From the oft-referenced numpy indexing page:
.... A single boolean index array is practically identical to x[obj.nonzero()] .... However, it is faster when obj.shape == x.shape.
np.where(cond)
is np.nonzero(cond)
.
But let's do some simple timing
In [239]: x = np.arange(10000)
In [240]: y = (x%2).astype(bool)
In [241]: x[y].shape
Out[241]: (5000,)
In [242]: idx = np.nonzero(y)
In [243]: x[idx].shape
Out[243]: (5000,)
In [244]: timeit x[y].shape
89.9 μs ± 726 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [245]: timeit x[idx].shape
13.3 μs ± 107 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [246]: timeit x[np.nonzero(y)].shape
34.2 μs ± 893 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
So array indexing is faster than boolean indexing, even when we use an explicit where
.
A[ind][k]=
does not work because A[ind]
is a copy, not a view.
In [251]: A = np.arange(100,110)
In [252]: ind = np.logical_and(A>104, A%2)
In [253]: ind
Out[253]:
array([False, False, False, False, False, True, False, True, False,
True])
In [254]: k = np.array([0,1,0], dtype=bool)
In [255]: A[ind]
Out[255]: array([105, 107, 109])
In [256]: A[ind][k]
Out[256]: array([107])
In [257]: A[ind][k] = 12
In [258]: A
Out[258]: array([100, 101, 102, 103, 104, 105, 106, 107, 108, 109])
But using the k
to select indices from np.where(ind)
works:
In [262]: A[np.where(ind)[0][k]]=12
In [263]: A
Out[263]: array([100, 101, 102, 103, 104, 105, 106, 12, 108, 109])
Timings for a fetch rather than a set:
In [264]: timeit A[np.where(ind)[0][k]]
1.94 μs ± 75.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [265]: timeit A[ind][k]
1.34 μs ± 13.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
So yes, the double masking is a bit faster in this case, but that doesn't matter if it doesn't work. Don't sweat the small time improvements.
A boolean indexing method
In [345]: ind1=ind.copy()
In [346]: ind1[ind] = k
In [348]: A[ind1]=3
In [349]: A
Out[349]: array([100, 101, 102, 103, 104, 105, 106, 3, 108, 109])
In this small example timeit is basically the same as for A[np.where(ind)[0][k]]=12
.