if 'a' or 'b' in L, where L is a list (Python) [duplicate]

Question

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 1

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')):

Answer 2

Use this instead:

 if 'a' in L or 'b' in L:

If we want to check if all these of this "items" are in the list, all and a generator comprehension is your friend:

items = 'a', 'b', 'c'
if all(i in L for i in items):

Or if any of these items are in the list, use any:

if any(i in L for i in items)