如何“扩展” Python 中的数组？

Question

我想以某种方式扩展 python 中的二维数组。

没有循环

Fe 如果是：

[[255, 255, 255],
     [255, 255, 255],
     [255, 255, 255]]

我会说我想将它扩展 2 倍并得到这样的结果：

[[255, 0, 255, 0, 255, 0],
  [0,  0,  0,  0,  0,  0],
 [255, 0, 255, 0, 255, 0],
  [0,  0,  0,  0,  0,  0],
 [255, 0, 255, 0, 255, 0],
  [0,  0,  0,  0,  0,  0]]

等等，如果是 4 因子。

有没有function？

Answer 1

您可以在没有 numpy 的情况下执行此操作，只需使用列表理解，但它很复杂：

lst = [[255, 255, 255],
       [255, 255, 255],
       [255, 255, 255]]
extend_list = [ [lst[j // 2][i // 2] if j % 2 == 0 and i % 2 == 0 else 0 for i in range( 2 * (len(lst[j // 2])) )] if j != len(2 * lst) else [0 for _ in range( (2 * len(lst)) -1)] for j in range(2 * (len(lst)) )]

print(extend_list)

output：

[[255, 0, 255, 0, 255, 0], [0, 0, 0, 0, 0, 0], [255, 0, 255, 0, 255, 0], [0, 0, 0], [255, 0, 255, 0, 255, 0], [0, 0, 0, 0, 0, 0]]

Answer 2

这是使用 numpy 的解决方案。 您没有提供 N=4 的示例，所以我猜想：

import numpy as np

arr = np.array([[255, 255, 255], [255, 255, 255], [255, 255, 255]])
factor = 4

print(arr)
nx, ny = arr.shape
if nx != ny:
    raise Exception("Array is not square")

step = 2 + factor//2 - 1
stop = nx * step
print('stop:', stop)
print('step:', step)

for x in range(1,stop,step):
    print()
    nx, ny = arr.shape
    print('x:', x)
    value = [[0]*nx]*(factor//2)
    print('Inserting columns:', value)
    arr = np.insert(arr, x, value, axis=1)
    nx, ny = arr.shape
    print(arr)
    value = [[0]*ny]*(factor//2)
    print('Inserting rows:', value)
    arr = np.insert(arr, x, value, axis=0)
    print(arr)

[[255 255 255]
 [255 255 255]
 [255 255 255]]
stop: 9
step: 3

x: 1
Inserting columns: [[0, 0, 0], [0, 0, 0]]
[[255   0   0 255 255]
 [255   0   0 255 255]
 [255   0   0 255 255]]
Inserting rows: [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
[[255   0   0 255 255]
 [  0   0   0   0   0]
 [  0   0   0   0   0]
 [255   0   0 255 255]
 [255   0   0 255 255]]

x: 4
Inserting columns: [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
[[255   0   0 255   0   0 255]
 [  0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255]
 [255   0   0 255   0   0 255]]
Inserting rows: [[0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0]]
[[255   0   0 255   0   0 255]
 [  0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255]
 [  0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255]]

x: 7
Inserting columns: [[0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0]]
[[255   0   0 255   0   0 255   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255   0   0]]
Inserting rows: [[0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
[[255   0   0 255   0   0 255   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [255   0   0 255   0   0 255   0   0]
 [  0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0]]

Answer 3

你可以用零创建新的更大的数组

factor = 2

h, w = arr.shape[:2]

new_w = h*factor
new_h = h*factor

new_arr = np.zeros((new_h, new_w), np.uint8)

然后您可以使用带有zip()和range(0, new_h, factor)的for -loops 来获取价值及其新的 position

for row, y in zip(arr, range(0, new_h, factor)):
    for value, x in zip(row, range(0, new_w, factor)):
        new_arr[y,x] = value

给

[[255   0 255   0 255   0]
 [  0   0   0   0   0   0]
 [255   0 255   0 255   0]
 [  0   0   0   0   0   0]
 [255   0 255   0 255   0]
 [  0   0   0   0   0   0]]

如果您在range使用不同的值而不是0 ，那么您可以获得offset

offset_y = 1
offset_x = 1
for row, y in zip(arr, range(offset_y, new_h, factor)):
    for value, x in zip(row, range(offset_x, new_w, factor)):
        new_arr[y,x] = value

给出：

[[  0   0   0   0   0   0]
 [  0 255   0 255   0 255]
 [  0   0   0   0   0   0]
 [  0 255   0 255   0 255]
 [  0   0   0   0   0   0]
 [  0 255   0 255   0 255]]

---

工作代码

import numpy as np

arr = np.array([[255, 255, 255],
     [255, 255, 255],
     [255, 255, 255]]
)

factor = 2

h, w = arr.shape[:2]

new_w = h*factor
new_h = h*factor

new_arr = np.zeros((new_h, new_w), np.uint8)

offset_x = 0
offset_y = 0
for row, y in zip(arr, range(offset_y, new_h, factor)):
    #print(row, y)
    for value, x in zip(row, range(offset_x, new_w, factor)):
        #print(y, x, value)
        new_arr[y,x] = value
        
print(new_arr)        
                  

---

顺便说一句：您甚至可以使用具有不同值的factor_x 、 factor_y 。

例如

factor_x = 4
factor_y = 2

在代码中

import numpy as np

arr = np.array([[255, 255, 255],
     [255, 255, 255],
     [255, 255, 255]]
)

factor_x = 4
factor_y = 2

h, w = arr.shape[:2]
new_w = h*factor_x
new_h = h*factor_y
new_arr = np.zeros((new_h, new_w), np.uint8)

offset_x = 0
offset_y = 0
for row, y in zip(arr, range(offset_y, new_h, factor_y)):
    #print(row, y)
    for value, x in zip(row, range(offset_x, new_w, factor_x)):
        #print(y, x, value)
        new_arr[y,x] = value
        
print(new_arr)        
                  

给

[[255   0   0   0 255   0   0   0 255   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0   0   0]
 [255   0   0   0 255   0   0   0 255   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0   0   0]
 [255   0   0   0 255   0   0   0 255   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0   0   0]]

Answer 4

使用切片进行扩展很简单：

array = np.array([[255, 255, 255], [255, 255, 255], [255, 255, 255]])
factor = 2

extended_array = np.zeros((array.shape[0]*factor, array.shape[1]*factor))
extended_array[::factor,::factor] = array