如何从字符串列表中检索部分匹配项? [英] How to retrieve partial matches from a list of strings?

问题描述

有关在 数字 列表中检索部分匹配项的方法，请访问:

• 如何返回匹配条件的列表的子集?

• Python:在列表中查找

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但是，如果您正在寻找如何检索 字符串 列表的部分匹配项，您将在以下答案中找到简要解释的最佳方法. /p>

SO:具有部分匹配项的Python列表查找显示了如何返回bool，如果list包含部分匹配某个字符串的元素(例如begins，ends或contains).但是如何 返回元素本身 ，而不是True或False

示例:

l = ['ones', 'twos', 'threes']
wanted = 'three'

此处，链接问题中的方法将使用以下命令返回True:

any(s.startswith(wanted) for s in l)

那么如何返回元素'threes'呢?

解决方案

• startswith 和 filter 创建一个filter对象，因此list()用于显示list中的所有匹配值.

 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 startswith
result = [v for v in l if v.startswith(wanted)]

# using in
result = [v for v in l if wanted in v]

print(result)
[out]:
['threes']
 

哪种实施速度更快?

• 使用nltk
中的words语料库
• 带'three'的词
  • ['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)
 

For approaches to retrieving partial matches in a numeric list, go to:

• How to return a subset of a list that matches a condition?

• Python: Find in list

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But if you're looking for how to retrieve partial matches for a list of strings, you'll find the best approaches concisely explained in the answer below.

SO: Python list lookup with partial match shows how to return a bool, if a list contains an element that partially matches (e.g. begins, ends, or contains) a certain string. But how can you return the element itself, instead of True or False

Example:

l = ['ones', 'twos', 'threes']
wanted = 'three'

Here, the approach in the linked question will return True using:

any(s.startswith(wanted) for s in l)

So how can you return the element 'threes' instead?

解决方案

• startswith and in, return a Boolean
• The in operator is a test of membership.
• This can be performed with a list-comprehension or filter
• Using a list-comprehension, with in, is the fastest implementation tested.
• If case is not an issue, consider mapping all the words to lowercase.
  • l = list(map(str.lower, l)).

filter:

• Using filter creates a filter object, so list() is used to show all the matching values in a 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 startswith
result = [v for v in l if v.startswith(wanted)]

# using in
result = [v for v in l if wanted in v]

print(result)
[out]:
['threes']

Which implementation is faster?

• Using the words corpus from nltk
• Words with '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)

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