在Python中提取一个包含字符串的列表项，而无需列表理解？

Question

Given a list:

my_list = ['abc', 'def', 'ghi']

I want to extract the element that contains 'ab' (of which there will only ever be one).

Rather than something a little convoluted like:

[i for i in my_list if 'ab' in i][0]

I'm looking for some way of applying something that does the equivalent of XPath contains() , but for a list.

Answer 1

If you are concerned about iterating through the entire list:

In [50]: L = ['abc', 'def', 'ghi']

In [51]: next(itertools.dropwhile(lambda s: 'ab' not in s, L))
Out[51]: 'abc'

Don't forget to import itertools

Answer 2

You can use filter

>>> my_list = ['abc', 'def', 'ghi']
>>> filter(lambda x: 'ab' in x, my_list)
['abc']
>>>

You can also create a function which can call a list:

>>> from functools import partial
>>> my_ab_func = partial(filter, lambda x: 'ab' in x)
>>> my_ab_func(my_list)
['abc']
>>> my_list2 = ['dabd', 'no','yes', 'zyx']
>>> my_ab_func(my_list2)
['dabd']
>>>

Answer 3

Use ifilter from itertools , it returns iterator instead of list:

from itertools import ifilter

my_list = ['abc', 'def', 'ghi']  

next(ifilter(lambda x: 'ab' in x, my_list))

Answer 4

Rather than something a little convoluted like:

 [i for i in my_list if 'ab' in i][0] 

List comprehension is very much idiomatic for Python. If your lists are small, what you have is probably sufficient with some additional error checking:

def lcontains(needle_s, haystack_l):
    try: return [i for i in haystack_l if needle_s in i][0]
    except IndexError: return None

# Example use:
lcontains('ab', my_list)

That being said, if your input lists (or more generally, iterables) are potentially large, you might consider using itertools :

import itertools

try:
    irange = xrange # Python 2
    ifilter = itertools.ifilter
except NameError:
    irange = range # Python 3
    ifilter = filter

def oneornone(iterable):
    try: return iterable.next()
    except StopIteration: return None

icontains = lambda predicate_f, haystack_i: oneornone(ifilter(predicate_f, haystack_i))

# Example use:
result = icontains(lambda i: i % 3323 == 0, irange(2, 1000000000000, 5))

# Or, in your case:
result = icontains(lambda i: 'ab' in i, mylist)