You probably want any
:
>>> list = [('foo', 'bar'), ('bar', 'foo')]
>>> any('foo' in e for e in list)
True
Some sort of loop is inevitable though.
Domanda
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!
Soluzione
You probably want any
:
>>> list = [('foo', 'bar'), ('bar', 'foo')]
>>> any('foo' in e for e in list)
True
Some sort of loop is inevitable though.
Altri suggerimenti
You could use itertools.chain like that :
from itertools import chain
nested__seq = [(1,2,3), (4,5,6)]
print 4 in chain(*nested__seq)
PS : you shouldn't override bultins like "list"
It's abusive, but you can do this in one line pretty easily.
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
If you have a list of iterables with arbitrary depth, flatten it first:
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)
Prints:
True
True
False
You could also do this with in
>>> list = [('foo', 'bar'), ('bar', 'foo')]
>>> 'foo' in (x[1] for x in list)
True
EDIT: this method check only if foo
is as fist element.
To search as 'foo' a element (any):
>>>'foo' in reduce(lambda x,y: x+y, list)
True
Some more try:
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