There is a 3 dimension list.
a = [[[1, 2], [2, 3], [3, 4]], [[4, 5], [5, 6], [6, 7]], [[7, 8], [8, 9], [9, 10]], [[10, 11], [11, 12], [12, 13]]]
so,
a[0][0][1] is 2
a[0][1][1] is 3
a[0][2][1] is 4
a[1][0][1] is 5
a[1][1][1] is 6
a[1][2][1] is 7
...
If I want to remove of particular element, such as a[:][:][1]
I haved tried that
a.pop([0][0][1])
Tnen, Traceback (most recent call last): File "", line 1, in
TypeError: 'int' object is not subscriptable
Try:
a[0][0].pop(1)
Given the list a in your question this would pop the number 2 from the list and a would look like this:
>>>a = [[[1], [2, 3], [3, 4]], [[4, 5], [5, 6], [6, 7]], [[7, 8], [8, 9], [9, 10]], [[10, 11], [11, 12], [12, 13]]]
See the note in the Python list.pop documentation.
It says:
The square brackets around the i in the method signature denote that the parameter is optional, not that you should type square brackets at that position. You will see this notation frequently in the Python Library Reference
