Question

我有10x10网格的一维位置列表：

[(0, 0), (0, 1), (0, 2), ..., (9, 9)]

我想像这样的numpy数组（10个长度列表的列表）：

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

如何在二维np数组中转换一维点列表？

我写了这个：

import numpy as np

l = []
for i in range(10):
    for ii in range(10):
        l.append((i, ii))

print(l)
a = np.array(l)
print(a)
a.shape = (a.size // 10, 10)
print(a)

但结果不是预期的结果：

python3.4 tmp/t.py
[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), ..., (9, 9)]
[[0 0]
 [0 1]
 [0 2]
 [0 3]
 [0 4]
 [0 5]
 [0 6]
 [0 7]
 [0 8]
 [0 9]
 ...
 [9 9]]
[[0 0 0 1 0 2 0 3 0 4]
 [0 5 0 6 0 7 0 8 0 9]
 [1 0 1 1 1 2 1 3 1 4]
 [1 5 1 6 1 7 1 8 1 9]
 [2 0 2 1 2 2 2 3 2 4]
 ...
 [9 0 9 1 9 2 9 3 9 4]
 [9 5 9 6 9 7 9 8 9 9]]

Answer 1

试试这个：

>>> z = [(i,j) for i in range(10) for j in range(10)]
>>> z
[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), ..., (9, 9)]
>>> np.array(z).reshape((10,10, 2))
array([[[0, 0],
        [0, 1],
        [0, 2],
        [0, 3],
        [0, 4],
        [0, 5],
        [0, 6],
        [0, 7],
        [0, 8],
        [0, 9]],

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

        ...

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

Answer 2

您需要numpy.reshape

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

对于您的情况，如果您正在处理单个元素，则可以执行以下操作：

>>> >>> a = np.array(range(100))
>>> a.reshape(10,10)
array([[ 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],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
       [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
       [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
       [70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
       [80, 81, 82, 83, 84, 85, 86, 87, 88, 89],
       [90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])

但是，你的案例每个元素都有2个元素，所以你可以这样做：

>>> a = np.array([[i,j] for i in range(10) for j in range(10)])
>>> a.reshape(10,10, 2)

Answer 3

在我看来，大多数pythonic方式应该是这样的：

a = np.zeros(100,dtype=('i4,i4'))
a[:] = [(i,j) for i in range(10) for j in range(10)]
In [29]: a.reshape(10,10)
Out[29]: 
array([[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7),
        (0, 8), (0, 9)],
       [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7),
        (1, 8), (1, 9)],
       [(2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7),
        (2, 8), (2, 9)],
       [(3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7),
        (3, 8), (3, 9)],
       [(4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7),
        (4, 8), (4, 9)],
       [(5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (5, 7),
        (5, 8), (5, 9)],
       [(6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (6, 7),
        (6, 8), (6, 9)],
       [(7, 0), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7),
        (7, 8), (7, 9)],
       [(8, 0), (8, 1), (8, 2), (8, 3), (8, 4), (8, 5), (8, 6), (8, 7),
        (8, 8), (8, 9)],
       [(9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5), (9, 6), (9, 7),
        (9, 8), (9, 9)]], 
      dtype=[('f0', '<i4'), ('f1', '<i4')])

编辑：如果要保存元组，请选中：

{{1}}

在二维np数组中转换一维点列表

3 个答案: