Python: 返回传递函数为 true 的列表第一个元素的索引

One possibility is the built-in enumerate function:

def index_of_first(lst, pred):
for i,v in enumerate(lst):
if pred(v):
return i
return None

It's typical to refer a function like the one you describe as a "predicate"; it returns true or false for some question. That's why I call it pred in my example.

I also think it would be better form to return None, since that's the real answer to the question. The caller can choose to explode on None, if required.

you could do this with a list-comprehension:

l = [8,10,4,5,7]
filterl = [a for a in l if a % 2 != 0]

Then filterl will return all members of the list fulfilling the expression a % 2 != 0. I would say a more elegant method...

Not one single function, but you can do it pretty easily:

>>> test = lambda c: c == 'x'
>>> data = ['a', 'b', 'c', 'x', 'y', 'z', 'x']
>>> map(test, data).index(True)
3
>>>

If you don't want to evaluate the entire list at once you can use itertools, but it's not as pretty:

>>> from itertools import imap, ifilter
>>> from operator import itemgetter
>>> test = lambda c: c == 'x'
>>> data = ['a', 'b', 'c', 'x', 'y', 'z']
>>> ifilter(itemgetter(1), enumerate(imap(test, data))).next()[0]
3
>>>

Just using a generator expression is probably more readable than itertools though.

Note in Python3, map and filter return lazy iterators and you can just use:

from operator import itemgetter
test = lambda c: c == 'x'
data = ['a', 'b', 'c', 'x', 'y', 'z']
next(filter(itemgetter(1), enumerate(map(test, data))))[0]  # 3

You could do that in a one-liner using generators:

next(i for i,v in enumerate(l) if is_odd(v))

The nice thing about generators is that they only compute up to the requested amount. So requesting the first two indices is (almost) just as easy:

y = (i for i,v in enumerate(l) if is_odd(v))
x1 = next(y)
x2 = next(y)

Though, expect a StopIteration exception after the last index (that is how generators work). This is also convenient in your "take-first" approach, to know that no such value was found --- the list.index() function would throw ValueError here.

@Paul's accepted answer is best, but here's a little lateral-thinking variant, mostly for amusement and instruction purposes...:

>>> class X(object):
...   def __init__(self, pred): self.pred = pred
...   def __eq__(self, other): return self.pred(other)
...
>>> l = [8,10,4,5,7]
>>> def is_odd(x): return x % 2 != 0
...
>>> l.index(X(is_odd))
3

essentially, X's purpose is to change the meaning of "equality" from the normal one to "satisfies this predicate", thereby allowing the use of predicates in all kinds of situations that are defined as checking for equality -- for example, it would also let you code, instead of if any(is_odd(x) for x in l):, the shorter if X(is_odd) in l:, and so forth.

Worth using? Not when a more explicit approach like that taken by @Paul is just as handy (especially when changed to use the new, shiny built-in next function rather than the older, less appropriate .next method, as I suggest in a comment to that answer), but there are other situations where it (or other variants of the idea "tweak the meaning of equality", and maybe other comparators and/or hashing) may be appropriate. Mostly, worth knowing about the idea, to avoid having to invent it from scratch one day;-).

A variation on Alex's answer. This avoids having to type X every time you want to use is_odd or whichever predicate

>>> class X(object):
...     def __init__(self, pred): self.pred = pred
...     def __eq__(self, other): return self.pred(other)
...
>>> L = [8,10,4,5,7]
>>> is_odd = X(lambda x: x%2 != 0)
>>> L.index(is_odd)
3
>>> less_than_six = X(lambda x: x<6)
>>> L.index(less_than_six)
2