[英]How to flip half of a numpy array
I have a numpy array:我有一个 numpy 阵列:
arr=np.array([[1., 2., 0.],
[2., 4., 1.],
[1., 3., 2.],
[-1., -2., 4.],
[-1., -2., 5.],
[1., 2., 6.]])
I want to flip the second half of this array upward.我想向上翻转这个数组的后半部分。 I mean I want to have:
我的意思是我想要:
flipped_arr=np.array([[-1., -2., 4.],
[-1., -2., 5.],
[1., 2., 6.],
[1., 2., 0.],
[2., 4., 1.],
[1., 3., 2.]])
When I try this code:当我尝试这段代码时:
fliped_arr=np.flip(arr, 0)
It gives me:它给了我:
fliped_arr= array([[1., 2., 6.],
[-1., -2., 5.],
[-1., -2., 4.],
[1., 3., 2.],
[2., 4., 1.],
[1., 2., 0.]])
In advance, I do appreciate any help.在此之前,我非常感谢任何帮助。
You can simply concatenate rows below the n
th row (included) with np.r_ for instance, with row index n
of your choice, at the top and the other ones at the bottom:您可以简单地将第
n
行(包括)下方的行与np.r_ 连接起来,例如,您选择的行索引n
位于顶部,其他行位于底部:
import numpy as np
n = 3
arr_flip_n = np.r_[arr[n:],arr[:n]]
>>> array([[-1., -2., 4.],
[-1., -2., 5.],
[ 1., 2., 6.],
[ 1., 2., 0.],
[ 2., 4., 1.],
[ 1., 3., 2.]])
you can do this by slicing the array using the midpoint:您可以通过使用中点对数组进行切片来做到这一点:
ans = np.vstack((arr[int(arr.shape[0]/2):], arr[:int(arr.shape[0]/2)]))
to break this down a little:把它分解一下:
find the midpoint of arr, by finding its shape, the first index of which is the number of rows, dividing by two and converting to an integer:找到 arr 的中点,通过找到它的形状,第一个索引是行数,除以 2 并转换为 integer:
midpoint = int(arr.shape[0]/2)
the two halves of the array can then be sliced like so:然后可以像这样切片数组的两半:
a = arr[:midpoint]
b = arr[midpoint:]
then stack them back together using np.vstack
:然后使用
np.vstack
将它们堆叠在一起:
ans = np.vstack((a, b))
(note vstack takes a single argument, which is a tuple containing a and b: (a, b)
) (注意 vstack 接受一个参数,它是一个包含 a 和 b 的元组:
(a, b)
)
You can do this with array slicing and vstack -您可以使用数组切片和 vstack 来做到这一点 -
arr=np.array([[1., 2., 0.],
[2., 4., 1.],
[1., 3., 2.],
[-1., -2., 4.],
[-1., -2., 5.],
[1., 2., 6.]])
mid = arr.shape[0]//2
np.vstack([arr[mid:],arr[:mid]])
array([[-1., -2., 4.],
[-1., -2., 5.],
[ 1., 2., 6.],
[ 1., 2., 0.],
[ 2., 4., 1.],
[ 1., 3., 2.]])
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