将 1d 掩码数组表示为切片列表

Question

I have a 1d numpy array with booleans values ( mask ) which I would like to convert into a list of slices where the mask is True, eg:

mask = [False, True, True, True, False, False, True, True]

and I would like to obtain

[slice(1, 4, None), slice(6, 8, None)]

The numpy masked array operations (in particular np.ma.clump_masked() ) can do that, but the only way I found to use it would be to do the following:

np.ma.clump_masked(np.ma.masked_array(np.ones_like(mask), mask))

which yields exactly what I'm looking for:

[slice(1, 4, None), slice(6, 8, None)]

ie, generating an array with the same shape as mask , applying the mask to it, and then computing mask_clumped() on that.

However, the np.ma.masked_array(np.ones_like(mask), mask) -step seems unnecessary to me. Is there any way to obtain the list of slices from a simplified operation which I would imagine to look like the following?

np.ma.clump_masked(mask)

Answer 1

np.ma.masked_array requires a masked array as input, not an ndarray . One approach is to do what you're currently doing and create a masked array

import numpy as np
mask = np.asarray([False, True, True, True, False, False, True, True])
masked_array = np.ma.masked_array(data=mask, mask=mask)
np.ma.clump_masked(masked_array)

However, I assume you're generating mask based on some condition? In which case, you can use np.ma.masked_where . For example, to get all the slices of each even number from 0 to 9:

import numpy as np
arr = np.arange(10)
masked_arr = np.ma.masked_where(arr % 2 == 0, arr)
np.ma.clump_masked(masked_arr)

which outputs:

[slice(0, 1, None),
 slice(2, 3, None),
 slice(4, 5, None),
 slice(6, 7, None),
 slice(8, 9, None)]

There are other functions such as np.ma.masked_inside which will create a masked array and mask all elements within some interval. Check the 'see also' of the masked_where docs for a list of the related funcitons.