How to Efficiently Check for Element Existence in a List: Shortcuts and Best Practices

Checking for Element Existence in a List Using Shortcuts

It's a common task to verify if one or more elements reside within a list. Instead of devising an elaborate function, you can utilize the following concise approaches.

The Python or operator evaluates its arguments sequentially, returning the first truthy or non-empty value. While this may seem like a solution, it falls short in the case of lists. As demonstrated above, (1 or 2) in a evaluates to False, while (2 or 1) in a evaluates to True. This happens because 1 evaluates to False in a Boolean context, resulting in the expression being equivalent to False in a.

A more efficient and readable method is to employ a list comprehension or set intersection. Using list comprehension, you can filter the elements of the first list based on their presence in the second list. For instance:

L1 = [2, 3, 4]
L2 = [1, 2]
[i for i in L1 if i in L2]  # Returns [2]

Alternatively, you can convert the lists to sets, perform set intersection, and utilize the resulting set's Boolean value. This approach is advantageous when handling duplicate elements efficiently:

S1 = set(L1)
S2 = set(L2)
S1.intersection(S2)  # Returns set([2])

Both empty lists and empty sets evaluate to False, allowing you to assess their presence directly using Boolean logic.

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