如果列表中的元素具有满足的特定索引和条件，如何更改它？

Question

I want to be able to take a list of lists (lst) and a list of indexes and those elements in lst that have that have those indexes and also meet the condition ( == '1') to be changed to '0' .

If I input

lst = [['1','2','3'],[],['4','2','1']]

and

specific_indexes = [(0, 0), (0, 2), (2, 0), (2, 2)]

I get [['0', '2', '3'], [], ['4', '2', '0']]

but I would like faster way to do this.

def change(lst, specific_indexes):

    for (x,y) in specific_indexes:
        if lst[y][x] == '1':
            lst[y][x] = '0'
    return lst

Answer 1

...but I would like faster way to do this.

If you are interested in performance, you can use a specialist 3rd party library such as NumPy . This does mean you have to define a regular 2d array as an input, or transform it into one as shown below.

import numpy as np

lst = [['1','2','3'],[],['4','2','1']]
idx = [(0, 0), (0, 2), (2, 0), (2, 2)]

# calculate column number and construct NumPy array
colnum = max(map(len, lst))
arr = np.array([sublst if sublst else ['0'] * colnum for sublst in lst]).astype(int)
idx = np.array(idx)

# calculate indexer and mask array conditionally
mask = np.ix_(idx[:, 1], idx[:, 0])
arr[mask] = np.where(arr[mask] == 1, 0, arr[mask])

print(arr)
# array([[0, 2, 3],
#        [0, 0, 0],
#        [4, 2, 0]])