关于字符串:在Python中查找列表中的子字符串

Finding a substring within a list in Python

本问题已经有最佳答案,请猛点这里访问。

背景:

示例列表:EDOCX1[0]

我想检索一个元素,如果有一个匹配的子字符串,如abc

代码:

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sub = 'abc'
print any(sub in mystring for mystring in mylist)

如果列表中的任何元素包含模式,则上面打印True

我想打印与子字符串匹配的元素。所以如果我检查'abc',我只想从列表中打印'abc123'


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print [s for s in list if sub in s]

如果要用换行符分隔它们:

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print"
"
.join(s for s in list if sub in s)

完整示例,不区分大小写:

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mylist = ['abc123', 'def456', 'ghi789', 'ABC987', 'aBc654']
sub = 'abc'

print"
"
.join(s for s in mylist if sub.lower() in s.lower())


所有答案都有效,但它们总是贯穿整个列表。如果我理解你的问题,你只需要第一场比赛。因此,如果你找到了第一个匹配项,就不必考虑列表的其余部分:

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mylist = ['abc123', 'def456', 'ghi789']
sub = 'abc'
next((s for s in mylist if sub in s), None) # returns 'abc123'

如果匹配在列表的末尾或非常小的列表中,则不会有什么不同,但请考虑以下示例:

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import timeit

mylist = ['abc123'] + ['xyz123']*1000
sub = 'abc'

timeit.timeit('[s for s in mylist if sub in s]', setup='from __main__ import mylist, sub', number=100000)
# for me 7.949463844299316 with Python 2.7, 8.568840944994008 with Python 3.4
timeit.timeit('next((s for s in mylist if sub in s), None)', setup='from __main__ import mylist, sub', number=100000)
# for me 0.12696599960327148 with Python 2.7, 0.09955992100003641 with Python 3.4


使用一个简单的for循环:

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seq = ['abc123', 'def456', 'ghi789']
sub = 'abc'

for text in seq:
    if sub in text:
        print(text)

产量

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abc123

我只需要一个简单的正则表达式,你可以做这样的事情

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import re
old_list = ['abc123', 'def456', 'ghi789']
new_list = [x for x in old_list if re.search('abc', x)]
for item in new_list:
    print item


这将打印包含Sub的所有元素:

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for s in filter (lambda x: sub in x, list): print (s)