that column one, row one matched column one row two, three, and four, and so on for the subsequent columns.
It may be able to be optimized further, and re-worded to be shorter, but I tried to keep the logic fairly easy to read. Enjoy!
UPDATE / EDIT
I thought about it and realized that though my existing method does exactly what your original question requested, it did it in O(n^2) time, which is comparatively slow. But if we use hash keys for each column's letters (one letter per key), and do a count of how many times each letter appears in the column (as the value of the hash element), we could do our summations in O(1) time, and our traversal of the list in O(n*c) time (where c is the number of columns, and n is the number of words). There's some setup time too (creation of the hash). But we still have a big improvement. Here is a new version of each technique, as well as a benchmark comparison of each.
use strict;
use warnings;
use List::Util qw/ max sum /;
use Benchmark qw/ cmpthese /;
my #words = qw/
PARTNERSHIPS
REPRIMANDING
CIVILIZATION
APPRECIATION
CONVERSATION
CIRCUMSTANCE
PURIFICATION
SECLUSIONIST
CONSTRUCTION
DISAPPEARING
TRANSMISSION
APPREHENSIVE
ENCOUNTERING
/;
# Just a test run for each solution.
my( $top, $indexes_ref );
($top, $indexes_ref ) = find_top_matches_force( \#words );
print "Testing force method: $top matches.\n";
print "#words[#$indexes_ref]\n";
( $top, $indexes_ref ) = find_top_matches_hash( \#words );
print "Testing hash method: $top matches.\n";
print "#words[#$indexes_ref]\n";
my $count = 20000;
cmpthese( $count, {
'Hash' => sub{ find_top_matches_hash( \#words ); },
'Force' => sub{ find_top_matches_force( \#words ); },
} );
sub find_top_matches_hash {
my $words = shift;
my #scores;
my $columns;
my $max_col = max( map { length $_ } #{$words} ) - 1;
foreach my $col_idx ( 0 .. $max_col ) {
$columns->[$col_idx]{ substr $_, $col_idx, 1 }++
for #{$words};
}
foreach my $word ( #{$words} ) {
my $score = sum(
map{
$columns->[$_]{ substr $word, $_, 1 } - 1
} 0 .. $max_col
);
push #scores, $score;
}
my $max = max( #scores );
my ( #max_ixs ) = grep { $scores[$_] == $max } 0 .. $#scores;
return( $max, \#max_ixs );
}
sub find_top_matches_force {
my $words = shift;
my #scores;
foreach my $word ( #{$words} ) {
my $score;
foreach my $comp_word ( #{$words} ) {
next if $comp_word eq $word;
foreach my $pos ( 0 .. ( length $word ) - 1 ) {
$score++ if
substr( $word, $pos, 1 ) eq substr( $comp_word, $pos, 1);
}
}
push #scores, $score;
}
my $max = max( #scores );
my ( #max_ixs ) = grep { $scores[$_] == $max } 0 .. $#scores;
return( $max, \#max_ixs );
}
The output is:
Testing force method: 39 matches.
APPRECIATION
Testing hash method: 39 matches.
APPRECIATION
Rate Force Hash
Force 2358/s -- -74%
Hash 9132/s 287% --
I realize your original spec changed after you saw some of the other options provided, and that's sort of the nature of innovation to a degree, but the puzzle was still alive in my mind. As you can see, my hash method is 287% faster than the original method. More fun in less time!
As a starting point, you can efficiently check how many letters they have in common with:
$count = ($word1 ^ $word2) =~ y/\0//;
But that's only useful if you loop through all possible pairs of words, something that isn't necessary in this case:
use strict;
use warnings;
my #words = qw/
BAKER
SALER
BALER
CARER
RUFFR
/;
# you want a hash to indicate which letters are present how many times in each position:
my %count;
for my $word (#words) {
my #letters = split //, $word;
$count{$_}{ $letters[$_] }++ for 0..$#letters;
}
# then for any given word, you get the count for each of its letters minus one (because the word itself is included in the count), and see if it is a maximum (so far) for any position or for the total:
my %max_common_letters_count;
my %max_common_letters_words;
for my $word (#words) {
my #letters = split //, $word;
my $total;
for my $position (0..$#letters, 'total') {
my $count;
if ( $position eq 'total' ) {
$count = $total;
}
else {
$count = $count{$position}{ $letters[$position] } - 1;
$total += $count;
}
if ( ! $max_common_letters_count{$position} || $count >= $max_common_letters_count{$position} ) {
if ( $max_common_letters_count{$position} && $count == $max_common_letters_count{$position} ) {
push #{ $max_common_letters_words{$position} }, $word;
}
else {
$max_common_letters_count{$position} = $count;
$max_common_letters_words{$position} = [ $word ];
}
}
}
}
# then show the maximum words for each position and in total:
for my $position ( sort { $a <=> $b } grep $_ ne 'total', keys %max_common_letters_count ) {
printf( "Position %s had a maximum of common letters of %s in words: %s\n",
$position,
$max_common_letters_count{$position},
join(', ', #{ $max_common_letters_words{$position} })
);
}
printf( "The maximum total common letters was %s in words(s): %s\n",
$max_common_letters_count{'total'},
join(', ', #{ $max_common_letters_words{'total'} })
);
Here's a complete script. It uses the same idea that ysth mentioned (although I had it independently). Use bitwise xor to combine the strings, and then count the number of NULs in the result. As long as your strings are ASCII, that will tell you how many matching letters there were. (That comparison is case sensitive, and I'm not sure what would happen if the strings were UTF-8. Probably nothing good.)
