带余数抛出错误的Numpy整形

### Two Examples Without Padding

x = np.array([0,1,2,3,4,5,6,7,8,9,10])
desired_length = 3
num_splits = np.ceil(x.shape[0]/desired_length)

print(np.array_split(x, num_splits))

# Prints:
# [array([0, 1, 2]), array([3, 4, 5]), array([6, 7, 8]), array([ 9, 10])]

x = np.arange(13)
desired_length = 3
num_splits = np.ceil(x.shape[0]/desired_length)

print(np.array_split(x, num_splits))

# Prints:
# [array([0, 1, 2]), array([3, 4, 5]), array([6, 7, 8]), array([ 9, 10]), array([11, 12])]

### One Example With Padding

x = np.arange(13)
desired_length = 3
padding = int(num_splits*desired_length - x.shape[0])
x_pad = np.pad(x, (0,padding), 'constant', constant_values=0)

print(np.split(x_pad, num_splits))

# Prints:
# [array([0, 1, 2]), array([3, 4, 5]), array([6, 7, 8]), array([ 9, 10, 11]), array([12,  0,  0])]

如果要避免用零填充,最优雅的方法可能是在列表理解中切片:

>>> import numpy as np
>>> x = np.arange(11)
>>> [x[i:i+3] for i in range(0, x.size, 3)]
[array([0, 1, 2]), array([3, 4, 5]), array([6, 7, 8]), array([ 9, 10])]

如果你想用零填充, ndarray.resize() 这样做是为了您,但您必须自己计算出预期数组的大小:

import numpy as np

x = np.array([0,1,2,3,4,5,6,7,8,9,10])

cols = 3
rows = np.ceil(x.size / cols).astype(int)

x.resize((rows, cols))
print(x)

结果是:

[[ 0  1  2]
 [ 3  4  5]
 [ 6  7  8]
 [ 9 10  0]]

据我所知,这比列表理解方法快数百倍(见我的另一个答案)。

注意如果你做了什么 x 在调整大小之前,您可能会遇到“引用”问题。要么继续工作 x.copy() 或通过 refcheck=False 到 resize() .