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)
new_array = [tuple(row) for row in data] uniques = np.unique(new_array) but it still output uniques array([1, 3, 4, 8, 9]) @ThePredatorimport numpy as np data = np.array([[1,8,3,3,4], [1,8,9,9,4], [1,8,3,3,4]]) new_array = [tuple(row) for row in data] uniques = np.unique(new_array) uniques Out[30]: array([1, 3, 4, 8, 9]) Is that anything about the numpy version? my numpy version is 1.9.2
np.unique(data, axis=0)
lexsort solution is still the fastest presented here (at least for this example).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
unique_based is about twice as fast as np.unique(data, axis=0), so lexsort is still preferable in 2022.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
np.uniqueover an axis, sonp.unique(data, axis = 0)works.