use strict;
use warnings;
use 5.010;
use List::Util qw(max);
sub findMatches
{
my ($words) = #_;
# Compare each word to every other word:
my #matches = (0) x #$words;
for my $i (0 .. $#$words-1) {
for my $j ($i+1 .. $#$words) {
my $m = ($words->[$i] ^ $words->[$j]) =~ tr/\0//;
$matches[$i] += $m;
$matches[$j] += $m;
}
}
# Find how many matches in the best word:
my $max = max(#matches);
# Find the words with that many matches:
my #wanted = grep { $matches[$_] == $max } 0 .. $#matches;
wantarray ? #$words[#wanted] : $words->[$wanted[0]];
} # end findMatches
my #words = qw(
BAKER
SALER
BALER
CARER
RUFFR
);
say for findMatches(\#words);
Haven't touched perl in a while, so pseudo-code it is. This isn't the fastest algorithm, but it will work fine for a small amount of words.
totals = new map #e.g. an object to map :key => :value
for each word a
for each word b
next if a equals b
totals[a] = 0
for i from 1 to a.length
if a[i] == b[i]
totals[a] += 1
end
end
end
end
return totals.sort_by_key.last
Sorry about the lack of perl, but if you code this into perl, it should work like a charm.
A quick note on run-time: this will run in time number_of_words^2 * length_of_words, so on a list of 100 words, each of length 10 characters, this will run in 100,000 cycles, which is adequate for most applications.
Here's a version that relies on transposing the words in order to count the identical characters. I used the words from your original comparison, not the code.
This should work with any length words, and any length list. Output is:
Word score
---- -----
BALER 12
SALER 11
BAKER 11
CARER 10
RUFFR 4
The code:
use warnings;
use strict;
my #w = qw(BAKER SALER BALER CARER RUFFR);
my #tword = t_word(#w);
my #score;
push #score, str_count($_) for #tword;
#score = t_score(#score);
my %total;
for (0 .. $#w) {
$total{$w[$_]} = $score[$_];
}
print "Word\tscore\n";
print "----\t-----\n";
print "$_\t$total{$_}\n" for (sort { $total{$b} <=> $total{$a} } keys %total);
# transpose the words
sub t_word {
my #w = #_;
my #tword;
for my $word (#w) {
my $i = 0;
while ($word =~ s/(.)//) {
$tword[$i++] .= $1;
}
}
return #tword;
}
# turn each character into a count
sub str_count {
my $str = uc(shift);
while ( $str =~ /([A-Z])/ ) {
my $chr = $1;
my $num = () = $str =~ /$chr/g;
$num--;
$str =~ s/$chr/$num /g;
}
return $str;
}
# sum up the character counts
# while reversing the transpose
sub t_score {
my #count = #_;
my #score;
for my $num (#count) {
my $i = 0;
while( $num =~ s/(\d+) //) {
$score[$i++] += $1;
}
}
return #score;
}
Here is my attempt at an answer. This will also allow you to see each individual match if you need it. (ie. BALER matches 4 characters in BAKER). EDIT: It now catches all matches if there is a tie between words (I added "CAKER" to the list to test).
