I want to assert that two Python dictionaries are equal (that means: equal amount of keys, and each mapping from key to value is equal; order is not important). A simple way
Consider this code
>>> import numpy as np
>>> np.identity(5)
array([[ 1., 0., 0., 0., 0.],
[ 0., 1., 0., 0., 0.],
[ 0., 0., 1., 0., 0.],
[ 0., 0., 0., 1., 0.],
[ 0., 0., 0., 0., 1.]])
>>> np.identity(5)+np.ones([5,5])
array([[ 2., 1., 1., 1., 1.],
[ 1., 2., 1., 1., 1.],
[ 1., 1., 2., 1., 1.],
[ 1., 1., 1., 2., 1.],
[ 1., 1., 1., 1., 2.]])
>>> np.identity(5) == np.identity(5)+np.ones([5,5])
array([[False, False, False, False, False],
[False, False, False, False, False],
[False, False, False, False, False],
[False, False, False, False, False],
[False, False, False, False, False]], dtype=bool)
>>>
Note the the result of the comparison is a matrix, not a boolean value. Dict comparisons will compare values using the values cmp methods, which means that when comparing matrix values, the dict comparison will get a composite result. What you want to do is use numpy.all to collapse the composite array result into a scalar boolean result
>>> np.all(np.identity(5) == np.identity(5)+np.ones([5,5]))
False
>>> np.all(np.identity(5) == np.identity(5))
True
>>>
You would need to write your own function to compare these dictionaries, testing value types to see if they are matricies, and then comparing using numpy.all, otherwise using ==. Of course, you can always get fancy and start subclassing dict and overloading cmp if you want too.