判断随机产生单纯词的另一种方法

判断随机产生单词的另一种方法

在上一篇中我介绍了判断随机产生单词的3种方法，大致都是用了外在程序spell。现在本猫又在Mac OS X系统上找到了如下文件：/usr/share/dict/words ，其中放置了N多个英语单词啊：

apple@kissAir: dict$ls -ldh words

lrwxr-xr-x 1 root wheel 4B 10 18 14:00 words -> web2

apple@kissAir: dict$ls -lh web2

-r--r--r-- 1 root wheel 2.4M 9 10 04:47 web2

apple@kissAir: dict$wc -n words

wc: illegal option -- n

usage: wc [-clmw] [file ...]

apple@kissAir: dict$wc -l words

235886 words

一行一个单词，即一共23万多个单词，我们可以抛掉spell程序，自己写一个is_spell?方法来判断单词是否可拼写啦，以下是增加way4后的代码，放弃了命令行参数的方式，而是用benchmark包来测试性能：

#!/usr/bin/ruby
#code by hopy 2014.12.08
#random create some words and check if a valid word!

require 'tempfile'
require 'benchmark'

words_path = "/usr/share/dict/words"
f = File.open(words_path,"r")
$lines = f.readlines
$lines.map! {|word|word.chomp!}
f.close

def rand_words(n=10000,min_len=2,max_len=12)
	chars = (("a".."z").to_a * max_len).freeze
	words = []
	srand
	n.times do |x|
		len = min_len + (rand*1000).to_i % max_len
		idxes = []
		len.times {idxes<<(rand*100)%26}
		chars.shuffle
		words << chars.values_at(*idxes).join
		idxes.clear
	end 
	words
end

#ret word that can spell or ret nil. (way1)
def spell_word(word)
	cmd = `echo #{word}|spell`.chomp
	if cmd == word
		return nil
	else
		return word
	end
end

#spell all words by tmpfile. (way2)
def spell_words(words)
	puts "using spell_words..."
	f = Tempfile.new("#{$$}_spell_blablabla")
	#f = File.open("spell_test","w+")
	#f.write Marshal.dump(words)
	f.write words.join(" ")
	f.close

	cmd = `spell #{f.path}`
	no_spell_words = cmd.split("\n")
	words - no_spell_words
end

#spell all words by tmpfile and spell ret is also use tmpfile. (way3)
def spell_words2(words)
	puts "using spell_words2..."
	f_words = Tempfile.new("#{$$}_spell_words")
	f_ret = Tempfile.new("#{$$}_spell_ret")
	f_ret.close

	f_words.write words.join(" ")
	f_words.close

	cmd = `spell #{f_words.path} > #{f_ret.path}`
	f=File.open(f_ret.path)
	no_spell_words = f.read.split("\n")
	f.close
	words - no_spell_words
end

def is_spell?(word)
	$lines.include? word
end

#利用is_spell?判断word是否可拼写的方法。(way4)
def spell_words3(words)
=begin
	words.each do |word|
		printf "#{word} " if is_spell?(word)
	end
=end
	words.select {|word|is_spell?(word)}
end

def sh_each_spell_word(spell_words)
	spell_words.each {|word|printf "#{word} "}
end

words_count = 2000
$words = nil
puts "words_count is 2000,now test..."
Benchmark.bm do |bc|
	bc.report("rand_words:\n") {$words = rand_words(words_count)};puts ""
	bc.report("way1:spell_word:\n") {$words.each {|w|printf "#{w} " if spell_word(w)}};puts ""
	bc.report("way2:spell_words:\n") {sh_each_spell_word(spell_words($words))};puts ""
	bc.report("way3:spell_words2:\n") {sh_each_spell_word(spell_words2($words))};puts ""
	bc.report("way4:spell_words3:\n") {sh_each_spell_word(spell_words3($words))};puts ""
end

不过Mac OS X自身不带spell程序，用brew不知要安装哪一个；而虚拟机中的ubuntu的spell死活无法升级。等明天用本猫的x61来测试吧！

现在已经是明天鸟!发现ubuntu中自带的words文件包含单词比Mac下的要少,只有9万多个单词啊,遂将其用Mac下的文件替换,可以看到他比spell程序实际枚举的单词要多哦:

wisy@wisy-ThinkPad-X61:~/src/ruby_src$ ./x.rb
words_count is 2000,now test...
       user     system      total        real
rand_words:
  0.050000   0.000000   0.050000 (  0.069850)

way1:spell_word:
ho of ts mu so or wag us to lo um ts pa pip mid hip vs no of oboe iv yr re so   0.330000   3.170000  13.480000 ( 29.903239)

way2:spell_words:
using spell_words...
ho of ts mu so or wag us to lo um ts pa pip mid hip vs no of oboe iv yr re so   0.000000   0.000000   0.080000 (  5.485613)

way3:spell_words2:
using spell_words2...
ho of ts mu so or wag us to lo um ts pa pip mid hip vs no of oboe iv yr re so   0.010000   0.010000   0.100000 (  4.854248)

way4:spell_words3:
ho of pob dob mu bo so sa or wag us jo aw to lo um li ca se pa ava bo sho pip mid til tue ya en hip no of di ug oboe io en yr re da eer so ym  36.580000   0.290000  36.870000 ( 37.444370)

我们写的新的方法(way4)竟然是最慢的!!!不试不知道,一试吓一跳啊!

判断随机产生单纯词的另一种方法

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