仅使用python列表将2D数组的一部分替换为另一个2D数组？

Question

How can I slice a smaller array into an N x M array if I know the point of insertion?

ie,

# Larger array
[1,1,1,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,1,1,1]

# Smaller array
[1,2,3,4]
[5,6,7,8]

# Insert at [1,6] gives:
[1,1,1,1,1,1,1,1,1,1]
[1,1,1,1,1,1,1,2,3,4]
[1,1,1,1,1,1,5,6,7,8]

And using just list comprehensions?

Answer 1

l = [[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1]]
s = [[1,2,3,4],
[5,6,7,8]]
def insert(large, small, row, col):
    for i, r in enumerate(small):
        large[row + i][col:col + len(r)] = r
insert(l, s, 1, 6)
print(l)

This outputs:

[[1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 2, 3, 4], [1, 1, 1, 1, 1, 1, 5, 6, 7, 8]]

Answer 2

Edit

I just realized now you have a using just list comprehensions condition

For that, you can use iter and enumerate

So given

a1 = [[1,1,1,1,1,1,1,1,1,1],
     [1,1,1,1,1,1,1,1,1,1],
     [1,1,1,1,1,1,1,1,1,1],
     [1,1,1,1,1,1,1,1,1,1]]

a2=[[1,2,3,4],
    [5,6,7,8]]

x,y = (1,6)

You can do

cols = len(a2[0])
ran = range(x,x+len(a2))
it = iter(a2)

final = [r if i not in ran else r[:y] + next(it) + r[(y+ncols):] for i,r in enumerate(a1)]

Use cases:

x,y = (2,6)

Output

[[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
 [1, 1, 1, 1, 1, 1, 1, 2, 3, 4],
 [1, 1, 1, 1, 1, 1, 5, 6, 7, 8]]

x,y = (1,4)
[[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
 [1, 1, 1, 1, 1, 2, 3, 4, 1, 1],
 [1, 1, 1, 1, 5, 6, 7, 8, 1, 1],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]


x,y = (0,0)

[[1, 2, 3, 4, 1, 1, 1, 1, 1, 1],
 [5, 6, 7, 8, 1, 1, 1, 1, 1, 1],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
 [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]

---

Another solution is to use numpy

If you have

import numpy as np

x = np.array([[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1]])

a = np.array([[1,2,3,4],
[5,6,7,8]])

You can use numpy indexing

x[1:,6:] = a

to get

array([[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 2, 3, 4],
       [1, 1, 1, 1, 1, 1, 5, 6, 7, 8]])

Answer 3

If you are happy to use a 3rd party library, NumPy offers a generic solution for arbitrary coordinates:

i, j = (1, 6)

x[i:i+a.shape[0], j:j+a.shape[1]] = a

print(x)

array([[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 2, 3, 4],
       [1, 1, 1, 1, 1, 1, 5, 6, 7, 8]])