1 
import numpy as np

temp_array = np.array(
  [[33.5, 35.3, 33.6, 33.6, 33.5, 33.9, 32.3, 33.2, 53.8, 54.6, 53.4, 54.2],
   [33.1, 34.2, 34.1, 34.3, 34.7, 31.3, 32.3, 33.4, 57.5, 55. , 53.5, 56.1],
   [35.3, 35.4, 35.6, 32.6, 33.2, 34.3, 32.8, 33.1, 54.7, 55.4, 54.6, 55.1],
   [34.2, 36.1, 33.5, 32.4, 32.1, 33.5, 34.5, 35. , 53.8, 56.9, 54.5, 54.7],
   [33.4, 33.8, 36.2, 33. , 35. , 34.2, 33.8, 33.8, 55.7, 55.2, 56. , 54.5],
   [34.3, 35.9, 34.4, 34.2, 53.5, 54.2, 55.7, 54. , 56.3, 54.4, 55.5, 53.8],
   [34.7, 35.4, 34.7, 33.1, 53.6, 54.5, 54.4, 55.5, 54.7, 55.4, 55.1, 55.6],
   [33.3, 34.3, 33.6, 33.1, 55.4, 55.7, 55.4, 55.4, 55.8, 55. , 55.3, 54.1],
   [33.7, 33.5, 37. , 34.9, 57.6, 54.2, 54.9, 54.6, 56. , 55.7, 55.1, 55.9],
   [34. , 35.1, 33.6, 34.5, 56.2, 55.3, 55.2, 54. , 54.1, 54.5, 54.4, 56. ]])


    
cell_shape = (5,4)

I want to find the average for the average the section for example the first section:

[33.5, 35.3, 33.6, 33.6]
[33.1, 34.2, 34.1, 34.3]
[35.3, 35.4, 35.6, 32.6]
[34.2, 36.1, 33.5, 32.4]
[33.4, 33.8, 36.2, 33. ]

Second section:

[33.5, 33.9, 32.3, 33.2]
[34.7, 31.3, 32.3, 33.4]
[33.2, 34.3, 32.8, 33.1]
[32.1, 33.5, 34.5, 35. ]
[35. , 34.2, 33.8, 33.8]

and etc.

Improve this question
1

3 Answers 3

Reset to default
2

The current answers works fine, but they may be slow if you want to do this often on large arrays. The operation you want to do is called "strided convolutions using a mean/uniform kernel". There exists many libraries to do said operation a lot faster than using for-loops in Python (e.g. PyTorch, scikit-image, or a more advanced numpy way using stride tricks).

Here is an example using skimage which avoids for-loops in Python:

import numpy as np
from skimage.util.shape import view_as_windows

temp_array = np.array(
  [[33.5, 35.3, 33.6, 33.6, 33.5, 33.9, 32.3, 33.2, 53.8, 54.6, 53.4, 54.2],
   [33.1, 34.2, 34.1, 34.3, 34.7, 31.3, 32.3, 33.4, 57.5, 55. , 53.5, 56.1],
   [35.3, 35.4, 35.6, 32.6, 33.2, 34.3, 32.8, 33.1, 54.7, 55.4, 54.6, 55.1],
   [34.2, 36.1, 33.5, 32.4, 32.1, 33.5, 34.5, 35. , 53.8, 56.9, 54.5, 54.7],
   [33.4, 33.8, 36.2, 33. , 35. , 34.2, 33.8, 33.8, 55.7, 55.2, 56. , 54.5],
   [34.3, 35.9, 34.4, 34.2, 53.5, 54.2, 55.7, 54. , 56.3, 54.4, 55.5, 53.8],
   [34.7, 35.4, 34.7, 33.1, 53.6, 54.5, 54.4, 55.5, 54.7, 55.4, 55.1, 55.6],
   [33.3, 34.3, 33.6, 33.1, 55.4, 55.7, 55.4, 55.4, 55.8, 55. , 55.3, 54.1],
   [33.7, 33.5, 37. , 34.9, 57.6, 54.2, 54.9, 54.6, 56. , 55.7, 55.1, 55.9],
   [34. , 35.1, 33.6, 34.5, 56.2, 55.3, 55.2, 54. , 54.1, 54.5, 54.4, 56. ]])

cell_shape = (5,4)
sections: np.ndarray = view_as_windows(temp_array, cell_shape, cell_shape)
print(sections.mean((-2,-1)))

result:

[[34.16  33.495 54.96 ]
 [34.365 54.965 55.135]]
Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

1 
for startx, endx in [ ( n * cell_shape[0], (n+1) * cell_shape[0] ) for n in range( temp_array.shape[0] // cell_shape[0] ) ]:
        for starty, endy in [ ( n * cell_shape[1], (n+1) * cell_shape[1] ) for n in range( temp_array.shape[1] // cell_shape[1] ) ]:
                np.average(temp_array[startx:endx, starty:endy])

Uses np slicing which is in the format, [startx:endx, starty:endy]

This only prints the arrays but making it an average is just a matter of changing it to np.mean

Improve this answer

Comments

1 
sh = temp_array.shape
for y in range(0, sh[0], cell_shape[0]):
    for x in range(0, sh[1], cell_shape[1]):
        print(np.average(temp_array[y:y + cell_shape[0], x:x + cell_shape[1]]))

Prints:

34.16
33.49499999999999
54.96
34.365
54.964999999999996
55.135000000000005
Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.