检查一个字符串中的单词是否在另一个字符串中的最快方法是什么? [英] What's the fastest way to check if a word from one string is in another string?

问题描述

我有一串单词；让我们称它们为 bad:

I have a string of words; let's call them bad:

bad = "foo bar baz"

我可以将此字符串保留为空格分隔的字符串或列表:

I can keep this string as a whitespace separated string, or as a list:

bad = bad.split(" ");

如果我有另一个字符串，就像这样:

If I have another string, like so:

str = "This is my first foo string"

检查 bad 字符串中是否有任何单词在我的比较字符串中的快速方法是什么?和如果找到该单词，删除该单词的最快方法是什么?

What's the fasted way to check if any word from the bad string is within my comparison string, and what's the fastest way to remove said word if it's found?

#Find if a word is there
bad.split(" ").each do |word|
  found = str.include?(word)
end

#Remove the word
bad.split(" ").each do |word|
  str.gsub!(/#{word}/, "")
end

推荐答案

如果坏词列表很大，哈希会快很多:

If the list of bad words gets huge, a hash is a lot faster:

    require 'benchmark'

    bad = ('aaa'..'zzz').to_a    # 17576 words
    str= "What's the fasted way to check if any word from the bad string is within my "
    str += "comparison string, and what's the fastest way to remove said word if it's "
    str += "found" 
    str *= 10

    badex = /\b(#{bad.join('|')})\b/i

    bad_hash = {}
    bad.each{|w| bad_hash[w] = true}

    n = 10
    Benchmark.bm(10) do |x|

      x.report('regex:') {n.times do 
        str.gsub(badex,'').squeeze(' ')
      end}

      x.report('hash:') {n.times do
        str.gsub(/\b\w+\b/){|word| bad_hash[word] ? '': word}.squeeze(' ')
      end}

    end
                user     system      total        real
regex:     10.485000   0.000000  10.485000 ( 13.312500)
hash:       0.000000   0.000000   0.000000 (  0.000000)

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