4 ms·

> It's quite possible that a sorted word count was easiest to do with bash and unix utilities in 1992. Now, it's easier and more comprehensible in your favorit

> It's quite possible that a sorted word count was easiest to do with bash and unix utilities in 1992. Now, it's easier and more comprehensible in your favorite scripting language It'd be interesting to see how true this is in reality. I'm not at all convinced that most scripting languages can do this quite as concisely as bash and the Unix userland. (But, being honest, this problem seems very well aligned to that tooling's strenghts.)

sobellian 8y ago
Well, what is easy or comprehensible isn't necessarily concise. Bash is great for code golf but I know few people who prefer to code in it.
- sobellian 8y ago
  And for completeness, a Ruby implementation: puts STDIN.read .split(/[^A-Za-z]+/) .group_by(&:downcase).transform_values(&:size) .sort_by(&:last).reverse .take(ARGV[0].to_i) .map{ |l| l.join("\t") }.join("\n") Wordier than the bash script, but still pretty short and fairly trivial. Anyone could make this in five minutes.
btilly 8y ago
It is easy to do in your favorite scripting language. But is unlikely to be as short. Comprehensibility is in the eye of the beholder. Here is a Perl solution for comparison. use strict; use warnings; my $limit = shift @ARGV; my %count; for my $line (<>) { while ($line =~ /(\w+)/g) { $count{lc $1}++; } } @words = sort {$count{$b} <=> $count{$a} or $a cmp $b} keys %count; $limit = @words if @words < $limit; print "$_\t$count{$_}\n" for @words[0..($limit-1)];
- mkl 8y ago
  Python is relatively short, batteries included and all that: import re import sys import collections words = re.findall('[a-z]+', sys.stdin.read().lower()) c = collections.Counter(words) for word, count in c.most_common(int(sys.argv[1])): print(f'{count:>7} {word}') Three times the character count of the Unix pipe version, but IMHO a lot more readable (and generalisable).
  - mkesper 8y ago
    You don't want to read all input into memory first.
    - mkl 8y ago
      Fair enough. import re import sys import collections c = collections.Counter() for line in sys.stdin: words = re.findall('[a-z]+', line.lower()) c.update(words) for word, count in c.most_common(int(sys.argv[1])): print(f'{count:>7} {word}')
- lizmat 8y ago
  Just for reference: a Perl 6 version: sub MAIN($limit = Inf) { my %bag is Bag = words.map(&lc); say "{.key}\t{.value}" for %bag.sort( { $^b.value cmp $^a.value || $^a.key cmp $^b.key } )[^$limit] } And for what it's worth, the `words` function is lazy, so it won't read all of the words into memory first.
  - b2gills 8y ago
    I would have written the sort differently %bag.sort({ -.value, .key })

sobellian 8y ago

Well, what is easy or comprehensible isn't necessarily concise. Bash is great for code golf but I know few people who prefer to code in it.

sobellian 8y ago
And for completeness, a Ruby implementation: puts STDIN.read .split(/[^A-Za-z]+/) .group_by(&:downcase).transform_values(&:size) .sort_by(&:last).reverse .take(ARGV[0].to_i) .map{ |l| l.join("\t") }.join("\n") Wordier than the bash script, but still pretty short and fairly trivial. Anyone could make this in five minutes.

btilly 8y ago

It is easy to do in your favorite scripting language. But is unlikely to be as short. Comprehensibility is in the eye of the beholder. Here is a Perl solution for comparison. use strict; use warnings; my $limit = shift @ARGV; my %count; for my $line (<>) { while ($line =~ /(\w+)/g) { $count{lc $1}++; } } @words = sort {$count{$b} <=> $count{$a} or $a cmp $b} keys %count; $limit = @words if @words < $limit; print "$_\t$count{$_}\n" for @words[0..($limit-1)];

mkl 8y ago
Python is relatively short, batteries included and all that: import re import sys import collections words = re.findall('[a-z]+', sys.stdin.read().lower()) c = collections.Counter(words) for word, count in c.most_common(int(sys.argv[1])): print(f'{count:>7} {word}') Three times the character count of the Unix pipe version, but IMHO a lot more readable (and generalisable).
- mkesper 8y ago
  You don't want to read all input into memory first.
  - mkl 8y ago
    Fair enough. import re import sys import collections c = collections.Counter() for line in sys.stdin: words = re.findall('[a-z]+', line.lower()) c.update(words) for word, count in c.most_common(int(sys.argv[1])): print(f'{count:>7} {word}')
lizmat 8y ago
Just for reference: a Perl 6 version: sub MAIN($limit = Inf) { my %bag is Bag = words.map(&lc); say "{.key}\t{.value}" for %bag.sort( { $^b.value cmp $^a.value || $^a.key cmp $^b.key } )[^$limit] } And for what it's worth, the `words` function is lazy, so it won't read all of the words into memory first.
- b2gills 8y ago
  I would have written the sort differently %bag.sort({ -.value, .key })