如何从字符串列表中检索部分匹配？[重复]

如何从字符串列表中检索部分匹配？[重复]

• startswith

  和

  in

  ，返回布尔值
• 该

  in

  运营商成员的考验。
• 可以使用

  list-comprehension

  或来执行

  filter

• list-comprehension

  与一起使用

  in

  是经过测试的最快实施。
• 如果大小写不是问题，请考虑将所有单词映射为小写。

  • l = list(map(str.lower, l))

    。

filter

：

• 使用

  filter

  会创建一个

  filter

  对象，因此

  list()

  用于显示中的所有匹配值

  list

  。

  l = [‘ones’, ‘twos’, ‘threes’]
  wanted = ‘three’
  

  using startswith

  result = list(filter(lambda x: x.startswith(wanted), l))

  using in

  result = list(filter(lambda x: wanted in x, l))

  print(result)
  [out]:
  [‘threes’]
  
  

list-comprehension

l = ['ones', 'twos', 'threes']wanted = 'three'# using startswithresult = [v for v in l if v.startswith(wanted)]# using inresult = [v for v in l if wanted in v]print(result)[out]:['threes']

哪个实现更快？

• 使用

  words

  语料库

  nltk

• 用词

  'three'

  • ['three', 'threefold', 'threefolded', 'threefoldedness', 'threefoldly', 'threefoldness', 'threeling', 'threeness', 'threepence', 'threepenny', 'threepennyworth', 'threescore', 'threesome']

  from nltk.corpus import words

  %timeit list(filter(lambda x: x.startswith(wanted), words.words()))
  [out]:
  47.4 ms ± 1.9 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

  %timeit list(filter(lambda x: wanted in x, words.words()))
  [out]:
  27 ms ± 1.78 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

  %timeit [v for v in words.words() if v.startswith(wanted)]
  [out]:
  34.1 ms ± 768 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

  %timeit [v for v in words.words() if wanted in v]
  [out]:
  14.5 ms ± 63.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)