Let's say I got the following array:
array = [[1, 2, 3, 1],
[4, 5, 6, 4],
[7, 8, 9, 7],
[7, 8, 9, 7]]
I want to remove the first and last list in the array and than the first and last element of the middle lists (return should basically be: [[5, 6], [8, 9]]).
I tried the following:
array.remove(array[0])
array.remove(array[-1])
for i in array:
array.remove(i[0])
array.remove(i[-1])
But I always get ValueError: list.remove(x): x not in list. Why?
You should remove the items from the sublist, not the parent list:
for i in array:
i.remove(i[0])
i.remove(i[-1])
You can also remove both items in one line using del:
>>> for i in array:
... del i[0], i[-1]
>>> array
[[5, 6], [8, 9]]
Using numpy
import numpy as np
array = np.array(array)
array[1:3, 1:3]
return
array([[5, 6],
[8, 9]])
Simple way to achive this is to slice the array list using list comprehension expression like:
array = [[1, 2, 3, 1],
[4, 5, 6, 4],
[7, 8, 9, 7],
[7, 8, 9, 7]]
array = [a[1:-1]for a in array[1:-1]]
Final value hold by array will be:
[[5, 6], [8, 9]]
Here array[1:-1] returns the list skipping the first and last element from the array list
Or, with <list>.pop:
array = [[1, 2, 3, 1],
[4, 5, 6, 4],
[7, 8, 9, 7],
[7, 8, 9, 7]]
for i in range(len(array)): # range(len(<list>)) converts 'i' to the list inedx number.
array[i].pop(0)
array[i].pop(-1)
To answer your question,
.remove removes the first matching value, not a specific index. Where as .pop removes the it by the index.
Hope this helps!
