Question

我想将一个大小为2 * 2的数组插入一个更大的数组中。此外，我想将零数组的位置从左到右，从上到下迭代地移位。

zero_array =[0 0
             0 0]

large_array =[ 1  2  3  4
               5  6  7  8
               9 10 11 12
               13 14 15 16]

必填结果：

Ist iteration
             [ 0  0  3  4
               0  0  7  8
               9 10 11 12
               13 14 15 16]
2nd iteration
             [ 1  0  0  4
               5  0  0  8
               9 10 11 12
               13 14 15 16]
3rd iteration 
             [ 1  2  0  0
               5  6  0  0
               9 10 11 12
               13 14 15 16]

4th Iteration 
             [ 1  2  3  4
               0  0  7  8
               0  0 11 12
               13 14 15 16]

等等......

Answer 1

import copy
import numpy as np
la=np.array(<insert array here>)
za=np.zeros((2,2))
ma=copy.deepcopy(la)
for i in range(len(la)-len(za)+1):
    for j in range(len(la)-len(za)+1):
        la=copy.deepcopy(ma)
        la[i:i+len(za),j:j+len(za)]=za
        print la
#la=large array
#za=zero array

Answer 2

在较大的数组中插入较小的数组非常简单，只需要找出它应插入的位置（例如，通过左上角坐标）：

def insert_array(bigger_arr, smaller_arr, pos):
    out = np.array(bigger_arr, copy=True) 
    if bigger_arr.ndim != smaller_arr.ndim or bigger_arr.ndim != len(pos):
        raise ValueError('incompatible dimension')
    slicer = [slice(p, p+extend) for p, extend in zip(pos, smaller_arr.shape)]
    out[tuple(slicer)] = smaller_arr
    return out

>>> insert_array(np.arange(16).reshape(4, 4), np.zeros((2, 2)), pos=(0, 0))
array([[ 0,  0,  2,  3],
       [ 0,  0,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15]])

然后剩下的就是计算循环中的坐标。使用生成器函数可以轻松完成：

from itertools import product
def looping_inserting(bigger_arr, smaller_arr):
    for startpos in product(*[range(big_dim - small_dim + 1) 
                              for big_dim, small_dim 
                              in zip(bigger_arr.shape, smaller_arr.shape)]):
        yield insert_array(bigger_arr, smaller_arr, startpos)

for newarr in looping_inserting(np.arange(1, 17).reshape(4, 4), np.zeros((2, 2))):
    print(newarr)

打印：

[[ 0  0  3  4]
 [ 0  0  7  8]
 [ 9 10 11 12]
 [13 14 15 16]]
[[ 1  0  0  4]
 [ 5  0  0  8]
 [ 9 10 11 12]
 [13 14 15 16]]
[[ 1  2  0  0]
 [ 5  6  0  0]
 [ 9 10 11 12]
 [13 14 15 16]]
[[ 1  2  3  4]
 [ 0  0  7  8]
 [ 0  0 11 12]
 [13 14 15 16]]
[[ 1  2  3  4]
 [ 5  0  0  8]
 [ 9  0  0 12]
 [13 14 15 16]]
[[ 1  2  3  4]
 [ 5  6  0  0]
 [ 9 10  0  0]
 [13 14 15 16]]
[[ 1  2  3  4]
 [ 5  6  7  8]
 [ 0  0 11 12]
 [ 0  0 15 16]]
[[ 1  2  3  4]
 [ 5  6  7  8]
 [ 9  0  0 12]
 [13  0  0 16]]
[[ 1  2  3  4]
 [ 5  6  7  8]
 [ 9 10  0  0]
 [13 14  0  0]]

如果您只处理2D数组，可以简化很多代码，也许更好理解：

from itertools import product

def insert_array(bigger_arr, smaller_arr, pos):
    out = np.array(bigger_arr, copy=True) 
    if bigger_arr.ndim != smaller_arr.ndim or bigger_arr.ndim != len(pos):
        raise ValueError('incompatible dimension')
    out[pos[0]: pos[0] + smaller_arr.shape[0],
        pos[1]: pos[1] + smaller_arr.shape[1]] = smaller_arr
    return out

def looping_inserting(bigger_arr, smaller_arr):
    for pos0 in range(bigger_arr.shape[0] - smaller_arr.shape[0] + 1):
        for pos1 in range(bigger_arr.shape[1] - smaller_arr.shape[1] + 1):
            yield insert_array(bigger_arr, smaller_arr, (pos0, pos1))

for newarr in looping_inserting(np.arange(1, 17).reshape(4, 4), np.zeros((2, 2))):
    print(newarr)

将预定义的数组插入大型数组并迭代地移动较小数组的位置

2 个答案: