python numpy matrix row wise operations: columns in each row

Question

How can I loop over a matrix and get the number of columns in each row? If I have a matrix, and some elements are NaN (empty) in the matrix, e.g.: [[4,2,9,4],[3,4,8,6],[5,NaN,7,7],[Nan,8,Nan,Nan]], how can I compute the row-wise length?

I have tried:

len(matrix) # number of rows
=len(matrix[0]) # number of columns

But that gives me the total number.

So I want to get a vector saying the number of columns in each row: [4,4,3,1] e.g.

My idea is to make a loop like this:

for i in matrix:

And then a loop where it searches. But I'm not sure how to do this

EDIT: I tried @wavy's method and it worked. Can I do like this:

# empty list
Final=[]
for i in range(matrix):
 columns=np.isnan(matrix).sum(axis=1)
 result=-columns+matrix.shape[1]
 if result==1:
 Final.append(matrix[i])
 
 
 print(Final)
I also need to put other conditions, when result==2, and when result>2

• David Wierichs: Do you want to use this on large matrices?
• user11658272: No I'm trying now with a 3 (columns)x8 (rows) matrix
• hpaulj: In a numpy array all columns have the same length. np.nan doesn't count as "empty". Sounds instead like you want to count the number of non-nan values. Which is fine, but the description should be clearer.

Answer 1

This might be faster than David Wierichs suggestion:

import numpy as np 
x = np.array([[4, 2, 9, 4], [3, 4, 8, 6], [5, np.nan, 7, 7], [np.nan, 8, np.nan, np.nan]])
y = np.isnan(x).sum(axis=1)
result = -y + x.shape[1]

• user11658272: Thank you! Now that I have this, can I then somehow get it to say "if length of that row is 1 (so only where there is 1 number) then the number in the new vector should be that number"?
• user11658272: Something like this: for i in range(matrix): columns=np.isnan(matrix).sum(axis=1) result=-columns+matrix.shape[1] if result==1: # then here it should keep that number in the vector where result is 1 print(result)
• Wavy: After having done what I posted there you can try this: x[np.isnan(x)] = 0 for i in range(len(result)): if result[i] == 1: result[i] = np.max(x[i, :]). There's likely a faster way than looping over the result array, but I cannot immediately think of it.
• user11658272: Is it possible somehow I can ask you some more questions in any way?
• Wavy: I'm online for a bit longer, we can chat here I think: chat.stackoverflow.com/rooms/216320/discuss-numpy-question

Answer 2

You could loop over the rows and for each row use the (negated) numpy.isnan method:

lengths = [np.sum(~np.isnan(row)) for row in matrix]

As this builds up a boolean array in the np.isnan, there might be faster approaches.