python - get item number from list

Question

I just asked the following question:

Python - find integer closest to 0 in list

The best answer was to use: min(lst, key=abs).

That code returns the item from the list.

How do I get the item number from the list? (i.e. 2 instead of -18)

Answer 1

You'd need to augment your list with indices:

min(enumerate(lst), key=lambda x: abs(x[1]))

It'll return the index and closest-to-zero value, use [0] if you only need the index.

On your example:

>>> example = [237, 72, -18, 237, 236, 237, 60, -158, -273, -78, 492, 243]
>>> min(enumerate(example), key=lambda x: abs(x[1]))
(2, -18)
>>> min(enumerate(example), key=lambda x: abs(x[1]))[0]
2

This is very efficient (worst-case O(n)); you can use example.index(min(example, key=abs)) as well, but that has to go through the list twice in the worst case (O(2n)).

Answer 2

Try:

lst.index(min(lst, key=abs))

Answer 3

One way is after finding the integer you want to find, you can use "index"...

result = min(lst, key=abs)
index = lst.index(result)
print(index) # prints 2

Answer 4

>>> lst = [237, 72, -18, 237, 236, 237, 60, -158, -273, -78, 492, 243]
>>> lst.index(min(lst, key=abs))
2