Shape of array python

Question

Suppose I create a 2 dimensional array

 m = np.random.normal(0, 1, size=(1000, 2))
 q = np.zeros(shape=(1000,1))
 print m[:,0] -q

When I take m[:,0].shape I get (1000,) as opposed to (1000,1) which is what I want. How do I coerce m[:,0] to a (1000,1) array?

I simply want a one column vector with 1000 rows as I am trying to subtract two column vectors — lord12
– lord12, Commented Mar 27, 2013 at 20:16
@MattDMo: they're sometimes interchangeable because of how numpy broadcasting works, but they're not quite the same thing. — DSM
– DSM, Commented Mar 27, 2013 at 20:20
You can see that if you just enter m[:,0] (prepare for a bit of scrolling, though :) — MattDMo
– MattDMo, Commented Mar 27, 2013 at 20:21
@MattDMo: well, the two objects have different dimensionality, for one. For a good broadcasting explanation you can read this tutorial. — DSM
– DSM, Commented Mar 27, 2013 at 20:27

DSM · Accepted Answer · 2013-03-27 20:18:44Z

8

By selecting the 0th column in particular, as you've noticed, you reduce the dimensionality:

>>> m = np.random.normal(0, 1, size=(5, 2))
>>> m[:,0].shape
(5,)

You have a lot of options to get a 5x1 object back out. You can index using a list, rather than an integer:

>>> m[:, [0]].shape
(5, 1)

You can ask for "all the columns up to but not including 1":

>>> m[:,:1].shape
(5, 1)

Or you can use None (or np.newaxis), which is a general trick to extend the dimensions:

>>> m[:,0,None].shape
(5, 1)
>>> m[:,0][:,None].shape
(5, 1)
>>> m[:,0, None, None].shape
(5, 1, 1)

Finally, you can reshape:

>>> m[:,0].reshape(5,1).shape
(5, 1)

but I'd use one of the other methods for a case like this.

answered Mar 27, 2013 at 20:18

DSM

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2 Comments

seberg Over a year ago

Just to note, using m[:,[0]] is subtly different since it triggers fancy indexing, which means the result is not a view.

DSM Over a year ago

@seberg: true enough. I'm fortunate in that I seldom have to worry, either on memory or performance fronts, and so I've probably grown a little sloppy..

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