On 3/22/07, Stefan van der Walt wrote:

On Thu, Mar 22, 2007 at 08:13:22PM -0400, Brian Blais wrote:

...

Hello,

I'd like to concatenate a couple of 1D arrays to make it a 2D array, with two columns (one for each of the original 1D arrays). I thought this would work:

In [47]:a=arange(0,10,1)

In [48]:a Out[48]:array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [49]:b=arange(-10,0,1)

In [51]:b Out[51]:array([-10, -9, -8, -7, -6, -5, -4, -3, -2, -1])

In [54]:concatenate((a,b)) Out[54]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1])

In [55]:concatenate((a,b),axis=1) Out[55]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1])

but it never expands the dimensions. Do I have to do this...

In [65]:concatenate((a.reshape(10,1),b.reshape(10,1)),axis=1) Out[65]: array([[ 0, -10], [ 1, -9], [ 2, -8], [ 3, -7], [ 4, -6], [ 5, -5], [ 6, -4], [ 7, -3], [ 8, -2], [ 9, -1]])

?

I thought there would be an easier way. Did I overlook something?

How about

N.vstack((a,b)).T

Also mentioned here should be the use of newaxis. As in a[:,newaxis] However I never got a "good feel" for how to use it, so I can't complete the code you would need. -Sebastian Haase

On 3/23/07, Eric Firing wrote:

Sebastian Haase wrote:

...

On 3/22/07, Stefan van der Walt wrote:

...

On Thu, Mar 22, 2007 at 08:13:22PM -0400, Brian Blais wrote:

...

Hello,

I'd like to concatenate a couple of 1D arrays to make it a 2D array, with two columns (one for each of the original 1D arrays). I thought this would work:

In [47]:a=arange(0,10,1)

In [48]:a Out[48]:array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [49]:b=arange(-10,0,1)

In [51]:b Out[51]:array([-10, -9, -8, -7, -6, -5, -4, -3, -2, -1])

In [54]:concatenate((a,b)) Out[54]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1])

In [55]:concatenate((a,b),axis=1) Out[55]: array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1])

but it never expands the dimensions. Do I have to do this...

In [65]:concatenate((a.reshape(10,1),b.reshape(10,1)),axis=1) Out[65]: array([[ 0, -10], [ 1, -9], [ 2, -8], [ 3, -7], [ 4, -6], [ 5, -5], [ 6, -4], [ 7, -3], [ 8, -2], [ 9, -1]])

?

I thought there would be an easier way. Did I overlook something? How about

N.vstack((a,b)).T

Also mentioned here should be the use of newaxis. As in a[:,newaxis]

However I never got a "good feel" for how to use it, so I can't complete the code you would need.

n [9]:c = N.concatenate((a[:,N.newaxis], b[:,N.newaxis]), axis=1)

In [10]:c Out[10]: array([[ 0, -10], [ 1, -9], [ 2, -8], [ 3, -7], [ 4, -6], [ 5, -5], [ 6, -4], [ 7, -3], [ 8, -2], [ 9, -1]])

Then of course, there's r_ and c_: c = numpy.c_[a,b] c = numpy.r_[a[None],b[None]].T --bb

On 3/24/07, Sebastian Haase wrote:

...

Then of course, there's r_ and c_:

c = numpy.c_[a,b]

c = numpy.r_[a[None],b[None]].T

--bb

So, None is the same as newaxis - right?

Yes, newaxis is None. None is newaxis. Same thing. I just don't see much advantage in spelling it numpy.newaxis, since it's pretty common and not likely to ever change.

But what is a[None] vs. a[:,N.newaxis] ?

a[None] is the same as a[None,:]. It prepends the new axis, so a shape of (5,) becomes (1,5), a "row vector" a[:,None] puts the new axis after the first axis, so shape of (5,) becomes (5,1) a "column vector" a[None,:,None] puts a new axis both before and after, so (5,) becomes (1,5,1). If 'a' is higher dimensional, the same kind of thing goes. Wherever None/newaxis appears in the index, insert a new axis there in the result. Say A is shape (2,3,4), then A[:,None,:,None] is shape (2,1,3,1,4). (Same as A[:,None,:,None,:] since unspecified trailing indices are always taken to be ':') --bb

On Fri, Mar 23, 2007 at 11:09:03AM -0400, Robert Pyle wrote:

...

In [65]:concatenate((a.reshape(10,1),b.reshape(10,1)),axis=1) Out[65]: array([[ 0, -10], [ 1, -9], [ 2, -8], [ 3, -7], [ 4, -6], [ 5, -5], [ 6, -4], [ 7, -3], [ 8, -2], [ 9, -1]])

?

I thought there would be an easier way. Did I overlook something?

What's wrong with zip? Or did *I* miss the point? (I'm just getting the hang of numpy.)

If you use 'zip' you don't make use of numpy's fast array mechanisms. I attach some code you can run as a benchmark. From my ipython session: In [1]: run vsbench.py In [2]: timeit using_vstack(x,y) 1000 loops, best of 3: 997 µs per loop In [3]: timeit using_zip(x,y) 10 loops, best of 3: 503 ms per loop In [4]: timeit using_custom_iteration(x,y) 1000 loops, best of 3: 1.64 ms per loop Cheers Stéfan