Re: [Numpy-discussion] Question regarding concatenate/vstack.

2011-03-30 Thread Ralf Gommers
On Wed, Mar 30, 2011 at 1:42 PM, gary ruben wrote: > You're right, they are not equivalent. vstack will happily create an > array of higher rank than the parts it is stacking, whereas > concatenate requires the arrays it is working with to already be at > least 2d, so the equivalent is > np.concat

Re: [Numpy-discussion] Question regarding concatenate/vstack.

2011-03-30 Thread gary ruben
You're right, they are not equivalent. vstack will happily create an array of higher rank than the parts it is stacking, whereas concatenate requires the arrays it is working with to already be at least 2d, so the equivalent is np.concatenate((np.arange(5.)[newaxis],np.arange(5.)[newaxis]), axis=0)

[Numpy-discussion] Question regarding concatenate/vstack.

2011-03-30 Thread andrew nelson
Dear List, I have a quick question regarding vstack and concatenate. In the docs for vstack it says that: np.concatenate(tup, axis=0) should be equivalent to: np.vstack(tup) However, I tried this out and it doesn't seem to be case, i.e. >>> np.vstack((np.arange(5.), np.arange(5.))) array([[ 0.