How to test if all rows are equal in a numpy

Question

In numpy, is there a nice idiomatic way of testing if all rows are equal in a 2d array?

I can do something like

np.all([np.array_equal(M[0], M[i]) for i in xrange(1,len(M))]) 

This seems to mix python lists with numpy arrays which is ugly and presumably also slow.

Is there a nicer/neater way?

Answer 1

One way is to check that every row of the array arr is equal to its first row arr[0]:

(arr == arr[0]).all() 

Using equality == is fine for integer values, but if arr contains floating point values you could use np.isclose instead to check for equality within a given tolerance:

np.isclose(a, a[0]).all() 

If your array contains NaN and you want to avoid the tricky NaN != NaN issue, you could combine this approach with np.isnan:

(np.isclose(a, a[0]) | np.isnan(a)).all() 

Answer 2

Simply check if the number if unique items in the array are 1:

>>> arr = np.array([[1]*10 for _ in xrange(5)]) >>> len(np.unique(arr)) == 1 True 

A solution inspired from unutbu's answer:

>>> arr = np.array([[1]*10 for _ in xrange(5)]) >>> np.all(np.all(arr == arr[0,:], axis = 1)) True 

One problem with your code is that you're creating an entire list first before applying np.all() on it. Due to that there's no short-circuiting happening in your version, instead of that it would be better if you use Python's all() with a generator expression:

Timing comparisons:

>>> M = arr = np.array([[3]*100] + [[2]*100 for _ in xrange(1000)]) >>> %timeit np.all(np.all(arr == arr[0,:], axis = 1)) 1000 loops, best of 3: 272 µs per loop >>> %timeit (np.diff(M, axis=0) == 0).all() 1000 loops, best of 3: 596 µs per loop >>> %timeit np.all([np.array_equal(M[0], M[i]) for i in xrange(1,len(M))]) 100 loops, best of 3: 10.6 ms per loop >>> %timeit all(np.array_equal(M[0], M[i]) for i in xrange(1,len(M))) 100000 loops, best of 3: 11.3 µs per loop  >>> M = arr = np.array([[2]*100 for _ in xrange(1000)]) >>> %timeit np.all(np.all(arr == arr[0,:], axis = 1)) 1000 loops, best of 3: 330 µs per loop >>> %timeit (np.diff(M, axis=0) == 0).all() 1000 loops, best of 3: 594 µs per loop >>> %timeit np.all([np.array_equal(M[0], M[i]) for i in xrange(1,len(M))]) 100 loops, best of 3: 9.51 ms per loop >>> %timeit all(np.array_equal(M[0], M[i]) for i in xrange(1,len(M))) 100 loops, best of 3: 9.44 ms per loop