Efficiently sorting a numpy array in descending order?
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Chapters
00:00 Efficiently Sorting A Numpy Array In Descending Order?
00:59 Accepted Answer Score 220
01:21 Answer 2 Score 21
01:39 Answer 3 Score 181
01:48 Answer 4 Score 13
02:49 Thank you
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Full question
https://stackoverflow.com/questions/2698...
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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
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Tags
#python #arrays #sorting #numpy
#avk47
    Hire the world's top talent on demand or became one of them at Toptal: https://topt.al/25cXVn
--------------------------------------------------
Music by Eric Matyas
https://www.soundimage.org
Track title: Puzzle Game 5 Looping
--
Chapters
00:00 Efficiently Sorting A Numpy Array In Descending Order?
00:59 Accepted Answer Score 220
01:21 Answer 2 Score 21
01:39 Answer 3 Score 181
01:48 Answer 4 Score 13
02:49 Thank you
--
Full question
https://stackoverflow.com/questions/2698...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #arrays #sorting #numpy
#avk47
ACCEPTED ANSWER
Score 220
temp[::-1].sort() sorts the array in place, whereas np.sort(temp)[::-1] creates a new array.
In [25]: temp = np.random.randint(1,10, 10)
In [26]: temp
Out[26]: array([5, 2, 7, 4, 4, 2, 8, 6, 4, 4])
In [27]: id(temp)
Out[27]: 139962713524944
In [28]: temp[::-1].sort()
In [29]: temp
Out[29]: array([8, 7, 6, 5, 4, 4, 4, 4, 2, 2])
In [30]: id(temp)
Out[30]: 139962713524944
ANSWER 2
Score 181
>>> a=np.array([5, 2, 7, 4, 4, 2, 8, 6, 4, 4])
>>> np.sort(a)
array([2, 2, 4, 4, 4, 4, 5, 6, 7, 8])
>>> -np.sort(-a)
array([8, 7, 6, 5, 4, 4, 4, 4, 2, 2])
ANSWER 3
Score 21
For short arrays I suggest using np.argsort() by finding the indices of the sorted negatived array, which is slightly faster than reversing the sorted array:
In [37]: temp = np.random.randint(1,10, 10)
In [38]: %timeit np.sort(temp)[::-1]
100000 loops, best of 3: 4.65 µs per loop
In [39]: %timeit temp[np.argsort(-temp)]
100000 loops, best of 3: 3.91 µs per loop
ANSWER 4
Score 13
Be careful with dimensions.
Let
x  # initial numpy array
I = np.argsort(x) or I = x.argsort() 
y = np.sort(x)    or y = x.sort()
z  # reverse sorted array
Full Reverse
z = x[I[::-1]]
z = -np.sort(-x)
z = np.flip(y)
First Dimension Reversed
z = y[::-1]
z = np.flipud(y)
z = np.flip(y, axis=0)
Second Dimension Reversed
z = y[::-1, :]
z = np.fliplr(y)
z = np.flip(y, axis=1)
Testing
Testing on a 100×10×10 array 1000 times.
Method       | Time (ms)
-------------+----------
y[::-1]      | 0.126659  # only in first dimension
-np.sort(-x) | 0.133152
np.flip(y)   | 0.121711
x[I[::-1]]   | 4.611778
x.sort()     | 0.024961
x.argsort()  | 0.041830
np.flip(x)   | 0.002026
This is mainly due to reindexing rather than argsort.
# Timing code
import time
import numpy as np
def timeit(fun, xs):
    t = time.time()
    for i in range(len(xs)):  # inline and map gave much worse results for x[-I], 5*t
        fun(xs[i])
    t = time.time() - t
    print(np.round(t,6))
I, N = 1000, (100, 10, 10)
xs = np.random.rand(I,*N)
timeit(lambda x: np.sort(x)[::-1], xs)
timeit(lambda x: -np.sort(-x), xs)
timeit(lambda x: np.flip(x.sort()), xs)
timeit(lambda x: x[x.argsort()[::-1]], xs)
timeit(lambda x: x.sort(), xs)
timeit(lambda x: x.argsort(), xs)
timeit(lambda x: np.flip(x), xs)