Efficiently sorting a numpy array in descending order?

Question:

I am surprised this specific question hasn’t been asked before, but I really didn’t find it on SO nor on the documentation of np.sort.

Say I have a random numpy array holding integers, e.g:

> temp = np.random.randint(1,10, 10)    
> temp
array([2, 4, 7, 4, 2, 2, 7, 6, 4, 4])

If I sort it, I get ascending order by default:

> np.sort(temp)
array([2, 2, 2, 4, 4, 4, 4, 6, 7, 7])

but I want the solution to be sorted in descending order.

Now, I know I can always do:

reverse_order = np.sort(temp)[::-1]

but is this last statement efficient? Doesn’t it create a copy in ascending order, and then reverses this copy to get the result in reversed order? If this is indeed the case, is there an efficient alternative? It doesn’t look like np.sort accepts parameters to change the sign of the comparisons in the sort operation to get things in reverse order.

Answers:

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
Answered By: Padraic Cunningham

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
Answered By: Mazdak

Unfortunately when you have a complex array, only np.sort(temp)[::-1] works properly. The two other methods mentioned here are not effective.

Answered By: anishtain4
>>> 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])
Answered By: Mike O'Connor

Hello I was searching for a solution to reverse sorting a two dimensional numpy array, and I couldn’t find anything that worked, but I think I have stumbled on a solution which I am uploading just in case anyone is in the same boat.

x=np.sort(array)
y=np.fliplr(x)

np.sort sorts ascending which is not what you want, but the command fliplr flips the rows left to right! Seems to work!

Hope it helps you out!

I guess it’s similar to the suggest about -np.sort(-a) above but I was put off going for that by comment that it doesn’t always work. Perhaps my solution won’t always work either however I have tested it with a few arrays and seems to be OK.

Answered By: Naz

i suggest using this …

np.arange(start_index, end_index, intervals)[::-1]

for example:

np.arange(10, 20, 0.5)
np.arange(10, 20, 0.5)[::-1]

Then your resault:

[ 19.5,  19. ,  18.5,  18. ,  17.5,  17. ,  16.5,  16. ,  15.5,
    15. ,  14.5,  14. ,  13.5,  13. ,  12.5,  12. ,  11.5,  11. ,
    10.5,  10. ]
Answered By: morteza omidipoor

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)
  • flip changed in 1.15, previous versions 1.14 required axis. Solution: pip install --upgrade numpy.

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)
Answered By: A. West

You could sort the array first (Ascending by default) and then apply np.flip()
(https://docs.scipy.org/doc/numpy/reference/generated/numpy.flip.html)

FYI It works with datetime objects as well.

Example:

    x = np.array([2,3,1,0]) 
    x_sort_asc=np.sort(x) 
    print(x_sort_asc)

    >>> array([0, 1, 2, 3])

    x_sort_desc=np.flip(x_sort_asc) 
    print(x_sort_desc)

    >>> array([3,2,1,0])
Answered By: maleckicoa

Here is a quick trick

In[3]: import numpy as np
In[4]: temp = np.random.randint(1,10, 10)
In[5]: temp
Out[5]: array([5, 4, 2, 9, 2, 3, 4, 7, 5, 8])

In[6]: sorted = np.sort(temp)
In[7]: rsorted = list(reversed(sorted))
In[8]: sorted
Out[8]: array([2, 2, 3, 4, 4, 5, 5, 7, 8, 9])

In[9]: rsorted
Out[9]: [9, 8, 7, 5, 5, 4, 4, 3, 2, 2]
Answered By: Don Coder

np.flip() and reversed indexed are basically the same. Below is a benchmark using three different methods. It seems np.flip() is slightly faster. Using negation is slower because it is used twice so reversing the array is faster than that.

** Note that np.flip() is faster than np.fliplr() according to my tests.

def sort_reverse(x):
    return np.sort(x)[::-1]

def sort_negative(x):
    return -np.sort(-x)

def sort_flip(x):
    return np.flip(np.sort(x)) 

arr=np.random.randint(1,10000,size=(1,100000))

%timeit sort_reverse(arr)
%timeit sort_negative(arr)
%timeit sort_flip(arr)

and the results are:

6.61 ms ± 67.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
6.69 ms ± 64.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
6.57 ms ± 58.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Answered By: Mehdi
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