Saturday 9 February 2019
if 'a' or 'b' in L, where L is a list (Python)
Answer
Answer
I am having trouble with the following logic:
Lets say I have a list L = ['a', 'b', 'c']
Both items are in the list...
if ('a' or 'b') in L:
print 'it\'s there!'
else:
print 'No sorry'
prints It's there!
Only the first item is in the list...
if ('a' or 'd') in L:
print 'it\'s there!'
else:
print 'No sorry'
prints It's there!
Neither item in the list...
if ('e' or 'd') in L:
print 'it\'s there!'
else:
print 'No sorry'
prints No sorry
Here's the confusing one Only the second item in the list...
if ('e' or 'a') in L:
print 'it\'s there!'
else:
print 'No sorry'
prints No sorry
I do not understand why this is not registering as a true statement. How does this generalize to an or statement with n conditionals?
Forehead-slapping easy answer in 3,2,1...
Answer
Let's break down the expression:
('e' or 'a') will first check if 'e' is True. If it is, the expression will return 'e'. If not, it will return 'a'.
Since all non-empty strings returns True, this expression will always return 'e'. This means that if ('e' or 'a') in L: can be translated to if 'e' in L, which in this case is False.
A more generic way to check if a list contains at least one value of a set of values, is to use the any function coupled with a generator expression.
if any(c in L for c in ('a', 'e')):
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