Numpy - slicing 2d row or column vector from array

Question

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

Answer 1

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]])

Answer 2

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)

Answer 3

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]])