检查列表列表的第一个元素

Question

I have a list which looks like this:

List = [['name1','surname1'], ['name2','surname2'],['name3','surname3']]

I would like to check if "name1" is in the object "List". I have tried:

if 'name1' in List:
    print True
else:
    print False

and the output is 'False'. Any idea how to create a sublist (or something similar) to check the first element of every sub-list without looping through all the elements of the main list?

POSSIBLE SOLUTION

What I have thought about is:

for i in range(0, len(List)):
    if List[i][0] == 'name1':
        print True

but I want to avoid exactly this iteration with something more optimized, if possible.

Answer 1

You can use a generator expression:

>>> 'name1' in (x[0] for x in List)
True

This will short-circuit as soon as 'name1' is found and won't create any unnecessary list in memory.

Related: List comprehension vs generator expression's weird timeit results?

Answer 2

I suggest using a dictionary, which seems more suitable here.

But if a list of lists is to be used, you can have such a code: 'name1' in [list[0] for list in List]

Answer 3

You can do it with:

if 'name1' in [l[0] for l in List]:

You can also add an if l at then end of the list comprehension just in case there's an empty list around:

if 'name1' in [l[0] for l in List if l]:  # safe if there's an empty list

Answer 4

More idiomatic way to do this, is to use any function

>>> any('name1' == current_list[0] for current_list in my_list)
True

This also short circuits on the first occurrence of name1 .

Edit : In case, your name1 can be anywhere in the sub-list, you can use the in operator

>>> any('name1' in current_list for current_list in my_list)
True