How to make integer index of unique values in Numpy array? [duplicate]

Question

I have a question regarding Numpy arrays (the answer should include them). I want to create and array where each element will be mapped into a specific integer. Here is the code I got until now:

array1 = np.array(['dog', 'cat', 'parrot', 'parrot', 'dog'])
array2 = (np.unique(array1))
for index, key in enumerate(array2):
     print(np.where(array1 == key, index+1, 0))

The output now is:

[1 0 0 0 1]
[0 2 0 0 0]
[0 0 3 3 0]

But I want it to be just

[1 2 3 3 1]

Note that you usually want to avoid iterating over a NumPy array, and use it's vectorised approach (including broadcasting where necessary). — 9769953
– 9769953, Commented Sep 20, 2021 at 15:26
Depending on the final goal, you could actually use scikit-learn's OrdinalEncoder. — 9769953
– 9769953, Commented Sep 20, 2021 at 15:29
Not sure the title matches the problem you want solved. How about 'How to make integer index of unique values in Numpy array?' — Bill
– Bill, Commented Sep 20, 2021 at 15:36

9769953 · Accepted Answer · 2021-09-20 15:32:19Z

1

If you're not too picky about the exact value of the integer (just that they are unique and match a string one-to-one):

array1 = np.array(['dog', 'cat', 'parrot', 'parrot', 'dog'])
result = np.unique(array1, return_inverse=True)[1] + 1

(The + 1 is there so that it matches your result, since you also have index+1 in the original code.)

answered Sep 20, 2021 at 15:32

9769953

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