search - Python: Find the position in a list of the first object that respect a given condition -

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i have list of following kind:

class any(object):     def __init__(self,a,b):         self.a=a         self.b=b  l=[any(1,3),any(2,4),any(1,2),any(none,6),any('hello',6), any(1,'chucknorris'),any(1,2)]

lis list contains instances of any. i'd find position of first of these instances attribute a equals 'none`.

as list long, algorithm should not explore whole list should stop condition (in example, attribute a equals none) found.

in above example answer of algorithm should 3.

use generator expression , next:

next((i i, item in enumerate(l) if item.a none), none)

this return none if no such item found.

demo:

>>> l=[any(1,3),any(2,4),any(1,2),any(none,6),any('hello',6), any(1,'chucknorris'),any(1,2)] >>> next((i i, item in enumerate(l) if item.a none), none) 3

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