如何在python中的嵌套列表中删除特定元素

Question

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?

Answer 1

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

Answer 2

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]]

Answer 3

Using numpy

import numpy as np

array = np.array(array)
array[1:3, 1:3]

return

array([[5, 6],
       [8, 9]])

Answer 4

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!