How to check if a list ONLY contains a certain item

Question

I have a list called bag. I want to be able to check if only a particular item is in it.

bag = ["drink"]
if only "drink" in bag:
    print 'There is only a drink in the bag'
else:
    print 'There is something else other than a drink in the bag'

Of course, where I put 'only' in the code there, it is wrong. Is there any simple replacement of that? I have tried a few similar words.

Sorry I'm not clear here, but do you mean that the list has only one item and it is "drink", or just that "drink" is in the list? In other words, would ["drink", "candy bar"] pass your if only test? — PTBNL, Mar 24 '16 at 15:30
You don't try "a few similar words" to do something in a programming language, but what you do is learn the basics of the language first, and then use your knowledge to do stuff. That's the natural process of learning. — Sнаđошƒаӽ, Mar 24 '16 at 15:37
@Sнаđошƒаӽ monte-carlo search for keywords that do what you want, that isn't how you program? You're missing out — en_Knight, Mar 24 '16 at 15:41
Possible dup. http://stackoverflow.com/questions/405516/if-all-in-list-something and Shadowfax just a joke (randomly guess words until sometihng works) — en_Knight, Mar 24 '16 at 15:44
@en_Knight I thought so to myself, I was just being cautious :D — Sнаđошƒаӽ, Mar 24 '16 at 15:47

Goodies · Accepted Answer · 2016-03-24T18:43:18.980

14

Use the builtin all() function.

if bag and all(elem == "drink" for elem in bag):
    print("Only 'drink' is in the bag")

The all() function is as follows:

def all(iterable):
    for element in iterable:
        if not element:
            return False
    return True

For this reason, an empty list will return True. Since there are no elements, it will skip the loop entirely and return True. Because this is the case, you must add an explicit and len(bag) or and bag to ensure that the bag is not empty (() and [] are false-like).

Also, you could use a set.

if set(bag) == {['drink']}:
    print("Only 'drink' is in the bag")

Or, similarly:

if len(set(bag)) == 1 and 'drink' in bag:
    print("Only 'drink' is in the bag")

All of these will work with 0 or more elements in the list.

edited Mar 24 '16 at 18:43

answered Mar 24 '16 at 15:28

Goodies

4,439
3
31
57

1

The first one is the only one that short circuits if there is something else in the bag (all good answers) – en_Knight Mar 24 '16 at 15:36
@en_Knight That's true. – Goodies Mar 24 '16 at 15:36
@en_Knight Good catch. I replaced it with `any` and `!=` which will not fail on an empty list :) – Goodies Mar 24 '16 at 15:44
for the any one you need to use if not, because the one you have there is true if any element does NOT equal drink – reemer9997 Mar 24 '16 at 15:49
(Or at least I assume your elem was short for element anyway) – reemer9997 Mar 24 '16 at 15:57
Hmm... Actually, this will still fail on an empty list. I suppose the only way is to add an explicit `and len(bag)` – Goodies Mar 24 '16 at 16:01
1

Looks good +1. I'd switch it so the and statement is first (for short circuiting again) but doesn't matter much – en_Knight Mar 24 '16 at 17:43

Byte Commander · Answer 2 · 2016-03-24T15:47:14.543

1

You could directly check for equality with a list that only contains this item:

if bag == ["drink"]:
    print 'There is only a drink in the bag'
else:
    print 'There is something else other than a drink in the bag'

Alternatively if you want to check whether the lists contains any number greater than zero of the same item "drink", you can count them and compare with the list length:

if bag.count("drink") == len(bag) > 0:
    print 'There are only drinks in the bag'
else:
    print 'There is something else other than a drink in the bag'

edited Mar 24 '16 at 15:47

answered Mar 24 '16 at 15:33

Byte Commander

6,506
6
44
71

Sam question as Lafada's - what about ["drink","drink"] ? Doesn't that qualfiy as having only drink in it? – en_Knight Mar 24 '16 at 15:34
Not sure. It isn't clearly described in the question. But I added an alternative for that as well. – Byte Commander Mar 24 '16 at 15:35
ok that second one seems almost right. Just to be pedantic, it fails on the empty list though – en_Knight Mar 24 '16 at 15:38
["drink","drink"] would indeed qualify as only having drink. I just realised that I wasn't clear in the question. The second one works best – reemer9997 Mar 24 '16 at 15:41
the second works best to test if there is nothing else, but as @ByteCommander indirectly pointed out `[].count("drink") == len([])` does evaluate as True. – Tadhg McDonald-Jensen Mar 24 '16 at 15:42
@en_Knight I corrected the second example and added a check that the length must be greater than zero. Now the empty list is no longer a false positive. – Byte Commander Mar 24 '16 at 15:48
and it doesn't have the double negative needed for `not any(elem != "drink" for elem in bag)` to work. – Tadhg McDonald-Jensen Mar 24 '16 at 15:49

score 0 · Answer 3 · answered Mar 24 '16 at 15:28

0

You can check length of list

if len(bag) == 1 and "drink" in bag:
    #do your operation.

answered Mar 24 '16 at 15:28

Nilesh

20,521
16
92
148

It seems like ['drink','drink'] should qualify as only having drink in it, but doesn't in your implementatino – en_Knight Mar 24 '16 at 15:31
1

yes you are right :), then have to check `all` second answer :) – Nilesh Mar 24 '16 at 15:32

How to check if a list ONLY contains a certain item

3 Answers3