#! usr/bin/perl
use strict;
use warnings;
my #wordlist = qw( BAKER SALER BALER CARER RUFFR CAKER);
my %wordcomparison;
#foreach word, break it into letters, then compare it against all other words
#break all other words into letters and loop through the letters (both words have same amount), adding to the count of matched characters each time there's a match
foreach my $word (#wordlist) {
my #letters = split(//, $word);
foreach my $otherword (#wordlist) {
my $count;
next if $otherword eq $word;
my #otherwordletters = split (//, $otherword);
foreach my $i (0..$#letters) {
$count++ if ( $letters[$i] eq $otherwordletters[$i] );
}
$wordcomparison{"$word"}{"$otherword"} = $count;
}
}
# sort (unnecessary) and loop through the keys of the hash (words in your list)
# foreach key, loop through the other words it compares with
#Add a new key: total, and sum up all the matched characters.
foreach my $word (sort keys %wordcomparison) {
foreach ( sort keys %{ $wordcomparison{$word} }) {
$wordcomparison{$word}{total} += $wordcomparison{$word}{$_};
}
}
#Want $word with highest total
my #max_match = (sort { $wordcomparison{$b}{total} <=> $wordcomparison{$a}{total} } keys %wordcomparison );
#This is to get all if there is a tie:
my $maximum = $max_match[0];
foreach (#max_match) {
print "$_\n" if ($wordcomparison{$_}{total} >= $wordcomparison{$maximum}{total} )
}
The output is simply: CAKER BALER and BAKER.
The hash %wordcomparison looks like:
'SALER'
{
'RUFFR' => 1,
'BALER' => 4,
'BAKER' => 3,
'total' => 11,
'CARER' => 3
};
You can do this, using a dirty regex trick to execute code if a letter matches in its place, but not otherwise, thankfully it's quite easy to build the regexes as you go:
An example regular expression is:
(?:(C(?{ $c++ }))|.)(?:(A(?{ $c++ }))|.)(?:(R(?{ $c++ }))|.)(?:(E(?{ $c++ }))|.)(?:(R(?{ $c++ }))|.)
This may or may not be fast.
use 5.12.0;
use warnings;
use re 'eval';
my #words = qw(BAKER SALER BALER CARER RUFFR);
my ($best, $count) = ('', 0);
foreach my $word (#words) {
our $c = 0;
foreach my $candidate (#words) {
next if $word eq $candidate;
my $regex_str = join('', map {"(?:($_(?{ \$c++ }))|.)"} split '', $word);
my $regex = qr/^$regex_str$/;
$candidate =~ $regex or die "did not match!";
}
say "$word $c";
if ($c > $count) {
$best = $word;
$count = $c;
}
}
say "Matching: first best: $best";
Using xor trick will be fast but assumes a lot about the range of characters you might encounter. There are many ways in which utf-8 will break with that case.
Many thanks to all the contributers! You've certainly shown me that I still have a lot to learn, but you have also helped me tremendously in working out my own answer. I'm just putting it here for reference and possible feedback, since there are probably better ways of doing it. To me this was the simplest and most straight forward approach I could find on my own. Enjøy! :)
#!/usr/bin/perl
use strict;
use warnings;
# a list of words for testing
my #list = qw(
BAKER
SALER
BALER
CARER
RUFFR
);
# populate two dimensional array with the list,
# so we can compare each letter with the other letters on the same row more easily
my $list_length = #list;
my #words;
for (my $i = 0; $i < $list_length; $i++) {
my #letters = split(//, $list[$i]);
my $letters_length = #letters;
for (my $j = 0; $j < $letters_length; $j++) {
$words[$i][$j] = $letters[$j];
}
}
# this gives a two-dimensionla array:
#
# #words = ( ["B", "A", "K", "E", "R"],
# ["S", "A", "L", "E", "R"],
# ["B", "A", "L", "E", "R"],
# ["C", "A", "R", "E", "R"],
# ["R", "U", "F", "F", "R"],
# );
# now, on to find the word with most letters in common with the other on the same row
# add up the score for each letter in each word
my $word_length = #words;
my #letter_score;
for my $i (0 .. $#words) {
for my $j (0 .. $#{$words[$i]}) {
for (my $k = 0; $k < $word_length; $k++) {
if ($words[$i][$j] eq $words[$k][$j]) {
$letter_score[$i][$j] += 1;
}
}
# we only want to add in matches outside the one we're testing, therefore
$letter_score[$i][$j] -= 1;
}
}
# sum each score up
my #scores;
for my $i (0 .. $#letter_score ) {
for my $j (0 .. $#{$letter_score[$i]}) {
$scores[$i] += $letter_score[$i][$j];
}
}
# find the highest score
my $max = $scores[0];
foreach my $i (#scores[1 .. $#scores]) {
if ($i > $max) {
$max = $i;
}
}
# and print it all out :D
for my $i (0 .. $#letter_score ) {
print "$list[$i]: $scores[$i]";
if ($scores[$i] == $max) {
print " <- best";
}
print "\n";
}
When run, the script yields the following:
BAKER: 11
SALER: 11
BALER: 12 <- best
CARER: 10
RUFFR: 4
I have two arrays. I need to check and see if the elements of one appear in the other one.
