Remove duplicate rows of a numpy array

How can I remove duplicate rows of a 2 dimensional numpy array?

data = np.array([[1,8,3,3,4],
                 [1,8,9,9,4],
                 [1,8,3,3,4]])

The answer should be as follows:

ans = array([[1,8,3,3,4],
             [1,8,9,9,4]])

If there are two rows that are the same, then I would like to remove one "duplicate" row.

You can use numpy unique. Since you want the unique rows, we need to put them into tuples:

import numpy as np

data = np.array([[1,8,3,3,4],
                 [1,8,9,9,4],
                 [1,8,3,3,4]])

just applying np.unique to the data array will result in this:

>>> uniques
array([1, 3, 4, 8, 9])

prints out the unique elements in the list. So putting them into tuples results in:

new_array = [tuple(row) for row in data]
uniques = np.unique(new_array)

which prints:

>>> uniques
array([[1, 8, 3, 3, 4],
       [1, 8, 9, 9, 4]])

UPDATE

In the new version, you need to set np.unique(data, axis=0)

One approach with lex-sorting -

# Perform lex sort and get sorted data
sorted_idx = np.lexsort(data.T)
sorted_data =  data[sorted_idx,:]

# Get unique row mask
row_mask = np.append([True],np.any(np.diff(sorted_data,axis=0),1))

# Get unique rows
out = sorted_data[row_mask]

Sample run -

In [199]: data
Out[199]: 
array([[1, 8, 3, 3, 4],
       [1, 8, 9, 9, 4],
       [1, 8, 3, 3, 4],
       [1, 8, 3, 3, 4],
       [1, 8, 0, 3, 4],
       [1, 8, 9, 9, 4]])

In [200]: sorted_idx = np.lexsort(data.T)
     ...: sorted_data =  data[sorted_idx,:]
     ...: row_mask = np.append([True],np.any(np.diff(sorted_data,axis=0),1))
     ...: out = sorted_data[row_mask]
     ...: 

In [201]: out
Out[201]: 
array([[1, 8, 0, 3, 4],
       [1, 8, 3, 3, 4],
       [1, 8, 9, 9, 4]])

Runtime tests -

This section times all approaches proposed in the solutions presented thus far.

In [34]: data = np.random.randint(0,10,(10000,10))

In [35]: def tuple_based(data):
    ...:     new_array = [tuple(row) for row in data]
    ...:     return np.unique(new_array)
    ...: 
    ...: def lexsort_based(data):                 
    ...:     sorted_data =  data[np.lexsort(data.T),:]
    ...:     row_mask = np.append([True],np.any(np.diff(sorted_data,axis=0),1))
    ...:     return sorted_data[row_mask]
    ...: 
    ...: def unique_based(a):
    ...:     a = np.ascontiguousarray(a)
    ...:     unique_a = np.unique(a.view([('', a.dtype)]*a.shape[1]))
    ...:     return unique_a.view(a.dtype).reshape((unique_a.shape[0], a.shape[1]))
    ...: 

In [36]: %timeit tuple_based(data)
10 loops, best of 3: 63.1 ms per loop

In [37]: %timeit lexsort_based(data)
100 loops, best of 3: 8.92 ms per loop

In [38]: %timeit unique_based(data)
10 loops, best of 3: 29.1 ms per loop

A simple solution can be:

import numpy as np
def unique_rows(a):
	a = np.ascontiguousarray(a)
	unique_a = np.unique(a.view([('', a.dtype)]*a.shape[1]))
	return unique_a.view(a.dtype).reshape((unique_a.shape[0], a.shape[1]))

data = np.array([[1,8,3,3,4],
				 [1,8,9,9,4],
				 [1,8,3,3,4]])


print unique_rows(data)
#prints:
[[1 8 3 3 4]
 [1 8 9 9 4]]

You can check this for many more solutions for this problem