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If I have one long list: myList = [0,2,1,0,2,1] that I split into two lists:

a = [0,2,1]
b = [0,2,1]

how can I compare these two lists to see if they are both equal/identical, with the constraint that they have to be in the same order?

I have seen questions asking to compare two lists by sorting them, but in my specific case, I am not checking for a sorted comparison, but identical list comparison.

Jeremy
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3 Answers3

217

Just use the classic == operator:

>>> [0,1,2] == [0,1,2]
True
>>> [0,1,2] == [0,2,1]
False
>>> [0,1] == [0,1,2]
False

Lists are equal if elements at the same index are equal. Ordering is taken into account then.

Maxime Lorant
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    This can return the following error with a numpy list: `ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()` – Alex Reynolds Feb 10 '20 at 00:11
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    What @AlexReynolds said. You have to test with `all(arr1 == arr2)` or `(arr1 == arr2).all()`. – Julio Batista Silva Jun 04 '20 at 15:36
  • @Alex That's an array, not a list. They're both ordered data types, but conceptually different. An action you apply to an array is applied to all of its elements, but the same isn't true for lists. – wjandrea Sep 07 '20 at 15:09
12

If you want to just check if they are identical or not, a == b should give you true / false with ordering taken into account.

In case you want to compare elements, you can use numpy for comparison

c = (numpy.array(a) == numpy.array(b))

Here, c will contain an array with 3 elements all of which are true (for your example). In the event elements of a and b don't match, then the corresponding elements in c will be false.

Vasanth
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4

The expression a == b should do the job.

Paul Rooney
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Abhiram
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