Extend different shapes/dimension matrices in python

Question

I want to make two matrices same dimension/shape padding with zeros..

for e.g. I have

>>> x array([[ 1., -1., 1.], [ 1., 1., - 1.]]) >>> >>> >>> y array([[ 2., 2.], [ 2., 2.], [ -2., 2.]])

Output I want

x1 = array([[ 1., -1., 1.], [ 1., 1., -1.], [ 0., 0., 0.]]) y1 =array([[ 2., 2., 0.], [ 2., 2., 0.], [ -2., 2., 0.]])

any help? I looked up the "pad" but the numpy version we are using is older so it does not have pad. ( numpy ver. 1.6.x)

I also looked up some solutions but they are specific to shape, in my case the shape is dynamic -- and the operation needs to be fast -- as I do this over a large matrices and many times

I would like to ... but it affects many other projects -- otherwise I have to get special exception to go solo for my project.. and q. will be raised why cant we do it in this version? — Sendi_t
– Sendi_t, Commented May 26, 2015 at 16:16
This is a common problem - you should read up on Conda environments and/or virtualenv which are designed for exactly this situation. — YXD
– YXD, Commented May 26, 2015 at 16:21

YXD · Accepted Answer · 2015-05-26 16:05:44Z

1

Every array has a shape:

>>> x = np.array([[ 1., -1., 1.], ... [ 1., 1., - 1.]]) >>> y = np.array([[ 2., 2.], ... [ 2., 2.], ... [ -2., 2.]]) >>> x.shape (2, 3) >>> y.shape (3, 2)

We just need to compute the "maximum shape" and use that for the new arrays:

>>> shape = np.maximum(x.shape, y.shape) >>> x1 = np.zeros(shape)

We can then copy the data from the original arrays into the corresponding parts of the new arrays:

>>> x1[:x.shape[0], :x.shape[1]] = x >>> y1 = np.zeros(shape) >>> y1[:y.shape[0], :y.shape[1]] = y

Results

>>> x1 array([[ 1., -1., 1.], [ 1., 1., -1.], [ 0., 0., 0.]]) >>> y1 array([[ 2., 2., 0.], [ 2., 2., 0.], [-2., 2., 0.]]) >>>

answered May 26, 2015 at 16:05

YXD

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6 Comments

Sendi_t Over a year ago

golden! , just quick q. here we are creating an empty array of zeros for x1, y1 of maximum shape, and than updating it --- I have matrix of 2k x 1k ( approximately ) since rows and column varies a little bit.. That means I will be holding 4 matrix at a time in memory.. do you think it will light enough? or should I loop?

YXD Over a year ago

I doubt you will run out of memory as each of those arrays will be ~16MB. Also the code y1[:y.shape[0], :y.shape[1]] = y will also just copy data into the previously allocated/zeroed memory rather than reallocating memory.

Sendi_t Over a year ago

Thanks! ~16MB ? as they are all (same) zeros (floats), wouldn't it optimize memory space? and only updated matrix will be actual size? is there any other lib that can do that?

hpaulj Over a year ago

This is basically what pad does, but with more options - create the destination array of the correct size, and copy old onto the new.

Sendi_t Over a year ago

Thank you hpaulj -- I need to create habit of looking into python/lib implementations.. :)

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