Check if something is not in a list in Python

Question

I have a list of tuples in Python  and I have a conditional where I want to take the branch ONLY if the tuple is not in the list  if it is in the list  then I don t want to take the if branch   if curr x -1  gt  0 and  curr x-1   curr y  not in myList          Do Something   This is not really working for me though  What have I done wrong

User · Accepted Answer

The bug is probably somewhere else in your code, because it should work fine:

>>> 3 not in [2, 3, 4]
False
>>> 3 not in [4, 5, 6]
True

Or with tuples:

>>> (2, 3) not in [(2, 3), (5, 6), (9, 1)]
False
>>> (2, 3) not in [(2, 7), (7, 3), "hi"]
True

User · Answer

How do I check if something is  not  in a list in Python    The cheapest and most readable solution is using the in operator  or in your specific case  not in   As mentioned in the documentation      The operators in and not in test for membership  x in s evaluates to   True if x is a member of s  and False otherwise  x not in s returns   the negation of x in s    Additionally      The operator not in is defined to have the inverse true value of in    y not in x is logically the same as not y in x   Here are a few examples    a  in  1  2  3    False   c  in   a    b    c     True   a  not in  1  2  3    True   c  not in   a    b    c     False   This also works with tuples  since tuples are hashable  as a consequence of the fact that they are also immutable     1  2  in   3  4    1  2      True   If the object on the RHS defines a   contains     method  in will internally call it  as noted in the last paragraph of the Comparisons section of the docs          in and not in    are supported by types that are iterable or implement the     contains     method  For example  you could  but shouldn t  do this     3  2  1    contains   1    True   in short-circuits  so if your element is at the start of the list  in evaluates faster   lst   list range 10001    timeit 1 in lst  timeit 10000 in lst    Expected to take longer time   68 9 ns    0 613 ns per loop  mean    std  dev  of 7 runs  10000000 loops each  178   s    5 01   s per loop  mean    std  dev  of 7 runs  10000 loops each    If you want to do more than just check whether an item is in a list  there are options    list index can be used to retrieve the index of an item  If that element does not exist  a ValueError is raised  list count can be used if you want to count the occurrences      The XY Problem  Have you considered sets   Ask yourself these questions    do you need to check whether an item is in a list more than once  Is this check done inside a loop  or a function called repeatedly  Are the items you re storing on your list hashable  IOW  can you call hash on them    If you answered  yes  to these questions  you should be using a set instead  An in membership test on lists is O n  time complexity  This means that python has to do a linear scan of your list  visiting each element and comparing it against the search item  If you re doing this repeatedly  or if the lists are large  this operation will incur an overhead    set objects  on the other hand  hash their values for constant time membership check  The check is also done using in   1 in  1  2  3     True   a  not in   a    b    c     False   1  2  in    a    c     1  2     True   If you re unfortunate enough that the element you re searching not searching for is at the end of your list  python will have scanned the list upto the end  This is evident from the timings below   l   list range 100001   s   set l    timeit 100000 in l  timeit 100000 in s  2 58 ms    58 9   s per loop  mean    std  dev  of 7 runs  100 loops each  101 ns    9 53 ns per loop  mean    std  dev  of 7 runs  10000000 loops each    As a reminder  this is a suitable option as long as the elements you re storing and looking up are hashable  IOW  they would either have to be immutable types  or objects that implement   hash

[python] Check if something is (not) in a list in Python

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