Count values in numpy array and return index by result

Question

I have a 2d numpy array my_array that starts out like this:

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

But after some processing which is irrelevant now looks like this:

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

As you can see, some of the items have been "zeroed out" quite randomly, but only the value of 3 was left with only 1 item that isn't zero. I'm looking for a function that takes this array and returns the index / row number that has the value 3 (or any other value that only appears once and only once in the array).

To explain this differently:

I first have to figure out if there is such an item that only appears once (in this example the answer is yes and that item is the number 3), and then I need to return its row number (in this case 4 since the only line with 3 in it is: my_array[4])

I have successfully done that with iterating over the array, item by item, and counting the number of times each number appears (and returning only the item whose count is 1) and then iterating over everything a second time to find the correct index / row number of where that item is located.

This seems very inefficient, especially if the array will be larger. Is there a better way in numpy to do this?

EDIT: if the number that appears only once is 0 that shouldn't count, i'm only looking for the "column" that was zeroed-out completely except 1 item in it

Create a columnwise histogram and then return any elements whose counts are 1. Here's a starting point: docs.scipy.org/doc/numpy-1.13.0/reference/generated/… — sshashank124
– sshashank124, Commented Feb 20, 2018 at 13:42
@sshashank124 that sounds good but I have no idea how to do what you just said, also, I don't need the element itself in the end, I need the index of that row — Ofer Sadan
– Ofer Sadan, Commented Feb 20, 2018 at 13:43

sshashank124 · Accepted Answer · 2018-02-20 13:47:39Z

3

Try using the numpy.count_nonzero method

numpy.count_nonzero(arr, axis=0)

This will count the non-zero values columnwise
Here's a Demo

I will leave the rest to you. Good Luck

answered Feb 20, 2018 at 13:47

sshashank124

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1 Comment

Ofer Sadan Over a year ago

Great answer and thank you, you're right I can keep going from here

user3483203 · Accepted Answer · 2018-02-20 13:57:08Z

2

Edit I wasn't even using the mask, you can just use the first and last lines:

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

res = (x == 3)
print(np.where(res * x)[0])

Output:

[4]

The full response to np.where() is:

(array([4], dtype=int64), array([2], dtype=int64))

So if you wanted both the column and the row number, you could use both of these.

edited Feb 20, 2018 at 13:57

answered Feb 20, 2018 at 13:51

user3483203

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2 Comments

Ofer Sadan Over a year ago

I don't fully understand how this works but it does, however I've chosen the other answer since it's more flexible for me. Thanks so much

user3483203 Over a year ago

Multiplying x by the condition x == 3 results in a matrix where all values of 3 are True, and everything else is false. Then using np.where() returns the index of everywhere that evaluates to true.

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