python - Alternative for 'in' operator for nested lists

Question

If I want to find something in a list in python I can use the 'in' operator:

list = ['foo', 'bar']
'foo' in list #returns True

But what should I do if I want to find something in a nested list?

list = [('foo', 'bar'), ('bar', 'foo')]
'foo' in list #returns False

Is it possible to do it in one row without a for loop for example?

Thanks!

Answer 1

你可能想要any：

>>> list = [('foo', 'bar'), ('bar', 'foo')]
>>> any('foo' in e for e in list)
True

不过，某种循环是不可避免的。

Answer 2

您可以像这样使用 itertools.chain ：

from itertools import chain

nested__seq = [(1,2,3), (4,5,6)]

print 4 in chain(*nested__seq)

PS：你不应该覆盖像“list”这样的bultins

Answer 3

这是滥用，但你可以很容易地在一行中做到这一点。

mainlist = [('foo', 'bar'), ('bar', 'foo')]
[elem for elem in sublist for sublist in mainlist] #['bar', 'bar', 'foo', 'foo']

'foo' in [elem for elem in sublist for sublist in mainlist] # True

Answer 4

如果您有任意深度的可迭代列表，请先将其展平：

import collections 

li= [('foo', 'bar'), ('bar', 'foo'),[[('deeper',('foobar'))]]]

def flatten(l):
    for el in l:
        if isinstance(el, collections.Iterable) and not isinstance(el, basestring):
            for sub in flatten(el):
                yield sub
        else:
            yield el

print 'deeper' in flatten(li)  
print 'foo' in flatten(li) 
print 'nope' in flatten(li)

印刷：

True
True
False

Answer 5

你也可以这样做in

>>> list = [('foo', 'bar'), ('bar', 'foo')]
>>> 'foo' in (x[1] for x in list)
True

编辑：此方法仅检查是否foo作为第一个元素。

将元素（任意）搜索为“foo”：

 >>>'foo' in reduce(lambda x,y: x+y, list)   
 True

更多尝试：

In [7]: list
Out[7]: [('foo', 'me', 'bar'), ('bar', 'foo', 'you')]
In [8]: 'me' in reduce(lambda x,y: x+y, list)
Out[8]: True

In [9]: 'you' in reduce(lambda x,y: x+y, list)
Out[9]: True