Is there a more efficient way to do it than nested loops? I have a few thousand elements in each and need to run the program frequently.
Another way to do it is to use Array::Utils
use Array::Utils qw(:all);
my #a = qw( a b c d );
my #b = qw( c d e f );
# symmetric difference
my #diff = array_diff(#a, #b);
# intersection
my #isect = intersect(#a, #b);
# unique union
my #unique = unique(#a, #b);
# check if arrays contain same members
if ( !array_diff(#a, #b) ) {
# do something
}
# get items from array #a that are not in array #b
my #minus = array_minus( #a, #b );
perlfaq4 to the rescue:
How do I compute the difference of two arrays? How do I compute the intersection of two arrays?
Use a hash. Here's code to do both and more. It assumes that each element is unique in a given array:
#union = #intersection = #difference = ();
%count = ();
foreach $element (#array1, #array2) { $count{$element}++ }
foreach $element (keys %count) {
push #union, $element;
push #{ $count{$element} > 1 ? \#intersection : \#difference }, $element;
}
If you properly declare your variables, the code looks more like the following:
my %count;
for my $element (#array1, #array2) { $count{$element}++ }
my ( #union, #intersection, #difference );
for my $element (keys %count) {
push #union, $element;
push #{ $count{$element} > 1 ? \#intersection : \#difference }, $element;
}
You need to provide a lot more context. There are more efficient ways of doing that ranging from:
Go outside of Perl and use shell (sort + comm)
map one array into a Perl hash and then loop over the other one checking hash membership. This has linear complexity ("M+N" - basically loop over each array once) as opposed to nested loop which has "M*N" complexity)
Example:
my %second = map {$_=>1} #second;
my #only_in_first = grep { !$second{$_} } #first;
# use a foreach loop with `last` instead of "grep"
# if you only want yes/no answer instead of full list
Use a Perl module that does the last bullet point for you (List::Compare was mentioned in comments)
Do it based on timestamps of when elements were added if the volume is very large and you need to re-compare often. A few thousand elements is not really big enough, but I recently had to diff 100k sized lists.
You can try Arrays::Utils, and it makes it look nice and simple, but it's not doing any powerful magic on the back end. Here's the array_diffs code:
sub array_diff(\#\#) {
my %e = map { $_ => undef } #{$_[1]};
return #{[ ( grep { (exists $e{$_}) ? ( delete $e{$_} ) : ( 1 ) } #{ $_[0] } ), keys %e ] };
}
Since Arrays::Utils isn't a standard module, you need to ask yourself if it's worth the effort to install and maintain this module. Otherwise, it's pretty close to DVK's answer.
There are certain things you must watch out for, and you have to define what you want to do in that particular case. Let's say:
#array1 = qw(1 1 2 2 3 3 4 4 5 5);
#array2 = qw(1 2 3 4 5);
Are these arrays the same? Or, are they different? They have the same values, but there are duplicates in #array1 and not #array2.
What about this?
#array1 = qw( 1 1 2 3 4 5 );
#array2 = qw( 1 1 2 3 4 5 );
I would say that these arrays are the same, but Array::Utils::arrays_diff begs to differ. This is because Array::Utils assumes that there are no duplicate entries.
And, even the Perl FAQ pointed out by mob also says that It assumes that each element is unique in a given array. Is this an assumption you can make?
No matter what, hashes are the answer. It's easy and quick to look up a hash. The problem is what do you want to do with unique values.
Here's a solid solution that assumes duplicates don't matter:
sub array_diff {
my #array1 = #{ shift() };
my #array2 = #{ shift() };
my %array1_hash;
my %array2_hash;
# Create a hash entry for each element in #array1
for my $element ( #array1 ) {
$array1_hash{$element} = #array1;
}
# Same for #array2: This time, use map instead of a loop
map { $array_2{$_} = 1 } #array2;
for my $entry ( #array2 ) {
if ( not $array1_hash{$entry} ) {
return 1; #Entry in #array2 but not #array1: Differ
}
}
if ( keys %array_hash1 != keys %array_hash2 ) {
return 1; #Arrays differ
}
else {
return 0; #Arrays contain the same elements
}
}
If duplicates do matter, you'll need a way to count them. Here's using map not just to create a hash keyed by each element in the array, but also count the duplicates in the array:
my %array1_hash;
my %array2_hash;
map { $array1_hash{$_} += 1 } #array1;
map { $array2_hash{$_} += 2 } #array2;
Now, you can go through each hash and verify that not only do the keys exist, but that their entries match
for my $key ( keys %array1_hash ) {
if ( not exists $array2_hash{$key}
or $array1_hash{$key} != $array2_hash{$key} ) {
return 1; #Arrays differ
}
}
You will only exit the for loop if all of the entries in %array1_hash match their corresponding entries in %array2_hash. Now, you have to show that all of the entries in %array2_hash also match their entries in %array1_hash, and that %array2_hash doesn't have more entries. Fortunately, we can do what we did before:
if ( keys %array2_hash != keys %array1_hash ) {
return 1; #Arrays have a different number of keys: Don't match
}
else {
return; #Arrays have the same keys: They do match
}
You can use this for getting diffrence between two arrays
#!/usr/bin/perl -w
use strict;
my #list1 = (1, 2, 3, 4, 5);
my #list2 = (2, 3, 4);
my %diff;
#diff{ #list1 } = undef;
delete #diff{ #list2 };
You want to compare each element of #x against the element of the same index in #y, right? This will do it.
