np.arange followed by reshape

Question

What is a more Pythonic way of doing this?

min_odds = np.arange( 1.05, 2.0, 0.01 )
min_odds = min_odds.reshape( len( min_ods ), -1 )

The code creates an ndarray of shape (95,) and converts it to shape (95,1).

Also, why does numpy sometimes create arrays of size (95,) with a blank last dimension?

Answer 1

I often follow arange with a reshape to generate test arrays, e.g.

 np.arange(12).reshape(3,4)

Use -1 to avoid taking len(), e.g.

 np.arange(10).reshape(-1,1).shape # (10, 1)

arange always returns a 1d array. numpy arrays can have any number of dimensions, including 0. Shape is expressed as a tuple. (10,) is just a 1 term tuple. (the , is needed to distinguish it from (10)).

Answer 2

You can slice with np.newaxis (which is just an fancy alias for None) if you'd like:

>>> np.arange( 1.05, 2.0, 0.01 )[:,np.newaxis].shape
(95, 1)

If you prefer what you've got, I'd get rid of the -1 and just use 1 (unless you want your users to have to look up what the -1 is supposed to mean like I just did...).

>>> arr = np.arange( 1.05, 2.0, 0.01 )
>>> arr = arr.reshape(len(arr), 1)
>>> arr.shape
(95, 1)

As for your second question,

"why does numpy sometimes create arrays of size (95,) with a blank last dimension?"

I'm not 100% sure I understand what you're asking. ndarray.shape is a tuple. A tuple with a single element's string representation looks like (something,).

^{Also note the comments below about preferring np.linspace to np.arange in this case.}