I want to check how many numpy array elements inside numpy array are different. The solution should not contain list comprehension. Something along these lines (note that a and b differ in the last array):
a = np.array( [[1,1,1],[2,2,2],[3,3,3],[4,4,4],[5,5,5]] )
b = np.array( [[1,1,1],[2,2,2],[3,3,3],[4,4,4],[5,0,0]] )
y = diff_count( a,b )
print y
>> 1
Approach #1
Perform element-wise comparison for non-equality and then get ANY reduction along last axis and finally count -
(a!=b).any(-1).sum()
Approach #2
Probably faster one with np.count_nonzero for counting booleans -
np.count_nonzero((a!=b).any(-1))
Approach #3
Much faster one with views -
# https://stackoverflow.com/a/45313353/ @Divakar
def view1D(a, b): # a, b are arrays
a = np.ascontiguousarray(a)
b = np.ascontiguousarray(b)
void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
return a.view(void_dt).ravel(), b.view(void_dt).ravel()
a1D,b1D = view1D(a,b)
out = np.count_nonzero(a1D!=b1D)
In [32]: np.random.seed(0)
...: m,n = 10000,100
...: a = np.random.randint(0,9,(m,n))
...: b = a.copy()
...:
...: # Let's set 10% of rows as different ones
...: b[np.random.choice(len(a), len(a)//10, replace=0)] = 0
In [33]: %timeit (a!=b).any(-1).sum() # app#1 from this soln
...: %timeit np.count_nonzero((a!=b).any(-1)) # app#2
...: %timeit np.any(a - b, axis=1).sum() # @Graipher's soln
1000 loops, best of 3: 1.14 ms per loop
1000 loops, best of 3: 1.08 ms per loop
100 loops, best of 3: 2.33 ms per loop
In [34]: %%timeit # app#3
...: a1D,b1D = view1D(a,b)
...: out = np.count_nonzero((a1D!=b1D).any(-1))
1000 loops, best of 3: 797 µs per loop
You can try it using np.ravel(). If you want element wise comparison.
(a.ravel()!=b.ravel()).sum()
(a-b).any(axis=0).sum()
above lines gives 2 as output.
If you want row wise comparison, you can use.
(a-b).any(axis=1).sum()
This gives 1 as output.
2. Think OP wants to compare on per row basis and not element-wise.Would this work?
y=sum(a[i]!=b[i]for i in range len(a))
Sorry that I can’t test this myself right now.