print "Index: $_ => \#x: $x[$_], \#y: $y[$_]\n"
for grep { $x[$_] != $y[$_] } 0 .. $#x;
...or...
foreach( 0 .. $#x ) {
print "Index: $_ => \#x: $x[$_], \#y: $y[$_]\n" if $x[$_] != $y[$_];
}
Which you choose kind of depends on whether you're more interested in keeping a list of indices to the dissimilar elements, or simply interested in processing the mismatches one by one. The grep version is handy for getting the list of mismatches. (original post)
n + n log n algorithm, if sure that elements are unique in each array (as hash keys)
my %count = ();
foreach my $element (#array1, #array2) {
$count{$element}++;
}
my #difference = grep { $count{$_} == 1 } keys %count;
my #intersect = grep { $count{$_} == 2 } keys %count;
my #union = keys %count;
So if I'm not sure of unity and want to check presence of the elements of array1 inside array2,
my %count = ();
foreach (#array1) {
$count{$_} = 1 ;
};
foreach (#array2) {
$count{$_} = 2 if $count{$_};
};
# N log N
if (grep { $_ == 1 } values %count) {
return 'Some element of array1 does not appears in array2'
} else {
return 'All elements of array1 are in array2'.
}
# N + N log N
my #a = (1,2,3);
my #b=(2,3,1);
print "Equal" if grep { $_ ~~ #b } #a == #b;
Not elegant, but easy to understand:
#!/usr/local/bin/perl
use strict;
my $file1 = shift or die("need file1");
my $file2 = shift or die("need file2");;
my #file1lines = split/\n/,`cat $file1`;
my #file2lines = split/\n/,`cat $file2`;
my %lines;
foreach my $file1line(#file1lines){
$lines{$file1line}+=1;
}
foreach my $file2line(#file2lines){
$lines{$file2line}+=2;
}
while(my($key,$value)=each%lines){
if($value == 1){
print "$key is in only $file1\n";
}elsif($value == 2){
print "$key is in only $file2\n";
}elsif($value == 3){
print "$key is in both $file1 and $file2\n";
}
}
exit;
__END__
Try to use List::Compare. IT has solutions for all the operations that can be performed on arrays.
I have two arrays, #a and #b. I want to do a compare among the elements of the two arrays.
my #a = qw"abc def efg ghy klm ghn";
my #b = qw"def ghy jgk lom com klm";
If any element matches then set a flag. Is there any simple way to do this?
First of all, your 2 arrays need to be written correctly.
#a = ("abc","def","efg","ghy","klm","ghn");
#b = ("def","efg","ghy","klm","ghn","klm");
Second of all, for arbitrary arrays (e.g. arrays whose elements may be references to other data structures) you can use Data::Compare.
For arrays whose elements are scalar, you can do comparison using List::MoreUtils pairwise BLOCK ARRAY1 ARRAY2, where BLOCK is your comparison subroutine. You can emulate pairwise (if you don't have List::MoreUtils access) via:
if (#a != #b) {
$equals = 0;
} else {
$equals = 1;
foreach (my $i = 0; $i < #a; $i++) {
# Ideally, check for undef/value comparison here as well
if ($a[$i] != $b[$i]) { # use "ne" if elements are strings, not numbers
# Or you can use generic sub comparing 2 values
$equals = 0;
last;
}
}
}
P.S. I am not sure but List::Compare may always sort the lists. I'm not sure if it can do pairwise comparisons.
List::Compare
if ( scalar List::Compare->new(\#a, \#b)->get_intersection ) {
…
}
Check to create an intersect function, which will return a list of items that are present in both lists. Then your return value is dependent on the number of items in the intersected list.
You can easily find on the web the best implementation of intersect for Perl. I remember looking for it a few years ago.
