1'' 1'' - 2 years ago 79
Python Question

List of indices of every value

Is there a single Numpy function that is equivalent to

[array == value for value in np.unique(array)]


[np.where(array == value) for value in np.unique(array)]

Or if not, a more efficient way to do this? These iterate through the array
times, when you can do it in a single pass:

indices = defaultdict(list)
for index, value in enumerate(array):

I'd like a solution that doesn't require an explicit for loop.

Answer Source

If what you want is a mask with multiple unique number conditions, you can do it using return_inverse from np.unique:

Having a sample array

>>> a = np.random.randint(5, 10, size=100) # 100 [5-10) random numbers

And its unique numbers and inverse mapping

>>> unique, inverse = np.unique(a, return_inverse=True)

We can create a container with extra dimensions (a.shape + 1 dimension foe each unique number)

>>> indexes = np.zeros((a.shape[0], unique.size), dtype=np.bool)

And finally, fill the array with the inverse mapping:

>>> indexes[np.arange(a.size), inverse]  = True

The indexes map contains True in the last dimension corresponding to the index of the unique number that it matches (the order in the unique array).

>>> indexes[:3, :]
array([[False, False, False, False,  True],
       [ True, False, False, False, False],
       [False, False, False,  True, False]], dtype=bool)

Each row corresponds to the index of your original array a and each column corresponds to a unique number.

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