Question

我正在寻找python oneliner，它将创建具有给定顺序的矩阵并填充从1到n的值。 n = r X c

我实现了这一点，

Matrix1 = [[x+1 for x in range(rowCount)] for x in range(columnCount)]
print Matrix1

但输出是

[[1, 2, 3], [1, 2, 3], [1, 2, 3]]

Answer 1

不，正确的方法是没有循环（并在一行中）：

import numpy as np

rowCount = 2
colCount = 5

np.array(range(1, 1+rowCount*colCount)).reshape(rowCount,colCount)

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

Answer 2

您对两个循环使用相同的变量。试试这个：

matrix1 = [[y*rowCount + x + 1 for x in range(rowCount)] for y in range(columnCount)]
print matrix1

Answer 3

>>> matrix1 = [[1+x+y*rowCount for x in range(rowCount)] for y in range(columnCount)]
>>> matrix1
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Answer 4

鉴于c（列数），n，这就是我想出的：

[range(0, n)[cx:cx+c] for cx in range(0, n, c)]

结果：

In [16]: n = 9

In [17]: c = 3

In [18]: [range(0, n)[cx:cx+c] for cx in range(0, n, c)]
Out[18]: [[0, 1, 2], [3, 4, 5], [6, 7, 8]]

编辑：非方形矩阵的更多示例：（mod是我创建的一项功能，可以快速更改r和c值，并从中计算n）

In [23]: mod(8, 2)  # 8 rows, 2 columns

In [24]: [range(0, n)[cx:cx+c] for cx in range(0, n, c)]
Out[24]: [[0, 1], [2, 3], [4, 5], [6, 7], [8, 9], [10, 11], [12, 13], [14, 15]]

In [25]: mod(3, 6)  # 3 rows, 6 columns

In [26]: [range(0, n)[cx:cx+c] for cx in range(0, n, c)]
Out[26]: [[0, 1, 2, 3, 4, 5], [6, 7, 8, 9, 10, 11], [12, 13, 14, 15, 16, 17]]

In [27]: mod(10, 3)  # 10 rows, 3 columns

In [28]: [range(0, n)[cx:cx+c] for cx in range(0, n, c)]
Out[28]: 
[[0, 1, 2],
 [3, 4, 5],
 [6, 7, 8],
 [9, 10, 11],
 [12, 13, 14],
 [15, 16, 17],
 [18, 19, 20],
 [21, 22, 23],
 [24, 25, 26],
 [27, 28, 29]]

它也适用于他们。

python oneline创建给定顺序的矩阵

4 个答案: