Python 在numpy数组中找到第n个最小元素
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Find nth smallest element in numpy array
提问by Akavall
I need to find just the the smallest nth element in a 1D numpy.array.
我只需要找到 1D 中最小的第 n 个元素numpy.array。
For example:
例如：
a = np.array([90,10,30,40,80,70,20,50,60,0])
I want to get 5th smallest element, so my desired output is 40.
我想获得第 5 个最小的元素，所以我想要的输出是40.
My current solution is this:
我目前的解决方案是这样的：
result = np.max(np.partition(a, 5)[:5])
However, finding 5 smallest elements and then taking the largest one them seems little clumsy to me. Is there a better way to do it? Am I missing a single function that would achieve my goal?
然而，找到 5 个最小的元素然后取最大的一个对我来说似乎并不笨拙。有没有更好的方法来做到这一点？我是否缺少一个可以实现我的目标的功能？
There are questions with similar titles to this one, but I did not see anything that answered my question.
有一些与此问题标题相似的问题，但我没有看到任何可以回答我的问题的问题。
Edit:
编辑：
I should've mentioned it originally, but performance is very important for me; therefore, heapqsolution though nice would not work for me.
我本来应该提到它的，但性能对我来说非常重要；因此，heapq虽然很好的解决方案对我不起作用。
import numpy as np
import heapq
def find_nth_smallest_old_way(a, n):
return np.max(np.partition(a, n)[:n])
# Solution suggested by Jaime and HYRY
def find_nth_smallest_proper_way(a, n):
return np.partition(a, n-1)[n-1]
def find_nth_smallest_heapq(a, n):
return heapq.nsmallest(n, a)[-1]
#
n_iterations = 10000
a = np.arange(1000)
np.random.shuffle(a)
t1 = timeit('find_nth_smallest_old_way(a, 100)', 'from __main__ import find_nth_smallest_old_way, a', number = n_iterations)
print 'time taken using partition old_way: {}'.format(t1)
t2 = timeit('find_nth_smallest_proper_way(a, 100)', 'from __main__ import find_nth_smallest_proper_way, a', number = n_iterations)
print 'time taken using partition proper way: {}'.format(t2)
t3 = timeit('find_nth_smallest_heapq(a, 100)', 'from __main__ import find_nth_smallest_heapq, a', number = n_iterations)
print 'time taken using heapq : {}'.format(t3)
Result:
结果：
time taken using partition old_way: 0.255564928055
time taken using partition proper way: 0.129678010941
time taken using heapq : 7.81094002724
采纳答案by Jaime
Unless I am missing something, what you want to do is:
除非我遗漏了什么，否则你想做的是：
>>> a = np.array([90,10,30,40,80,70,20,50,60,0])
>>> np.partition(a, 4)[4]
40
np.partition(a, k)will place the k-th smallest element of aat a[k], smaller values in a[:k]and larger values in a[k+1:]. The only thing to be aware of is that, because of the 0 indexing, the fifth element is at index 4.
np.partition(a, k)将放置at的k第 -th 个最小元素，将较小的值放入，将较大的值放入。唯一需要注意的是，由于 0 索引，第五个元素位于索引 4。aa[k]a[:k]a[k+1:]
回答by arshajii
You can use heapq.nsmallest:
您可以使用heapq.nsmallest：
>>> import numpy as np
>>> import heapq
>>>
>>> a = np.array([90,10,30,40,80,70,20,50,60,0])
>>> heapq.nsmallest(5, a)[-1]
40
回答by HYRY
you don't need call numpy.max():
你不需要打电话numpy.max()：
def nsmall(a, n):
return np.partition(a, n)[n]
