快速的numpy花式索引

Question

My code for slicing a numpy array (via fancy indexing) is very slow. It is currently a bottleneck in program.

a.shape
(3218, 6)

ts = time.time(); a[rows][:, cols]; te = time.time(); print('%.8f' % (te-ts));
0.00200009

What is the correct numpy call to get an array consisting of the subset of rows 'rows' and columns 'col' of the matrix a? (in fact, I need the transpose of this result)

Answer 1

Let my try to summarize the excellent answers by Jaime and TheodrosZelleke and mix in some comments.

1. Advanced (fancy) indexing always returns a copy, never a view.
2. a[rows][:,cols] implies two fancy indexing operations, so an intermediate copy a[rows] is created and discarded. Handy and readable, but not very efficient. Moreover beware that [:,cols] usually generates a Fortran contiguous copy form a C-cont. source.
3. a[rows.reshape(-1,1),cols] is a single advanced indexing expression basing on the fact that rows.reshape(-1,1) and cols are broadcast to the shape of the intended result.

4. A common experience is that indexing in a flattened array can be more efficient than fancy indexing, so another approach is

    indx = rows.reshape(-1,1)*a.shape[1] + cols a.take(indx) 

   or

    a.take(indx.flat).reshape(rows.size,cols.size) 

5. Efficiency will depend on memory access patterns and whether the starting array is C-countinous or Fortran continuous, so experimentation is needed.

6. Use fancy indexing only if really needed: basic slicing a[rstart:rstop:rstep, cstart:cstop:cstep] returns a view (although not continuous) and should be faster!

Answer 2

To my surprise this, kind of lenghty expression, which calculates first linear 1D-indices, is more than 50% faster than the consecutive array indexing presented in the question:

(a.ravel()[(
   cols + (rows * a.shape[1]).reshape((-1,1))
   ).ravel()]).reshape(rows.size, cols.size)

UPDATE: OP updated the description of the shape of the initial array. With the updated size the speedup is now above 99% :

In [93]: a = np.random.randn(3218, 1415)

In [94]: rows = np.random.randint(a.shape[0], size=2000)

In [95]: cols = np.random.randint(a.shape[1], size=6)

In [96]: timeit a[rows][:, cols]
10 loops, best of 3: 186 ms per loop

In [97]: timeit (a.ravel()[(cols + (rows * a.shape[1]).reshape((-1,1))).ravel()]).reshape(rows.size, cols.size)
1000 loops, best of 3: 1.56 ms per loop

INITAL ANSWER: Here is the transcript:

In [79]: a = np.random.randn(3218, 6)
In [80]: a.shape
Out[80]: (3218, 6)

In [81]: rows = np.random.randint(a.shape[0], size=2000)
In [82]: cols = np.array([1,3,4,5])

Time method 1:

In [83]: timeit a[rows][:, cols]
1000 loops, best of 3: 1.26 ms per loop

Time method 2:

In [84]: timeit (a.ravel()[(cols + (rows * a.shape[1]).reshape((-1,1))).ravel()]).reshape(rows.size, cols.size)
1000 loops, best of 3: 568 us per loop

Check that results are actually the same:

In [85]: result1 = a[rows][:, cols]
In [86]: result2 = (a.ravel()[(cols + (rows * a.shape[1]).reshape((-1,1))).ravel()]).reshape(rows.size, cols.size)

In [87]: np.sum(result1 - result2)
Out[87]: 0.0

Answer 3

You can get some speed up if you slice using fancy indexing and broadcasting:

from __future__ import division
import numpy as np

def slice_1(a, rs, cs) :
    return a[rs][:, cs]

def slice_2(a, rs, cs) :
    return a[rs[:, None], cs]

>>> rows, cols = 3218, 6
>>> rs = np.unique(np.random.randint(0, rows, size=(rows//2,)))
>>> cs = np.unique(np.random.randint(0, cols, size=(cols//2,)))
>>> a = np.random.rand(rows, cols)
>>> import timeit
>>> print timeit.timeit('slice_1(a, rs, cs)',
                        'from __main__ import slice_1, a, rs, cs',
                        number=1000)
0.24083110865
>>> print timeit.timeit('slice_2(a, rs, cs)',
                        'from __main__ import slice_2, a, rs, cs',
                        number=1000)
0.206566124519

If you think in term of percentages, doing something 15% faster is always good, but in my system, for the size of your array, this is taking 40 us less to do the slicing, and it is hard to believe that an operation taking 240 us will be your bottleneck.

Answer 4

Using np.ix_ you can a similar speed to ravel/reshape, but with code that is more clear:

a = np.random.randn(3218, 1415)
rows = np.random.randint(a.shape[0], size=2000)
cols = np.random.randint(a.shape[1], size=6)
a = np.random.randn(3218, 1415)
rows = np.random.randint(a.shape[0], size=2000)
cols = np.random.randint(a.shape[1], size=6)

%timeit (a.ravel()[(cols + (rows * a.shape[1]).reshape((-1,1))).ravel()]).reshape(rows.size, cols.size)
#101 µs ± 2.36 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)


%timeit ix_ = np.ix_(rows, cols); a[ix_]
#135 µs ± 7.47 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

ix_ = np.ix_(rows, cols)
result1 = a[ix_]
result2 = (a.ravel()[(cols + (rows * a.shape[1]).reshape((-1,1))).ravel()]).reshape(rows.size, cols.size)
​
np.sum(result1 - result2)
0.0