How can you turn an index array into a mask array in Numpy?

Question

Is it possible to convert an array of indices to an array of ones and zeros, given the range? i.e. [2,3] -> [0, 0, 1, 1, 0], in range of 5

I'm trying to automate something like this:

>>> index_array = np.arange(200,300)
array([200, 201, ... , 299])

>>> mask_array = ???           # some function of index_array and 500
array([0, 0, 0, ..., 1, 1, 1, ... , 0, 0, 0])

>>> train(data[mask_array])    # trains with 200~299
>>> predict(data[~mask_array]) # predicts with 0~199, 300~499

Solution 1

Here's one way:

In [1]: index_array = np.array([3, 4, 7, 9])

In [2]: n = 15

In [3]: mask_array = np.zeros(n, dtype=int)

In [4]: mask_array[index_array] = 1

In [5]: mask_array
Out[5]: array([0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0])

If the mask is always a range, you can eliminate index_array, and assign 1 to a slice:

In [6]: mask_array = np.zeros(n, dtype=int)

In [7]: mask_array[5:10] = 1

In [8]: mask_array
Out[8]: array([0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0])

If you want an array of boolean values instead of integers, change the dtype of mask_array when it is created:

In [11]: mask_array = np.zeros(n, dtype=bool)

In [12]: mask_array
Out[12]: 
array([False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False], dtype=bool)

In [13]: mask_array[5:10] = True

In [14]: mask_array
Out[14]: 
array([False, False, False, False, False,  True,  True,  True,  True,
        True, False, False, False, False, False], dtype=bool)

Solution 2

For a single dimension, try:

n = (15,)
index_array = [2, 5, 7]
mask_array = numpy.zeros(n)
mask_array[index_array] = 1

For more than one dimension, convert your n-dimensional indices into one-dimensional ones, then use ravel:

n = (15, 15)
index_array = [[1, 4, 6], [10, 11, 2]] # you may need to transpose your indices!
mask_array = numpy.zeros(n)
flat_index_array = np.ravel_multi_index(
    index_array,
    mask_array.shape)
numpy.ravel(mask_array)[flat_index_array] = 1

Solution 3

There's a nice trick to do this as a one-liner, too - use the numpy.in1d and numpy.arange functions like this (the final line is the key part):

>>> x = np.linspace(-2, 2, 10)
>>> y = x**2 - 1
>>> idxs = np.where(y<0)

>>> np.in1d(np.arange(len(x)), idxs)
array([False, False, False,  True,  True,  True,  True, False, False, False], dtype=bool)

The downside of this approach is that it's ~10-100x slower than the appropch Warren Weckesser gave... but it's a one-liner, which may or may not be what you're looking for.

Solution 4

As requested, here it is in an answer. The code:

[x in index_array for x in range(500)]

will give you a mask like you asked for, but it will use Bools instead of 0's and 1's.

View more solutions

Comments

[Efreeto]

+1 This is a very nice answer too, especially if someone wants their mask_array to be an np.array.

[JulienD]

And it is much more efficient than the list comprehension.

[AnnanFay]

Is there any advantage to using int instead of bool? I'm just wondering why the top part of the answer doesn't recommend bool when the question is asking for a mask.

[Efreeto]

This was the answer that op orignally marked. But marking it made other people downvote to like -3, so I had to change my mark...

[Johann Bzh]

Isn't the in1d() method far much expansive that the other proposes solutions ?

[Kyuuhachi]

This one is really slow: not only is it not vectorized, but it's also O(n²).