How do I select a row from a NxM numpy array as an array of size 1xM:
> import numpy
> a = numpy.array([[1,2], [3,4], [5,6]])
> a
array([[1, 2],
[3, 4],
[5, 6]])
> a.shape
(e, 2)
> a[0]
array([1, 2])
> a[0].shape
(2,)
I'd like
a[0].shape == (1,2)
I'm doing this because a library I want to use seems to require this.
If you have something of shape (2,) and you want to add a new axis so that the shape is (1,2), the easiest way is to use np.newaxis:
a = np.array([1,2])
a.shape
#(2,)
b = a[np.newaxis, :]
print b
#array([[1,2]])
b.shape
#(1,2)
If you have something of shape (N,2) and want to slice it with the same dimensionality to get a slice with shape (1,2), then you can use a range of length 1 as your slice instead of one index:
a = np.array([[1,2], [3,4], [5,6]])
a[0:1]
#array([[1, 2]])
a[0:1].shape
#(1,2)
Another trick is that some functions have a keepdims option, for example:
a
#array([[1, 2],
# [3, 4],
# [5, 6]])
a.sum(1)
#array([ 3, 7, 11])
a.sum(1, keepdims=True)
#array([[ 3],
# [ 7],
# [11]])
If you already have it, call .reshape():
>>> a = numpy.array([[1, 2], [3, 4]])
>>> b = a[0]
>>> c = b.reshape((1, -1))
>>> c
array([[1, 2]])
>>> c.shape
(1, 2)
You can also use a range to keep the array two-dimensional in the first place:
>>> b = a[0:1]
>>> b
array([[1, 2]])
>>> b.shape
(1, 2)
Note that all of these will have the same backing store.
