How to find min and max values in a 3d array in numpy, and group the results?

Question

I have a 3D NumPy array like so:

[[[4 1 5 2 5 5 7 8 9 7]
  [7 4 2 4 7 8 4 1 3 5]
  [6 1 2 1 1 1 2 3 7 6]
  [5 5 5 0 5 4 3 8 7 1]
  [2 8 6 7 4 7 5 5 5 1]]

 [[9 9 5 8 0 7 3 9 8 1]
  [9 1 9 5 7 4 5 4 7 0]
  [1 0 4 8 7 3 4 3 8 8]
  [8 1 3 1 7 0 9 9 3 8]
  [4 0 2 3 8 2 0 1 2 4]]

 [[1 6 2 4 4 0 2 3 0 3]
  [9 6 8 6 6 5 6 9 4 1]
  [0 4 0 2 9 1 1 2 4 6]
  [6 1 9 9 7 8 9 7 6 8]
  [9 3 9 0 7 0 0 0 7 0]]]

With it, I like to create a 2d ndarray like so:

[[6]
 [9]
 [9]]

Where each element of this 2d array is the max value of the third column on the original array:

I have been a couple of hours trying to puzzle this out but no luck...

I'm asking for a 2d array as the output because I have other calculations to make (say, I also need to min value of the second column in a similar fashion), but I think I can extrapolate those from this.

Any pointers are greatly appreciated!

Answer 1

Have you tried this:

import numpy as np
x = np.random.randint(0, 10, (3, 5, 10))
print(x)
maxes = x[:,:,2].max(axis=1)
print(maxes)

[[[5 0 6 6 4 7 5 0 4 8]
  [0 6 8 8 2 1 7 5 4 3]
  [2 7 5 5 0 2 6 8 6 3]
  [5 9 7 5 1 1 5 4 8 7]
  [0 2 3 7 8 1 9 1 2 6]]

 [[8 9 4 3 3 6 0 4 9 1]
  [1 5 6 4 3 2 7 7 0 2]
  [3 2 0 1 9 6 5 8 0 5]
  [6 1 5 9 1 6 4 7 4 5]
  [7 2 5 8 6 8 5 1 9 5]]

 [[9 4 0 9 0 6 3 7 4 1]
  [4 1 4 9 1 1 1 2 0 6]
  [7 3 3 2 5 2 0 6 9 1]
  [1 7 0 1 8 1 3 8 6 4]
  [6 9 0 2 6 0 2 1 7 7]]]
[8 6 4]

To understand how this works checkout:

• Numpy array indexing
• np.ndarray.max

And, to get the maximum of all the columns:

col_maximums = x.max(axis=1)
print(col_maximums)
                                                                                    
[[5 9 8 8 8 7 9 8 8 8]
 [8 9 6 9 9 8 7 8 9 5]
 [9 9 4 9 8 6 3 8 9 7]]

Answer 2

What you need to do is take the columns that you want(in this case it is arr[:,:,2] ). This will have the shape of (3,5) . arr=np.max(arr[:,:,2],axis=-1) then yields the array contents that you want with a shape of (3) . You can then call arr=np.reshape(arr,(3,1)) to get the shape you wanted.

Answer 3

Below I tried to do the operation step by step and printed the shape after every operation. the (2,) is important in order to preserve the dimension and stay with 2d like you wanted (otherwise it had auto-collapse the singleton dimension).

import numpy as np

a = np.array([
    [[4, 1, 5, 2, 5, 5, 7, 8, 9, 7],
     [7, 4, 2, 4, 7, 8, 4, 1, 3, 5],
     [6, 1, 2, 1, 1, 1, 2, 3, 7, 6],
     [5, 5, 5, 0, 5, 4, 3, 8, 7, 1],
     [2, 8, 6, 7, 4, 7, 5, 5, 5, 1]],

    [[9, 9, 5, 8, 0, 7, 3, 9, 8, 1],
     [9, 1, 9, 5, 7, 4, 5, 4, 7, 0],
     [1, 0, 4, 8, 7, 3, 4, 3, 8, 8],
     [8, 1, 3, 1, 7, 0, 9, 9, 3, 8],
     [4, 0, 2, 3, 8, 2, 0, 1, 2, 4]],

    [[1, 6, 2, 4, 4, 0, 2, 3, 0, 3],
     [9, 6, 8, 6, 6, 5, 6, 9, 4, 1],
     [0, 4, 0, 2, 9, 1, 1, 2, 4, 6],
     [6, 1, 9, 9, 7, 8, 9, 7, 6, 8],
     [9, 3, 9, 0, 7, 0, 0, 0, 7, 0]]
])

print(a.shape)  # --> (3, 5, 10)
col2 = a[:, :, (2,)]
print(col2.shape)  # --> (3, 5, 1)
max_col2 = col2.max(axis=1)
print(max_col2.shape)  # --> (3, 1)
print(max_col2)  # -->

[[6]
 [9]
 [9]]
 

Answer 4

You alternatively could use the keepdims argument of np.max like this ( a is your original array):

a[:,:,2].max(1, keepdims=True)

Output:

[[6]
 [9]
 [9]]