How can I count repetitive array in numpy?

Unless speed is really an issue, the simplest approach is to map the sub arrays to tuples and use a Counter dict:

X = [[1,2], [5,1], [1,2], [2,-1] , [5,1]]

from collections import Counter

cn = Counter(map(tuple, X))
print(cn)
print(list(filter(lambda x:x[1] > 1,cn.items())))
Counter({(1, 2): 2, (5, 1): 2, (2, -1): 1})
((1, 2), 2), ((5, 1), 2)]

If you consider [1, 2]equal to [2, 1] then you could use a frozenset Counter(map(frozenset, X)

Take a look at numpy.unique: http://docs.scipy.org/doc/numpy-1.10.1/reference/generated/numpy.unique.html

You can use the return_counts argument for getting the count of each item:

values, counts = numpy.unique(X, return_counts = True)
repeated = values[counts > 1]

Assuming I understand what you want:

Try to count each item in your list into a dictionary dict then select from dict items that its count > 1

The following code might help you:

freq = dict()

for item in x:
    if tuple(item) not in x:
        freq[tuple(item)] = 1
    else:
        freq[tuple(item)] += 1

print {k:v for(k,v) in freq.items() if v > 1}

That code will give you the output:

{(1, 2): 2}