Numpy shape method returns wrong dimensions

Question

Let's say I have two arrays like:

a = array([ 0.36981727,  0.06066488,  0.73031016])

b = array([[ 0.12375904,  0.11647815,  0.56665118],
       [ 0.9421819 ,  0.58797789,  0.26831203],
       [ 0.25769   ,  0.02517343,  0.76701222]])

where each element of a corresponds to one array of b. Now in order to sort 'a' and also keep track of its corresponding vectors in b I do:

ziped_and_sorted = np.array(sorted(zip(a,b), key=operation.itemgetter(0), reverese =True),'object')

then I need to split a and b again, so:

a = ziped_and_sorted[:,0]

In [158]: a
Out[158]: array([0.369817272838, 0.0606648844006, 0.730310164248], dtype=object)

b = ziped_and_sorted[:,1]

In [157]: b
Out[157]:
array([[ 0.12375904  0.11647815  0.56665118],
       [ 0.9421819   0.58797789  0.26831203],
       [ 0.25769     0.02517343  0.76701222]], dtype=object)

The problem is b.shape returns (3,)instead of (3,3). It is important because I need to do a matrix multiplication with b and the problem causes a dimension mismatched error.

P.S: If you have better solution, please suggest it.

Answer 1

This is because b is a ndarray of ndarray, but not a 2-dim ndarray.

You can use numpy.argsort to do this quickly:

import numpy as np
a = np.random.randint(0, 100, 5)
b = np.random.randint(0, 5, (5, 5))
print a
print b
idx = np.argsort(a)[::-1]
print a[idx]
print b[idx]

output is:

[27 65  8 19 32]

[[4 4 1 4 4]
 [1 3 4 3 3]
 [3 4 2 1 0]
 [1 0 1 0 4]
 [1 4 1 1 4]]

[65 32 27 19  8]

[[1 3 4 3 3]
 [1 4 1 1 4]
 [4 4 1 4 4]
 [1 0 1 0 4]
 [3 4 2 1 0]]

if you want to use sorted you can use numpy.vstack to convert a list of array to a 2-dim ndarray:

ziped_and_sorted = sorted(zip(a,b), key=operator.itemgetter(0), reverse=True)
np.vstack([row[1] for row in ziped_and_sorted])