Replacing Elements in an Array with its "Element Number" (Python)

Question

I have an array that looks something like this:

np.array([[0 , 5, 1], [0, 0, 3], [1, 7, 0]])

I want to check that each element is nonzero, and if it is nonzero replace it with a number that tracks how many elements it has checked. That is, I want the final product to look like

np.array([[0, 2, 3], [0, 0, 6], [7, 8, 0]])

where the first row reads [0, 2, 3] because the second element was checked second, passed the test, and then replaced (and so on). Can anyone think of any solutions? I imagine that numpy's indexing will be quite useful here. Thanks!

"I imagine that numpy's indexing will be quite useful here." Sure, so what did you try from that hunch? — roganjosh
– roganjosh, Commented Sep 26, 2019 at 10:52

Aryerez · Accepted Answer · 2019-09-26 11:02:22Z

3

You can do:

np.where(a == 0, a, np.arange(a.size).reshape(a.shape) + 1)

answered Sep 26, 2019 at 11:02

Aryerez

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1 Comment

Cameron F. Over a year ago

Incredibly elegant solution, I will have to spend some more time with the np.where() function in the future! Thanks :)

RomanPerekhrest · Accepted Answer · 2019-09-26 11:41:32Z

1

In case if you need to modify the initial array - additional approach using mask array:

(from IPython interactive console session)

In [211]: arr = np.array([[0, 5, 1], [0, 0, 3], [1, 7, 0]])

In [212]: m = arr.nonzero()

In [213]: arr[m] = np.arange(1, arr.size+1).reshape(arr.shape)[m]

In [214]: arr
Out[214]: 
array([[0, 2, 3],
       [0, 0, 6],
       [7, 8, 0]])

answered Sep 26, 2019 at 11:41

RomanPerekhrest

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