用Python中的regex优化查找两个列表之间的匹配子串

Question

下面是我的方法，通过一个包含“words”的列表来搜索包含“短语”的列表中的子字符串，并返回在包含短语的列表中的每个元素中找到的匹配子字符串。你知道吗

import re

def is_phrase_in(phrase, text):
    return re.search(r"\b{}\b".format(phrase), text, re.IGNORECASE) is not None

list_to_search = ['my', 'name', 'is', 'you', 'your']
list_to_be_searched = ['hello my', 'name is', 'john doe doe is last name', 'how are you', 'what is your name', 'my name is jane doe']

to_be_appended = []
for phrase in list_to_be_searched:
    searched = []
    for word in list_to_search:
        if is_phrase_in(word,phrase) is True:
            searched.append(word)
    to_be_appended.append(searched)
print(to_be_appended)

# (desired and actual) output
[['my'],
 ['name', 'is'],
 ['name', 'is'],
 ['you'],
 ['name', 'is', 'your'],
 ['my', 'name', 'is']]

由于“words”（或list-to-search）列表有约1700个单词，“phrases”（或list-to-be-search）列表有约26561个单词，完成代码需要30分钟。我不认为我上面的代码是考虑到Pythonic的编码方式和高效的数据结构实现的。：（

有谁能给我一些建议来优化或者加快速度？你知道吗

谢谢！你知道吗

实际上，我写错了上面的例子。 如果“列表到搜索”中的元素多于2个单词怎么办？你知道吗

import re

def is_phrase_in(phrase, text):
    return re.search(r"\b{}\b".format(phrase), text, re.IGNORECASE) is not None

list_to_search = ['hello my', 'name', 'is', 'is your name', 'your']
list_to_be_searched = ['hello my', 'name is', 'john doe doe is last name', 'how are you', 'what is your name', 'my name is jane doe']

to_be_appended = []
for phrase in list_to_be_searched:
    searched = []
    for word in list_to_search:
        if is_phrase_in(word,phrase) is True:
            searched.append(word)
    to_be_appended.append(searched)
print(to_be_appended)
# (desired and actual) output
[['hello my'],
 ['name', 'is'],
 ['name', 'is'],
 [],
 ['name', 'is', 'is your name', 'your'],
 ['name', 'is']]

时机 第一种方法：

%%timeit
def is_phrase_in(phrase, text):
    return re.search(r"\b{}\b".format(phrase), text, re.IGNORECASE) is not None

    list_to_search = ['hello my', 'name', 'is', 'is your name', 'your']
    list_to_be_searched = ['hello my', 'name is', 'john doe doe is last name', 'how are you', 'what is your name', 'my name is jane doe']
to_be_appended = []
for phrase in list_to_be_searched:
    searched = []
    for word in list_to_search:
        if is_phrase_in(word,phrase) is True:
            searched.append(word)
    to_be_appended.append(searched)
#43.2 µs ± 346 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

第二种方法（嵌套列表和关于芬德尔)你知道吗

%%timeit
[[j for j in list_to_search if j in re.findall(r"\b{}\b".format(j), i)] for i in list_to_be_searched]
#40.3 µs ± 454 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)\

时间安排确实有所改善，但会有更快的方法吗？或者，考虑到它的功能，这个任务在基因上是缓慢的？你知道吗

Answer 1

虽然最直接/清晰的方法是使用列表理解，但我想看看regex是否可以做得更好。你知道吗

在list_to_be_searched中的每个项目上使用regex似乎没有任何性能提升。但是将list_to_be_searched加入一个大的文本块，并将其与由list_to_search构造的正则表达式模式相匹配，性能略有提高：

In [1]: import re
   ...:
   ...: list_to_search = ['my', 'name', 'is', 'you', 'your']
   ...: list_to_be_searched = ['hello my', 'name is', 'john doe doe is last name', 'how are you', 'what is your name', 'my name is jane doe']
   ...:
   ...: def simple_method(to_search, to_be_searched):
   ...:   return [[j for j in to_search if j in i.split()] for i in to_be_searched]
   ...:
   ...: def regex_method(to_search, to_be_searched):
   ...:   word = re.compile(r'(\b(?:' + r'|'.join(to_search) + r')\b(?:\n)?)')
   ...:   blob = '\n'.join(to_be_searched)
   ...:   phrases = word.findall(blob)
   ...:   return [phrase.split(' ') for phrase in ' '.join(phrases).split('\n ')]
   ...:
   ...: def alternate_regex_method(to_search, to_be_searched):
   ...:   word = re.compile(r'(\b(?:' + r'|'.join(to_search) + r')\b(?:\n)?)')
   ...:   phrases = []
   ...:   for item in to_be_searched:
   ...:     phrases.append(word.findall(item))
   ...:   return phrases
   ...:

In [2]: %timeit -n 100 simple_method(list_to_search, list_to_be_searched)
100 loops, best of 3: 23.1 µs per loop

In [3]: %timeit -n 100 regex_method(list_to_search, list_to_be_searched)
100 loops, best of 3: 18.6 µs per loop

In [4]: %timeit -n 100 alternate_regex_method(list_to_search, list_to_be_searched)
100 loops, best of 3: 23.4 µs per loop

---

为了了解在大量输入下的表现，我使用了1000个英语中最常用的单词，一次一个单词作为^{cd3>}，而来自Gutenberg项目的David Copperfield的整个文本一次一行作为^{cd1>}：

In [5]: book = open('/tmp/copperfield.txt', 'r+')

In [6]: list_to_be_searched = [line for line in book]

In [7]: len(list_to_be_searched)
Out[7]: 38589

In [8]: words = open('/tmp/words.txt', 'r+')

In [9]: list_to_search = [word for word in words]

In [10]: len(list_to_search)
Out[10]: 1000

结果如下：

In [15]: %timeit -n 10 simple_method(list_to_search, list_to_be_searched)
10 loops, best of 3: 31.9 s per loop

In [16]: %timeit -n 10 regex_method(list_to_search, list_to_be_searched)
10 loops, best of 3: 4.28 s per loop

In [17]: %timeit -n 10 alternate_regex_method(list_to_search, list_to_be_searched)
10 loops, best of 3: 4.43 s per loop

因此，如果您对性能感兴趣，可以使用任意一种regex方法。希望有帮助！：）

Answer 2

可以使用嵌套列表：

list_to_search = ['my', 'name', 'is', 'you', 'your']
list_to_be_searched = ['hello my', 'name is', 'john doe doe is last name',
                       'how are you', 'what is your name', 'my name is jane doe']

[[j for j in list_to_search if j in i.split()] for i in list_to_be_searched]

[['my'],
 ['name', 'is'],
 ['name', 'is'],
 ['you'],
 ['name', 'is', 'your'],
 ['my', 'name', 'is']]