How to know the index of an element in a list

Question

If I have a list like

 [[a, b], [c, d], [e, f]]

How would I know that element a is in index 0 of the big list. I'm unsure on how to do this with a 2 dimensional array.

I tried to use index, but it not works

s = [['a', 'b'], ['c', 'd'], ['e', 'f']] s.index('a')

Traceback (most recent call last): File "C:/Users/xxy/PycharmProjects/tb/test.py", line 3, in <module> s.index('a') ValueError: 'a' is not in list

xingyu xia · Accepted Answer · 2022-11-30 16:20:19Z

You can solve it with a loop

s = [['a', 'b'], ['c', 'd'], ['e', 'f']] x = 'a' find = False for i, line in enumerate(s): if x in line: print(f'find {x} at', i, line.index(x)) find = True break if not find: print(f'{x} is not in list')

Bhargav · Accepted Answer · 2022-11-30 16:28:40Z

More simpler way you can use any()

s = [['a', 'b'], ['c', 'd'], ['e', 'f']] for i, v in enumerate(s): if any('a' in x for x in v): print(i)

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