问题:有效地对numpy数组进行降序排序?
令我惊讶的是,之前没有提出过这个具体问题,但是我真的没有在SO或文档中找到它np.sort
。
假设我有一个包含整数的随机numpy数组,例如:
> temp = np.random.randint(1,10, 10)
> temp
array([2, 4, 7, 4, 2, 2, 7, 6, 4, 4])
如果对它进行排序,则默认情况下我将获得升序:
> np.sort(temp)
array([2, 2, 2, 4, 4, 4, 4, 6, 7, 7])
但我希望解决方案按降序排序。
现在,我知道我可以永远做:
reverse_order = np.sort(temp)[::-1]
但这最后的陈述有效吗?它不是按升序创建副本,然后反转此副本以反转顺序获得结果吗?如果确实如此,是否有有效的选择?看起来好像不np.sort
接受参数来更改排序操作中的比较符号以使事情相反。
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.
回答 0
temp[::-1].sort()
对数组进行排序,然后np.sort(temp)[::-1]
创建一个新数组。
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
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
回答 1
>>> 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])
>>> 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])
回答 2
对于短数组,我建议np.argsort()
通过查找已排序的否定数组的索引来使用,这比反转已排序的数组要快一些:
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
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
回答 3
不幸的是,当您有一个复杂的数组时,只能np.sort(temp)[::-1]
正常工作。这里提到的其他两种方法无效。
Unfortunately when you have a complex array, only np.sort(temp)[::-1]
works properly. The two other methods mentioned here are not effective.
回答 4
注意尺寸。
让
x # initial numpy array
I = np.argsort(x) or I = x.argsort()
y = np.sort(x) or y = x.sort()
z # reverse sorted array
全反转
z = x[-I]
z = -np.sort(-x)
z = np.flip(y)
flip
更改1.15
,需要以前的版本。解决方案:。1.14
axis
pip install --upgrade numpy
第一维反转
z = y[::-1]
z = np.flipud(y)
z = np.flip(y, axis=0)
逆向二维
z = y[::-1, :]
z = np.fliplr(y)
z = np.flip(y, axis=1)
测试中
在100×10×10阵列上测试1000次。
Method | Time (ms)
-------------+----------
y[::-1] | 0.126659 # only in first dimension
-np.sort(-x) | 0.133152
np.flip(y) | 0.121711
x[-I] | 4.611778
x.sort() | 0.024961
x.argsort() | 0.041830
np.flip(x) | 0.002026
这主要是由于重新索引而不是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()], xs)
timeit(lambda x: x.sort(), xs)
timeit(lambda x: x.argsort(), xs)
timeit(lambda x: np.flip(x), xs)
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)
回答 5
您好,我在寻找一种对二维numpy数组进行反向排序的解决方案,但找不到任何有效的方法,但是我想我偶然发现了一个我上载的解决方案,以防万一有人在同一条船上。
x=np.sort(array)
y=np.fliplr(x)
np.sort对升序进行排序,这不是您想要的,但是命令fliplr将行从左向右翻转!似乎可以工作!
希望它可以帮助您!
我猜这与上面关于-np.sort(-a)的建议相似,但是由于评论它并不总是有效而推迟了我的建议。也许我的解决方案也不总是可行,但是我已经用几个阵列对其进行了测试,似乎还可以。
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.
回答 6
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])
回答 7
这是一个快速窍门
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]
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]
回答 8
我建议使用这个…
np.arange(start_index, end_index, intervals)[::-1]
例如:
np.arange(10, 20, 0.5)
np.arange(10, 20, 0.5)[::-1]
然后您的恢复:
[ 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. ]
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. ]