Numpy: Reshaping this array in the least number of operations

Question

I have an array with 112 lines and 40 columns.

The format I need to convert to is 40 sets of 56 points each with x, y.

So, the first line has the x coordinates of the first point in each set. The second line has the x of the second points in the set... until the 56th line. After that I have the y's.

1st line : 40 x's  
2nd line: 40 x's  
...  
56th line: 40 x's  
57th line: 40 y's  
...  
112th line: 40 y's

Initially I thought about doing data.reshape(40, 56, 2) but that doesn't work because the values for x come before the values for y. If instead I had one line with x's and another with y's that would work though.

Edit:

for i in xrange(len(data)/2):
    points.append(data[i])
    points.append(data[i+len(data)/2])
points = np.array(points).T.reshape(len(data[0]), len(data)/2, 2)
return points

Steve Tjoa · Accepted Answer · 2011-01-27 05:34:41Z

4

Just one idea:

[[(data[i,j], data[i+56,j]) for i in range(56)] for j in range(40)]

Returns a list of list of tuples.

EDIT: Your edit clarifies what you want. If you want pure Numpy, then does this work?

data.reshape(2, 56, 40).swapaxes(0,2)

edited Jan 27, 2011 at 5:34

answered Jan 27, 2011 at 5:00

Steve Tjoa

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4 Comments

pnodbnda Over a year ago

yeah that's what i was looking for. thanks. could you explain to me why you used 0 and 2 though?

Steve Tjoa Over a year ago

Trial and error. :-) reshape confuses me, especially since Matlab traverses column-wise by default, while Numpy traverses row-wise by default. As a sanity check, create numpy.arange(112*40).reshape(112,40). Then reshape it, and check the elements to see if they match the desired output.

pnodbnda Over a year ago

i got it. it has 3 dimensions, so when you swap you swap 0 and 2 you swap 2 and 40, meaning you get an array 40 x 56 x 2

Steve Tjoa Over a year ago

Yes, that part I understand. But if you did data.reshape(40, 2, 56).swapaxes(1,2), that would be wrong because the data is not reordered correctly, regardless of the fact that the dimensions are correct.

doug · Accepted Answer · 2011-01-27 06:09:38Z

I'll use a smaller array (8 x 5) so we can view the returned values easily.

import numpy as NP

# just create a smaller array to work with:
A = NP.random.randint(0, 10, 40).reshape(8, 5)

# split A in half, to separate x and y 
p, q = NP.vsplit(A, 2)

# create a 'template' array of the correct dimension
xy = NP.zeros(2, 4, 5)

# now just map the x and y values onto the template
xy[0:,:] = p
xy[1:,:] = q


# the transformed matrix:
array([[[ 8.,  5.,  2.,  5.,  7.],
        [ 2.,  6.,  0.,  7.,  2.],
        [ 4.,  4.,  7.,  5.,  5.],
        [ 8.,  5.,  2.,  0.,  5.]],

       [[ 4.,  8.,  6.,  9.,  2.],
        [ 2.,  6.,  5.,  8.,  1.],
        [ 3.,  2.,  6.,  2.,  2.],
        [ 1.,  8.,  0.,  7.,  3.]]])

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