How do I check if a list
xs contains an item?
if xs.contains(item): ...
if my_item in some_list: ...
Also, inverse operation:
if my_item not in some_list: ...
It works fine for lists, tuples, sets and dicts (check keys).
Note that this is an O(n) operation in lists and tuples, but an O(1) operation in sets and dicts.
The list method
index will return
-1 if the item is not present, and will return the index of the item in the list if it is present. Alternatively in an
if statement you can do the following:
if myItem in list: #do things
You can also check if an element is not in a list with the following if statement:
if myItem not in list: #do things
In addition to what other have said, you may also be interested to know that what
in does is to call the
list.__contains__ method, that you can define on any class you write and can get extremely handy to use python at his full extent.
A dumb use may be:
>>> class ContainsEverything: def __init__(self): return None def __contains__(self, *elem, **k): return True >>> a = ContainsEverything() >>> 3 in a True >>> a in a True >>> False in a True >>> False not in a False >>>
I came up with this one liner recently for getting
True if a list contains any number of occurrences of an item, or
False if it contains no occurrences or nothing at all. Using
next(...) gives this a default return value (
False) and means it should run significantly faster than running the whole list comprehension.
list_does_contain = next((True for item in list_to_test if item == test_item), False)
There is also the list method:
[2, 51, 6, 8, 3].__contains__(8) # Out: True [2, 51, 6, 3].__contains__(8) # Out: False
There is one another method that uses
index. But I am not sure if this has any fault or not.
list = [5,4,3,1] try: list.index(2) #code for when item is expected to be in the list print("present") except: #code for when item is not expected to be in the list print("not present")