Extract elements from numpy array, that are not in list of indexes

Question

I want to do something similar to what was asked here NumPy array, change the values that are NOT in a list of indices, but not quite the same.

Consider a numpy array:

> a = np.array([0.2, 5.6, 88, 12, 1.3, 6, 8.9])

I know I can access its elements via a list of indexes, like:

> indxs = [1, 2, 5] 
> a[indxs]
array([  5.6,  88. ,   6. ])

But I also need to access those elements which are not in the indxs list. Naively, this is:

> a[not in indxs]
> array([0.2, 12, 1.3, 8.9])

What is the proper way to do this?

Debrief

Answer 1

In [170]: a = np.array([0.2, 5.6, 88, 12, 1.3, 6, 8.9])
In [171]: idx=[1,2,5]
In [172]: a[idx]
Out[172]: array([  5.6,  88. ,   6. ])
In [173]: np.delete(a,idx)
Out[173]: array([  0.2,  12. ,   1.3,   8.9])

delete is more general than you really need, using different strategies depending on the inputs. I think in this case it uses the boolean mask approach (timings should be similar).

In [175]: mask=np.ones_like(a, bool)
In [176]: mask
Out[176]: array([ True,  True,  True,  True,  True,  True,  True], dtype=bool)
In [177]: mask[idx]=False
In [178]: mask
Out[178]: array([ True, False, False,  True,  True, False,  True], dtype=bool)
In [179]: a[mask]
Out[179]: array([  0.2,  12. ,   1.3,   8.9])

Dieppe

Answer 2

One way is to use a boolean mask and just invert the indices to be false:

mask = np.ones(a.size, dtype=bool)
mask[indxs] = False
a[mask]

Guria

Answer 3

One approach with np.in1d to create the mask of the ones from indxs present and then inverting it and indexing the input array with it for the desired output -

a[~np.in1d(np.arange(a.size),indxs)]

Dipsomania

Answer 4

TLDR: use mask.

Benchmark using given case:

import numpy as np


a = np.array([0.2, 5.6, 88, 12, 1.3, 6, 8.9])

%%timeit
idx = [1, 2, 5]
np.delete(a,idx)
# 2.06 µs ± 10.8 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

%%timeit
mask = np.ones(a.size, dtype=bool)
mask[idx] = False
a[mask]
# 1.51 µs ± 7.62 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)

%%timeit
a[~np.in1d(np.arange(a.size), idx)]
# 14.8 µs ± 187 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

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