How to flip half of a numpy array

Question

I have a numpy array:

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. I mean I want to have:

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:

fliped_arr=np.flip(arr, 0)

It gives me:

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.

What is the expected behaviour when the number of rows in the array is odd? — Kalpit
– Kalpit, Commented Dec 3, 2020 at 14:43
@Kalpit, Thanks for your hint. Then I want to say from that line until the end, bring rows with the same order up. Then that nth line will be the first one. — Link_tester
– Link_tester, Commented Dec 3, 2020 at 14:44

yacola · Accepted Answer · 2020-12-03 14:49:09Z

3

You can simply concatenate rows below the nth row (included) with np.r_ for instance, with row index n of your choice, at the top and the other ones at the bottom:

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.]])

answered Dec 3, 2020 at 14:49

yacola

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Comments

FinleyGibson · Accepted Answer · 2020-12-03 15:04:18Z

2

you can do this by slicing the array using the midpoint:

ans = np.vstack((arr[int(arr.shape[0]/2):], arr[:int(arr.shape[0]/2)]))

to break this down a little:

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:

midpoint = int(arr.shape[0]/2)

the two halves of the array can then be sliced like so:

a = arr[:midpoint]
b = arr[midpoint:]

then stack them back together using np.vstack:

ans = np.vstack((a, b))

(note vstack takes a single argument, which is a tuple containing a and b: (a, b))

edited Dec 3, 2020 at 15:04

answered Dec 3, 2020 at 14:58

FinleyGibson

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2 Comments

Akshay Sehgal Over a year ago

Hi @Finley, how is this different than my answer?

FinleyGibson Over a year ago

Doesn't seem to be. Your's wasn't there at the time of posting. (appeared upon the refresh that came with posting the asnwer)

FinleyGibson · Accepted Answer · 2020-12-03 15:45:33Z

1

You can do this with array slicing and vstack -

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.]])

edited Dec 3, 2020 at 15:45

FinleyGibson

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answered Dec 3, 2020 at 14:52

Akshay Sehgal

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