How to find mean of each two rows of a numpy array

Question

I have a big numpy array and want to take the mean of the second columns of each two rows and save the array as a new one. I want to take the mean of each two row, i.e. the mean of second column of of rows 1 and 2. Then, mean of second column of rows 3 and 4, and so on. Then, I want to merge each two rows as a single one. First and third columns of this paired rows are also the same. This is my simplified array:

input= np.array ([[1., 2., 5.],
                  [1., 4., 5.],
                  [4., 10., 3.],
                  [4., 2., 3.],
                  [1., 0., 0.],
                  [1., 1., 0.]])

Then, I want to get:

output= np.array ([[1., 3., 5.],
                   [4., 6., 3.],
                   [1., 0.5, 0.]])

I tried the following but it was not successful at all:

output=np.array([])
for i in range (len(input)-1):
    g=(input[i,1]+input[i+1,1])/2
    output=np.append(g,output)

In advance, I do appreciate any help.

Quang Hoang · Accepted Answer · 2020-11-18 15:17:51Z

2

For two rows, I find it easier to do:

(arr[::2] + arr[1::2])/2

answered Nov 18, 2020 at 15:17

Quang Hoang

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Daniel F · Accepted Answer · 2020-11-18 15:15:02Z

2

A little more robust method for reshape, using the input shape

i= np.array ([[1., 2., 5.],
              [1., 4., 5.],
              [4., 10., 3.],
              [4., 2., 3.],
              [1., 0., 0.],
              [1., 1., 0.]])

i.reshape(-1, 2, i.shape[-1]).mean(1)

array([[1. , 3. , 5. ],
       [4. , 6. , 3. ],
       [1. , 0.5, 0. ]])

answered Nov 18, 2020 at 15:15

Daniel F

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2 Comments

Link_tester Over a year ago

Dear @Daniel F, Thanks for the solution. It is exactly what I wanted. I do appreciate it.

Daniel F Over a year ago

Be sure to mark solved (the little checkmark) then.

Dani Mesejo · Accepted Answer · 2020-11-18 15:12:38Z

1

You could reshape and find the mean, as follows:

import numpy as np

ipt = np.array([[1., 2., 5.],
                [1., 4., 5.],
                [4., 10., 3.],
                [4., 2., 3.],
                [1., 0., 0.],
                [1., 1., 0.]])
result = np.mean(ipt.reshape((3, 2, 3)), axis=1)
print(result)

Output

[[1.  3.  5. ]
 [4.  6.  3. ]
 [1.  0.5 0. ]]

As a side note, avoid using input as a variable name as it shadows the built-in input.

answered Nov 18, 2020 at 15:12

Dani Mesejo

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2 Comments

Link_tester Over a year ago

Dear Dani Mesejo, Thanks for devoting time to my problem. I do appreciate it. But, I colud not run your solution for my big array. I was giving me a reshape error.

Dani Mesejo Over a year ago

You need to change the dimensions to fit it accordingly, in this regar the solution of @DanielF is better.

Pablo C · Accepted Answer · 2020-11-18 15:23:48Z

1

Take even rows (ipt[::2]), odd rows (ipt[1::2]), add them and divide by 2:

output = (ipt[::2] + ipt[1::2])/2

answered Nov 18, 2020 at 15:23

Pablo C

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