I'm trying to find a neat little trick for slicing a row/column from a 2d array and obtaining an array of (col_size x 1) or (1 x row_size).
Is there an easier way than to use numpy.reshape() after every slicing?
Cheers, Stephan
3 Answers
You can slice and insert a new axis in one single operation. For example, here's a 2D array:
>>> a = np.arange(1, 7).reshape(2, 3)
>>> a
array([[1, 2, 3],
[4, 5, 6]])
To slice out a single column (returning array of shape (2, 1)), slice with None as the third dimension:
>>> a[:, 1, None]
array([[2],
[5]])
To slice out a single row (returning array of shape (1, 3)), slice with None as the second dimension:
>>> a[0, None, :]
array([[1, 2, 3]])
2 Comments
Alex Riley
No problem! Indexing/reshaping takes a while to wrap your head around (at least it did for me), but makes sense after a bit of practice. The docs are pretty good for explaining what's happening here (the basic slicing section).
hpaulj
Curiously I never thought to combine the indexing with the
None addition. I guess it is just the force of habit, using None to expand the dimensions of an existing array. And it is faster.Make the index a slice, list or array
X[[0],:]
X[0:1,4]
But there's nothing wrong with reshape other than the fact that it requires typing. It isn't slow. [None,:] is a nice short hand for it.
Use of a list index may be the shortest, but it does produce a copy (a plus or minus?) and is slower
For (100,100) integer array:
In [487]: timeit x[[50],:]
100000 loops, best of 3: 10.3 µs per loop # slowest
In [488]: timeit x[50:51,:]
100000 loops, best of 3: 2.24 µs per loop # slice indexing is fast
In [489]: timeit x[50,:].reshape(1,-1)
100000 loops, best of 3: 3.29 µs per loop # minimal time penalty
In [490]: timeit x[50,:][None,:]
100000 loops, best of 3: 3.55 µs per loop
In [543]: timeit x[None,50,:] # **best**
1000000 loops, best of 3: 1.76 µs per loop
One test for copy is to compare the data buffer pointer with the original.
In [492]: x.__array_interface__['data']
Out[492]: (175920456, False)
In [493]: x[50,:].__array_interface__['data']
Out[493]: (175940456, False)
In [494]: x[[50],:].__array_interface__['data']
Out[494]: (175871672, False) # different pointer
In [495]: x[50:51,:].__array_interface__['data']
Out[495]: (175940456, False)
In [496]: x[50,:][None,:].__array_interface__['data']
Out[496]: (175940456, False)
2 Comments
neurotronix
The problem is that I'm implementing a scalable neural network, scalable in terms of layer sizes. With
reshape I had to access very often different instance attributes (e.g. the layer sizes). Finally to get it right: when I use a[0, None, :] to slice a row from array a, will it then return a copy?hpaulj
[0, None, :] returns a view.How about this nice and easy way?
In [73]: arr = (np.arange(5, 25)).reshape(5, 4)
In [74]: arr
Out[74]:
array([[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16],
[17, 18, 19, 20],
[21, 22, 23, 24]])
# extract column 1 as a column vector
In [79]: col1 = arr[:, [0]]
In [80]: col1.shape
Out[80]: (5, 1)
In [81]: col1
Out[81]:
array([[ 5],
[ 9],
[13],
[17],
[21]])
# extract row 1 as a row vector
In [82]: row1 = arr[[0], :]
In [83]: row1.shape
Out[83]: (1, 4)
In [84]: row1
Out[84]: array([[5, 6, 7, 8]])
3 Comments
neurotronix
Thank you for the clean and detailed answer! But haven't you noticed that the question is more than two years old? Cheers :)
kmario23
@neurotronix I just realized that. Anyway, for pedagogical reasons, time never matters ;)
neurotronix
Very true, thanks for taking the time to put up an answer!
np.ones((2,40)). From this array I want to slice a whole row in the form of annp.array((1,40)). The result should be a 2d array