Here's what I found :
my #array1 = (1, 2, 3);
my #array2 = (2, 3, 4);
my %original = ();
my #isect = ();
map { $original{$_} = 1 } #array1;
#isect = grep { $original{$_} } #array2;
This is one way:
use warnings;
use strict;
my #a = split /,/, "abc,def,efg,ghy,klm,ghn";
my #b = split /,/, "def,ghy,jgk,lom,com,klm";
my $flag = 0;
my %a;
#a{#a} = (1) x #a;
for (#b) {
if ($a{$_}) {
$flag = 1;
last;
}
}
print "$flag\n";
From the requirement that 'if any element matches', use the intersection of sets:
sub set{
my %set = map { $_, undef }, #_;
return sort keys %set;
}
sub compare{
my ($listA,$listB) = #_;
return ( (set(#$listA)-set(#$listB)) > 0)
}
my #a = qw' abc def efg ghy klm ghn ';
my #b = qw' def ghy jgk lom com klm ';
my $flag;
foreach my $item(#a) {
$flag = #b~~$item ? 0 : 1;
last if !$flag;
}
Note that you will need Perl 5.10, or later, to use the smart match operator (~~) .
Brute force should do the trick for small a n:
my $flag = 0;
foreach my $i (#a) {
foreach my $k (#b) {
if ($i eq $k) {
$flag = 1;
last;
}
}
}
For a large n, use a hash table:
my $flag = 0;
my %aa = ();
$aa{$_} = 1 foreach (#a);
foreach my $i (#b) {
if ($aa{$i}) {
$flag = 1;
last;
}
}
Where a large n is |#a| + |#b| > ~1000 items
IMHO, you should use List::MoreUtils::pairwise. However, if for some reason you cannot, then the following sub would return a 1 for every index where the value in the first array compares equal to the value in the second array. You can generalize this method as much as you want and pass your own comparator if you want to, but at that point, just installing List::MoreUtils would be a more productive use of your time.
use strict; use warnings;
my #a = qw(abc def ghi jkl);
my #b = qw(abc dgh dlkfj jkl kjj lkm);
my $map = which_ones_equal(\#a, \#b);
print join(', ', #$map), "\n";
sub which_ones_equal {
my ($x, $y, $compare) = #_;
my $last = $#$x > $#$y ? $#$x : $#$y;
no warnings 'uninitialized';
return [ map { 0 + ($x->[$_] eq $y->[$_]) } $[ .. $last ];
}
This is Perl. The 'obvious' solution:
my #a = qw"abc def efg ghy klm ghn";
my #b = qw"def ghy jgk lom com klm";
print "arrays equal\n"
if #a == #b and join("\0", #a) eq join("\0", #b);
given "\0" not being in #a.
But thanks for confirming that there is no other generic solution than rolling your own.
my #a1 = qw|a b c d|;
my #a2 = qw|b c d e|;
for my $i (0..$#a1) {
say "element $i of array 1 was not found in array 2"
unless grep {$_ eq $a1[$i]} #a2
}
If you would consider the arrays with different order to be different, you may use Array::Diff
if (Array::Diff->diff(\#a, \#b)->count) {
# not_same
} else {
# same
}
This question still could mean two things where it states "If any element matches then set a flag":
Elements at the same position, i.e $a[2] eq $b[2]
Values at any position, i.e. $a[3] eq $b[5]
For case 1, you might do this:
# iterate over all positions, and compare values at that position
my #matches = grep { $a[$_] eq $b[$_] } 0 .. $#a;
# set flag if there's any match at the same position
my $flag = 1 if #matches;
For case 2, you might do that:
# make a hash of #a and check if any #b are in there
my %a = map { $_ => 1 } #a;
my #matches = grep { $a{$_} } #b;
# set flag if there's matches at any position
my $flag = 1 if #matches;
Note that in the first case, #matches holds the indexes of where there are matching elements, and in the second case #matches holds the matching values in the order in which they appear in #b.
As per the title, I'm trying to find a way to programmatically determine the longest portion of similarity between several strings.
Example:
file:///home/gms8994/Music/t.A.T.u./
file:///home/gms8994/Music/nina%20sky/
file:///home/gms8994/Music/A%20Perfect%20Circle/
Ideally, I'd get back file:///home/gms8994/Music/, because that's the longest portion that's common for all 3 strings.
Specifically, I'm looking for a Perl solution, but a solution in any language (or even pseudo-language) would suffice.
From the comments: yes, only at the beginning; but there is the possibility of having some other entry in the list, which would be ignored for this question.
