Question

我有一个叫GTY的字典，其中有100个键，都是

的组合

range(10,101,10)

（即：（10,10），（10,20），...（90,100），（100,100））。我想对该字典进行“解压”，并得到一个10x10矩阵，其列数为0到100（带有步骤10），行的列数为0到100（带有步骤10）。

我希望通过运行一种简单的拆包技术来做到这一点

np.array([GTY[i] for i in itertools.combinations(range(10,101,10), 2)])

，但它在单列而不是10x10矩阵中产生数据。是否有一种有效的简单方法来获取矩阵，其中col为键元组nr1，行为键元组nr2？

非常感谢！

Answer 1

您可以使用numpy生成11x11的数组，并使用product生成所有组合：

import numpy as np
z = itertools.product(range(0,101,10),repeat= 2)
z1 = np.array([i for i in z]).reshape(11,11,2)
for i in range(11):
    print(f'z[{i}] = {z1[i]}')

输出：

z [0] = [[0 0] [0 10] [0 20] [0 30] [0 40] [0 50] [0 60] [0 70] [0 80] [0 90] [0 100]]

z [1] = [[10 0] [10 10] [10 20] [10 30] [10 40] [10 50] [10 60] [10 70] [10 80] [10 90] [10 100]]

z [2] = [[20 0] [20 10] [20 20] [20 30] [20 40] [20 50] [20 60] [20 70] [20 80] [20 90] [20 100]]

z [3] = [[30 0] [30 10] [30 20] [30 30] [30 40] [30 50] [30 60] [30 70] [30 80] [30 90] [30 100]]

z [4] = [[40 0] [40 10] [40 20] [40 30] [40 40] [40 50] [40 60] [40 70] [40 80] [40 90] [40 100]]

z [5] = [[50 0] [50 10] [50 20] [50 30] [50 40] [50 50] [50 60] [50 70] [50 80] [50 90] [50 100]]

z [6] = [[60 0] [60 10] [60 20] [60 30] [60 40] [60 50] [60 60] [60 70] [60 80] [60 90] [60 100]]

z [7] = [[70 0] [70 10] [70 20] [70 30] [70 40] [70 50] [70 60] [70 70] [70 80] [70 90] [70 100]]

z [8] = [[80 0] [80 10] [80 20] [80 30] [80 40] [80 50] [80 60] [80 70] [80 80] [80 90] [80 100]]

z [9] = [[90 0] [90 10] [90 20] [90 30] [90 40] [90 50] [90 60] [90 70] [90 80] [90 90] [90 100]]

z [10] = [[100 0] [100 10] [100 20] [100 30] [100 40] [100 50] [100 60] [100 70] [100 80] [100 90] [100 100]]

在nxn矩阵中以元组为键解压缩n字典

1 个答案: