Fill Numpy Array row wise

Question

I am experimenting with Numpy features and want to know if there is a way to achieve the desired behavior as explained. Given a numpy array as shown below

array = np.array([[1,3,3],[6,7,6],[9,9,4]])
print(array)

Output from print(array)

[[1 3 3]
 [6 7 6]
 [9 9 4]]

I am getting the maximum in each row

max_array =  array.max(axis=1,keepdims=1)
print(max_array)

Output from print(max_array)

[[3]
 [7]
 [9]]

I am applying the mask on all the maximum elements in the array using the code below

masked_array = np.ma.masked_equal(array,max_array)
print(masked_array)

Output from print(masked_array)

[[1 -- --]
 [6 -- 6]
 [-- -- 4]]

Now that I have masked the array, I want to perform operation to the non-max elements in the array like multiplication by 3 as show in the code below

 mul_array= np.multiply(masked_array,3)
 print(mul_array)

Output from print(mul_array)

[[3 -- --]
 [18 -- 18]
 [-- -- 12]]

I want to insert the maximum elements in the max_array which were masked earlier in the same position in the mul_array, but I couldn't find anything to achieve the desired behavior. Below is my expected matrix. I wanted to ask if there are any Numpy operation to achieve the desired behavior

Desired output

[[3 3 3]
 [18 7 18]
 [9 9 12]]

Thank you for helping!

Welcome to SO. This isn't a discussion forum or tutorial. Please take the tour and take the time to read How to Ask and the other links found on that page. — wwii
– wwii, Commented Oct 4, 2020 at 23:29
@wwii thank you for recommending the resources. I have updated my post. Please let me know if there is still something missing that can be updated to meet the standard — Swornim Baral
– Swornim Baral, Commented Oct 5, 2020 at 0:50
Sorry, it's my bad. I forgot to replace the a and b with array and max_array. I have updated my post. — Swornim Baral
– Swornim Baral, Commented Oct 5, 2020 at 18:01

wwii · Accepted Answer · 2020-10-05 16:21:36Z

import numpy as np
a = np.array([[1,3,3],[6,7,6],[9,9,4]])

First find you max values

max_array =  a.max(axis=1,keepdims=True)

In [19]: max_array
Out[19]: 
array([[3],
       [7],
       [9]])

Then make a boolean array for all the values that are not

mask = (a != max_array)

In [20]: mask
Out[20]: 
array([[ True, False, False],
       [ True, False,  True],
       [False, False,  True]], dtype=bool)

Then use the mask to filter the assignment on the left-hand-side and for the operation on the right-hand-side.

a[mask] = a[mask] * 3

In [21]: a
Out[21]: 
array([[ 3,  3,  3],
       [18,  7, 18],
       [ 9,  9, 12]])

Numpy Indexing

The same process with a masked array

b = np.array([[1,3,3],[6,7,6],[9,9,4]])
b_max =  b.max(axis=1,keepdims=True)
b_ma = np.ma.masked_equal(b,b_max)
q = b_ma * 3
b = q.data

# or
b_ma *= 3
b = b_ma.data

Those will result in b being a view of the data in b_ma.

In [80]: b.flags['OWNDATA']
Out[80]: False

If you don't want a view -

In [81]: b = b_ma.data.copy()

In [82]: b.flags['OWNDATA']
Out[82]: True

Biatrixia155 · Accepted Answer · 2020-10-04 23:30:22Z

First, I changed the syntax so that all the arrays and values match numpy syntax. Then you create an array 'indexes' to store indexes where A array has null values. Create a C array which is a copy of A, and use the np.take function to fill in the null values with B values.

import numpy as np
A = np.array([ [1, 2, np.nan],
      [4, np.nan, 6],
      [7, 8, np.nan] ])
B = np.array([ [3],
      [5],
      [9] ,
     ])

indexes = np.where(np.isnan(A))

C = np.copy(A)

C[indexes] = np.take(B, indexes[0])

print(C)

mathfux · Accepted Answer · 2020-10-04 23:36:06Z

One of the options is to use np.nan instead of -- and a flattened version of B:

A = np.array([[1, 2, np.nan], [4, np.nan, 6], [7, 8, np.nan]])
B = np.array([[3], [5], [9]])
B = B.flatten()

So B is of the form [3, 5, 9] now. Then you can identify locations of cells with these np.nan and reassign them:

x, y = np.nonzero(np.isnan(A))
A[x,y] = B

Output:

>>> A
array([[1., 2., 3.],
       [4., 5., 6.],
       [7., 8., 9.]])

Note that:

type of A is float because of np.nans were used before this
change np.nonzero is a substitute for np.where in 1D cases
You might like to work with np.copy(A) instead in order to avoid dynamical changes of A.

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