Python的正则表达式来计算句子中的多个匹配的字符串

Question

我试图让从大的文本中找到字符串hello awesome world生成模式计数。图案通过置换单词和在与之间替换*一个字生成。在这个例子中我仅使用4种模式，以简化的东西。我不是很熟悉的正则表达式，所以我的代码不符合我需要的一切呢。我可能会很快搞清楚，但我不知道这是否会很好地扩展，当我喂真实数据。

这些问题是如何解决我的代码，并有更好的/更快的方式来实现我的目标？这里是我的代码波纹管与解释相关。

import re
from collections import Counter


# Input text. Could consist of hundreds of thousands of sentences.
txt = """
Lorèm ipsum WORLD dolor AWESOME sit amèt, consectetur adipiscing elit. 
Duis id AWESOME HELLO lorem metus. Pràesent molestie malesuada finibus. 
Morbi non èx a WORLD HELLO AWESOME erat bibendum rhoncus. Quisque sit 
ametnibh cursus, tempor mi et, sodàles neque. Nunc dapibus vitae ligula at porta. 
Quisque sit amet màgna eù sem sagittis dignissim et non leo. 
Quisque WORLD, AWESOME dapibus et vèlit tristique tristique. Sed 
efficitur dui tincidunt, aliquet lèo eget, pellentesque felis. Donec 
venenatis elit ac aliquet varius. Vestibulum ante ipsum primis in faucibus
orci luctus et ultrices posuere cubilia Curae. Vestibulum sed ligula 
gravida, commodo neque at, mattis urna. Duis nisl neque, sollicitudin nec 
mauris sit amet, euismod semper massa. Curabitur sodales ultrices nibh, 
ut ultrices ante maximus sed. Donec rutrum libero in turpis gravida 
dignissim. Suspendisse potenti. Praesent eu tempor quam, id dictum felis. 
Nullam aliquam molestie tortor, at iaculis metus volutpat et. In dolor 
lacus, AWESOME sip HELLO volutpat ac convallis non, pulvinar eu massa.
"""

txt = txt.lower()

# Patterns generated from a 1-8 word input string. Could also consist of hundreds of 
# thousands of patterns
patterns = [
    'world',
    'awesome',
    'awesome hello', 
    'world hello awesome',
    'world (.*?) awesome'   # '*' - represents any word between
]

regex = '|'.join(patterns)
result = re.findall(regex, txt)
counter = Counter(result)
print(counter)
# >>> Counter({'awesome': 5, 'world': 3})

# For some reason i can't get strings with more than one word to match

# Expected output
found_pattern_counts = {
    'world': 3,
    'awesome': 5,
    'awesome hello': 1, 
    'world hello awesome': 1,
    'world * awesome': 2
}

Answer 1

你没有使用通配符正确，我固定它，现在它像你描述的，现在你可以创建这个操作附加功能的工作原理：

patterns = [
    'world',
    'awesome',
    'awesome hello', 
    'world hello awesome',
    'world (.*?) awesome'
]


result = {} 
for pattern in patterns:
   rex = re.compile(fr'{pattern}') 
   count = len(rex.findall(txt))   
   result[pattern] = result.get(pattern, 0) + count

print(result)

Answer 2

你可以看看

re.finditer()

迭代器为您节省大量的资源，如果你并不需要所有的数据一次（你很少做）。这样，您就不需要保存在内存这么多的信息。看看这个Do iterators save memory in Python?