How to check if all values in the columns of a numpy matrix are the same?

Question

I want to check if all values in the columns of a numpy array/matrix are the same. I tried to use reduce of the ufunc equal, but it doesn't seem to work in all cases:

In [55]: a = np.array([[1,1,0],[1,-1,0],[1,0,0],[1,1,0]])  In [56]: a Out[56]:  array([[ 1,  1,  0],        [ 1, -1,  0],        [ 1,  0,  0],        [ 1,  1,  0]])  In [57]: np.equal.reduce(a) Out[57]: array([ True, False,  True], dtype=bool)  In [58]: a = np.array([[1,1,0],[1,0,0],[1,0,0],[1,1,0]])  In [59]: a Out[59]:  array([[1, 1, 0],        [1, 0, 0],        [1, 0, 0],        [1, 1, 0]])  In [60]: np.equal.reduce(a) Out[60]: array([ True,  True,  True], dtype=bool) 

Why does the middle column in the second case also evaluate to True, while it should be False?

Thanks for any help!

Answer 1

In [45]: a Out[45]:  array([[1, 1, 0],        [1, 0, 0],        [1, 0, 0],        [1, 1, 0]]) 

Compare each value to the corresponding value in the first row:

In [46]: a == a[0,:] Out[46]:  array([[ True,  True,  True],        [ True, False,  True],        [ True, False,  True],        [ True,  True,  True]], dtype=bool) 

A column shares a common value if all the values in that column are True:

In [47]: np.all(a == a[0,:], axis = 0) Out[47]: array([ True, False,  True], dtype=bool) 

---

The problem with np.equal.reduce can be seen by micro-analyzing what happens when it is applied to [1, 0, 0, 1]:

In [49]: np.equal.reduce([1, 0, 0, 1]) Out[50]: True 

The first two items, 1 and 0 are tested for equality and the result is False:

In [51]: np.equal.reduce([False, 0, 1]) Out[51]: True 

Now False and 0 are tested for equality and the result is True:

In [52]: np.equal.reduce([True, 1]) Out[52]: True 

But True and 1 are equal, so the total result is True, which is not the desired outcome.

The problem is that reduce tries to accumulate the result "locally", while we want a "global" test like np.all.

Answer 2

Given ubuntu's awesome explanation, you can use reduce to solve your problem, but you have to apply it to bitwise_and and bitwise_or rather than equal. As a consequence, this will not work with floating point arrays:

In [60]: np.bitwise_and.reduce(a) == a[0] Out[60]: array([ True, False,  True], dtype=bool)  In [61]: np.bitwise_and.reduce(b) == b[0] Out[61]: array([ True, False,  True], dtype=bool) 

Basically, you are comparing the bits of each element in the column. Identical bits are unchanged. Different bits are set to zero. This way, any number that has a zero instead of a one bit will change the reduced value. bitwise_and will not trap the case where bits are introduced rather than removed:

In [62]: c = np.array([[1,0,0],[1,0,0],[1,0,0],[1,1,0]])  In [63]: c Out[63]:  array([[1, 0, 0],        [1, 0, 0],        [1, 0, 0],        [1, 1, 0]])  In [64]: np.bitwise_and.reduce(c) == c[0] Out[64]: array([ True,  True,  True], dtype=bool) 

The second coumn is clearly wrong. We need to use bitwise_or to trap new bits:

In [66]: np.bitwise_or.reduce(c) == c[0] Out[66]: array([ True, False,  True], dtype=bool) 

Final Answer

In [69]: np.logical_and(np.bitwise_or.reduce(a) == a[0], np.bitwise_and.reduce(a) == a[0]) Out[69]: array([ True, False,  True], dtype=bool)  In [70]: np.logical_and(np.bitwise_or.reduce(b) == b[0], np.bitwise_and.reduce(b) == b[0]) Out[70]: array([ True, False,  True], dtype=boo  In [71]: np.logical_and(np.bitwise_or.reduce(c) == c[0], np.bitwise_and.reduce(c) == c[0]) Out[71]: array([ True, False,  True], dtype=bool) 

This method is more restrictive and less elegant than ubunut's suggestion of using all, but it has the advantage of not creating enormous temporary arrays if your input is enormous. The temporary arrays should only be as big as the first row of your matrix.

EDIT

Based on this Q/A and the bug I filed with numpy, the solution provided only works because your array contains zeros and ones. As it happens, the bitwise_and.reduce() operations shown can only ever return zero or one because bitwise_and.identity is 1, not -1. I am keeping this answer in the hope that numpy gets fixed and the answer becomes valid.

Edit

Looks like there will in fact be a change to numpy soon. Certainly to bitwise_and.identity, and also possibly an optional parameter to reduce.

Edit

Good news everyone. The identity for np.bitwise_and has been set to -1 as of version 1.12.0.