我有一个像矩阵numpy这样的数组。import numpy as npa = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19, 20]])所需的数组是这样的：array([[ 9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19, 20], [ 1, 2, 3, 4], [ 5, 6, 7, 8],])描述：第一个和第二个数组移动到矩阵的末尾。我尝试了一些更改a为 alist并使用append和del函数，然后将其转换为 anumpy array但用 python 编写可能不是什么好东西。是否有任何函数可以替换更大的类数组矩阵中的数组位置numpy？
2 回答
眼眸繁星
TA贡献1873条经验 获得超9个赞
取转数的函数
In [5]: a
Out[5]:
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20]])
In [14]: def rotate(n):
...: n = n%len(a)
...: return np.concatenate([a[n:], a[:n]])
In [13]: rotate(2)
Out[13]:
array([[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20],
[ 1, 2, 3, 4],
[ 5, 6, 7, 8]])
如果你给出的n长度超过数组的长度怎么办？处理好了——n = n%len(a)
In [16]: rotate(9)
Out[16]:
array([[17, 18, 19, 20],
[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16]])
注释中给出的另一种解决方案是roll()method。
In [6]: a
Out[6]:
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20]])
In [7]: def rotate(n):
...: n = n % len(a)
...: return np.roll(a,-n,axis=0)
...:
In [8]: rotate(8)
Out[8]:
array([[13, 14, 15, 16],
[17, 18, 19, 20],
[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
In [9]: rotate(2)
Out[9]:
array([[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20],
[ 1, 2, 3, 4],
[ 5, 6, 7, 8]])
泛舟湖上清波郎朗
TA贡献1818条经验 获得超3个赞
如果您使用这行简单的代码，这将非常容易。不需要功能和其他东西。
只需使用numpy.roll. 请参阅此处的解释
# Assume your matrix is named a.
>>> a
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20]])
>>> np.roll(a,-(n % len(a)),axis=0)
array([[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20],
[ 1, 2, 3, 4],
[ 5, 6, 7, 8]])
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