在 Python 中的列表中查找子字符串
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Finding a substring within a list in Python
提问by frankV
Background:
背景：
Example list: mylist = ['abc123', 'def456', 'ghi789']
示例列表： mylist = ['abc123', 'def456', 'ghi789']
I want to retrieve an element if there's a match for a substring, like abc
如果子字符串匹配，我想检索一个元素，例如 abc
Code:
代码：
sub = 'abc'
print any(sub in mystring for mystring in mylist)
above prints Trueif any of the elements in the list contain the pattern.
True如果列表中的任何元素包含该模式，则上面打印。
I would like to print the element which matches the substring. So if I'm checking 'abc'I only want to print 'abc123'from list.
我想打印与子字符串匹配的元素。因此，如果我正在检查，'abc'我只想'abc123'从列表中打印。
采纳答案by David Robinson
print [s for s in list if sub in s]
If you want them separated by newlines:
如果您希望它们由换行符分隔：
print "\n".join(s for s in list if sub in s)
Full example, with case insensitivity:
完整示例，不区分大小写：
mylist = ['abc123', 'def456', 'ghi789', 'ABC987', 'aBc654']
sub = 'abc'
print "\n".join(s for s in mylist if sub.lower() in s.lower())
回答by unutbu
Use a simple forloop:
使用一个简单的for循环：
seq = ['abc123', 'def456', 'ghi789']
sub = 'abc'
for text in seq:
if sub in text:
print(text)
yields
产量
abc123
回答by Hyperboreus
This prints all elements that contain sub:
这将打印包含 sub 的所有元素：
for s in filter (lambda x: sub in x, list): print (s)
回答by oathead
I'd just use a simple regex, you can do something like this
我只是使用一个简单的正则表达式，你可以做这样的事情
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
回答by Frank Zalkow
All the answers work but they always traverse the whole list. If I understand your question, you only need the first match. So you don't have to consider the rest of the list if you found your first match:
所有答案都有效，但它们总是遍历整个列表。如果我理解你的问题，你只需要第一场比赛。因此，如果您找到第一个匹配项，则不必考虑列表的其余部分：
mylist = ['abc123', 'def456', 'ghi789']
sub = 'abc'
next((s for s in mylist if sub in s), None) # returns 'abc123'
If the match is at the end of the list or for very small lists, it doesn't make a difference, but consider this example:
如果匹配位于列表的末尾或非常小的列表，则没有区别，但请考虑以下示例：
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