Edit: I'm sorry for mistake. My pity that I overseen that using my variable inside countit(x, q{}) is big mistake. This string is evaluated inside Benchmark module and #str was empty there. This solution is not as fast as I presented. See correction below. I'm sorry again.
Perl can be fast:
use strict;
use warnings;
package LCP;
sub LCP {
return '' unless #_;
return $_[0] if #_ == 1;
my $i = 0;
my $first = shift;
my $min_length = length($first);
foreach (#_) {
$min_length = length($_) if length($_) < $min_length;
}
INDEX: foreach my $ch ( split //, $first ) {
last INDEX unless $i < $min_length;
foreach my $string (#_) {
last INDEX if substr($string, $i, 1) ne $ch;
}
}
continue { $i++ }
return substr $first, 0, $i;
}
# Roy's implementation
sub LCP2 {
return '' unless #_;
my $prefix = shift;
for (#_) {
chop $prefix while (! /^\Q$prefix\E/);
}
return $prefix;
}
1;
Test suite:
#!/usr/bin/env perl
use strict;
use warnings;
Test::LCP->runtests;
package Test::LCP;
use base 'Test::Class';
use Test::More;
use Benchmark qw(:all :hireswallclock);
sub test_use : Test(startup => 1) {
use_ok('LCP');
}
sub test_lcp : Test(6) {
is( LCP::LCP(), '', 'Without parameters' );
is( LCP::LCP('abc'), 'abc', 'One parameter' );
is( LCP::LCP( 'abc', 'xyz' ), '', 'None of common prefix' );
is( LCP::LCP( 'abcdefgh', ('abcdefgh') x 15, 'abcdxyz' ),
'abcd', 'Some common prefix' );
my #str = map { chomp; $_ } <DATA>;
is( LCP::LCP(#str),
'file:///home/gms8994/Music/', 'Test data prefix' );
is( LCP::LCP2(#str),
'file:///home/gms8994/Music/', 'Test data prefix by LCP2' );
my $t = countit( 1, sub{LCP::LCP(#str)} );
diag("LCP: ${\($t->iters)} iterations took ${\(timestr($t))}");
$t = countit( 1, sub{LCP::LCP2(#str)} );
diag("LCP2: ${\($t->iters)} iterations took ${\(timestr($t))}");
}
__DATA__
file:///home/gms8994/Music/t.A.T.u./
file:///home/gms8994/Music/nina%20sky/
file:///home/gms8994/Music/A%20Perfect%20Circle/
Test suite result:
1..7
ok 1 - use LCP;
ok 2 - Without parameters
ok 3 - One parameter
ok 4 - None of common prefix
ok 5 - Some common prefix
ok 6 - Test data prefix
ok 7 - Test data prefix by LCP2
# LCP: 22635 iterations took 1.09948 wallclock secs ( 1.09 usr + 0.00 sys = 1.09 CPU) # 20766.06/s (n=22635)
# LCP2: 17919 iterations took 1.06787 wallclock secs ( 1.07 usr + 0.00 sys = 1.07 CPU) # 16746.73/s (n=17919)
That means that pure Perl solution using substr is about 20% faster than Roy's solution at your test case and one prefix finding takes about 50us. There is not necessary using XS unless your data or performance expectations are bigger.
The reference given already by Brett Daniel for the Wikipedia entry on "Longest common substring problem" is very good general reference (with pseudocode) for your question as stated. However, the algorithm can be exponential. And it looks like you might actually want an algorithm for longest common prefix which is a much simpler algorithm.
Here's the one I use for longest common prefix (and a ref to original URL):
use strict; use warnings;
sub longest_common_prefix {
# longest_common_prefix( $|# ): returns $
# URLref: http://linux.seindal.dk/2005/09/09/longest-common-prefix-in-perl
# find longest common prefix of scalar list
my $prefix = shift;
for (#_) {
chop $prefix while (! /^\Q$prefix\E/);
}
return $prefix;
}
my #str = map {chomp; $_} <DATA>;
print longest_common_prefix(#ARGV), "\n";
__DATA__
file:///home/gms8994/Music/t.A.T.u./
file:///home/gms8994/Music/nina%20sky/
file:///home/gms8994/Music/A%20Perfect%20Circle/
If you truly want a LCSS implementation, refer to these discussions (Longest Common Substring and Longest Common Subsequence) at PerlMonks.org. Tree::Suffix would probably be the best general solution for you and implements, to my knowledge, the best algorithm. Unfortunately recent builds are broken. But, a working subroutine does exist within the discussions referenced on PerlMonks in this post by Limbic~Region (reproduced here with your data).
