Better way to pack numpy array?

Question

I need to pack a numpy 2D array with struct.pack, and I'm seeking a method which can do this in a batch. I tried:

X = numpy.array([[1,2,3],[4,5,6]])
b = struct.pack('=%sf' % X.size, *X)

but this doesn't work. It prompts:

struct.error: pack expected 6 items for packing (got 2)

Is there a better way to pack the NumPy array instead of looping over each element?

Answer 1

If you have a numpy array, you already have your data packed in memory, you don't need to use struct:

>>> a = np.arange(1, 7)
>>> struct.pack('=6f', *a)
'\x00\x00\x80?\x00\x00\x00@\x00\x00@@\x00\x00\x80@\x00\x00\xa0@\x00\x00\xc0@'

>>> a.astype('f').tostring()
'\x00\x00\x80?\x00\x00\x00@\x00\x00@@\x00\x00\x80@\x00\x00\xa0@\x00\x00\xc0@'

And if your array is multidimensional, .tostring takes a flattened view by default:

>>> a = np.arange(1, 7).reshape(2, 3)
>>> a.astype('f').tostring()
'\x00\x00\x80?\x00\x00\x00@\x00\x00@@\x00\x00\x80@\x00\x00\xa0@\x00\x00\xc0@'

Answer 2

Seemed the '*' operator combined with struct.pack only works with 1D array. So for ND array, it has to be flattened to 1D firstly.

X = numpy.array([[1,2,3],[4,5,6]])
b = struct.pack('=%sf' % X.size, *X.flatten('F'))

This works for me.

Answer 3

The * acts like an iterator. Iterating on an 2d array returns the rows of that array. For example:

In [9]: def foo(*args):
    print(args)    
In [10]: foo(*np.ones((2,3)))
(array([ 1.,  1.,  1.]), array([ 1.,  1.,  1.]))

args, inside the function, is 2 tuples, each a row of the array. That is why your pack got 2 items, rather than 6 (x.size). When you flatten the array, * produces all 6 items.

In [11]: foo(*np.ones((2,3)).flat)
(1.0, 1.0, 1.0, 1.0, 1.0, 1.0)

But what's the purpose of this pack?