#URLref: http://www.perlmonks.org/?node_id=549876
#by Limbic~Region
use Algorithm::Loops 'NestedLoops';
use List::Util 'reduce';
use strict; use warnings;
sub LCS{
my #str = #_;
my #pos;
for my $i (0 .. $#str) {
my $line = $str[$i];
for (0 .. length($line) - 1) {
my $char= substr($line, $_, 1);
push #{$pos[$i]{$char}}, $_;
}
}
my $sh_str = reduce {length($a) < length($b) ? $a : $b} #str;
my %map;
CHAR:
for my $char (split //, $sh_str) {
my #loop;
for (0 .. $#pos) {
next CHAR if ! $pos[$_]{$char};
push #loop, $pos[$_]{$char};
}
my $next = NestedLoops([#loop]);
while (my #char_map = $next->()) {
my $key = join '-', #char_map;
$map{$key} = $char;
}
}
my #pile;
for my $seq (keys %map) {
push #pile, $map{$seq};
for (1 .. 2) {
my $dir = $_ % 2 ? 1 : -1;
my #offset = split /-/, $seq;
$_ += $dir for #offset;
my $next = join '-', #offset;
while (exists $map{$next}) {
$pile[-1] = $dir > 0 ?
$pile[-1] . $map{$next} : $map{$next} . $pile[-1];
$_ += $dir for #offset;
$next = join '-', #offset;
}
}
}
return reduce {length($a) > length($b) ? $a : $b} #pile;
}
my #str = map {chomp; $_} <DATA>;
print LCS(#str), "\n";
__DATA__
file:///home/gms8994/Music/t.A.T.u./
file:///home/gms8994/Music/nina%20sky/
file:///home/gms8994/Music/A%20Perfect%20Circle/
It sounds like you want the k-common substring algorithm. It is exceptionally simple to program, and a good example of dynamic programming.
My first instinct is to run a loop, taking the next character from each string, until the characters are not equal. Keep a count of what position in the string you're at and then take a substring (from any of the three strings) from 0 to the position before the characters aren't equal.
In Perl, you'll have to split up the string first into characters using something like
#array = split(//, $string);
(splitting on an empty character sets each character into its own element of the array)
Then do a loop, perhaps overall:
$n =0;
#array1 = split(//, $string1);
#array2 = split(//, $string2);
#array3 = split(//, $string3);
while($array1[$n] == $array2[$n] && $array2[$n] == $array3[$n]){
$n++;
}
$sameString = substr($string1, 0, $n); #n might have to be n-1
Or at least something along those lines. Forgive me if this doesn't work, my Perl is a little rusty.
If you google for "longest common substring" you'll get some good pointers for the general case where the sequences don't have to start at the beginning of the strings.
Eg, http://en.wikipedia.org/wiki/Longest_common_substring_problem.
Mathematica happens to have a function for this built in:
http://reference.wolfram.com/mathematica/ref/LongestCommonSubsequence.html (Note that they mean contiguous subsequence, ie, substring, which is what you want.)
If you only care about the longest common prefix then it should be much faster to just loop for i from 0 till the ith characters don't all match and return substr(s, 0, i-1).
From http://forums.macosxhints.com/showthread.php?t=33780
my #strings =
(
'file:///home/gms8994/Music/t.A.T.u./',
'file:///home/gms8994/Music/nina%20sky/',
'file:///home/gms8994/Music/A%20Perfect%20Circle/',
);
my $common_part = undef;
my $sep = chr(0); # assuming it's not used legitimately
foreach my $str ( #strings ) {
# First time through loop -- set common
# to whole
if ( !defined $common_part ) {
$common_part = $str;
next;
}
if ("$common_part$sep$str" =~ /^(.*).*$sep\1.*$/)
{
$common_part = $1;
}
}
print "Common part = $common_part\n";
Faster than above, uses perl's native binary xor function, adapted from perlmongers solution (the $+[0] didn't work for me):
sub common_suffix {
my $comm = shift #_;
while ($_ = shift #_) {
$_ = substr($_,-length($comm)) if (length($_) > length($comm));
$comm = substr($comm,-length($_)) if (length($_) < length($comm));
if (( $_ ^ $comm ) =~ /(\0*)$/) {
$comm = substr($comm, -length($1));
} else {
return undef;
}
}
return $comm;
}
sub common_prefix {
my $comm = shift #_;
while ($_ = shift #_) {
$_ = substr($_,0,length($comm)) if (length($_) > length($comm));
$comm = substr($comm,0,length($_)) if (length($_) < length($comm));
if (( $_ ^ $comm ) =~ /^(\0*)/) {
$comm = substr($comm,0,length($1));
} else {
return undef;
}
}
return $comm;
}