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在 python 中按原始顺序查找列表的前 k 个最大项

[英]find top k largest item of a list in original order in python

say I have the following list:说我有以下列表:

my_list = [3.5, 1.6, 2.4, 8.9, 5.6]

I want to find the top 3 largest number in its original place, so the result should be:我想在原来的地方找到前3个最大的数,所以结果应该是:

[3.5, 8.9, 5.6]

How could I do that?我怎么能那样做? I think I can find the 3 largest number and use a filter, but I think it may not be a good idea to compare floats.我想我可以找到最大的 3 个数字并使用过滤器,但我认为比较浮点数可能不是一个好主意。 Any suggestions?有什么建议么?

Use a heap :使用

>>> import heapq
>>> heapq.nlargest(3, my_list)
[8.9, 5.6, 3.5]

Add a little polish to the same idea, to keep them in original order:为相同的想法添加一点润色,以保持它们的原始顺序:

>>> from operator import itemgetter
>>> i_val = heapq.nlargest(3, enumerate(my_list), key=itemgetter(1))
>>> [val for (i, val) in sorted(i_val)]
[3.5, 8.9, 5.6]

You can sort the index-value pairs (generated by enumerate ) by the value, get the last three pairs, and then sort those by the index (and then get just the values from the index-value pairs, all this in a one-liner list comprehension):您可以按值对索引值对(由enumerate生成)进行排序,获取最后三对,然后按索引对它们进行排序(然后仅从索引值对中获取值,所有这些都在一个-班轮列表理解):

from operator import itemgetter

my_list = [3.5, 1.6, 2.4, 5.6, 8.9]

result = [p[1] for p in sorted(sorted(enumerate(my_list), key = itemgetter(1))[-3:], key = itemgetter(0))]

print(result)

Output: Output:

[3.5, 5.6, 8.9]

Here you go, you can try with this function:在这里你go,你可以试试这个function:

my_list = [3.5, 1.6, 2.4, 5.6, 8.9]
def select_top(a,array):
    new_list = []
    extra_list = []
    for i in range(len(my_list)):
        extra_list.append(my_list[i])
    final_list = []
    for i in range(a):
        new_list.append(extra_list.index(max(extra_list)))
        extra_list.pop(extra_list.index(max(extra_list)))
    new_list = sorted(new_list,reverse=False)
    for i in new_list:
        final_list.append(array[i])
    return final_list
print(select_top(3,my_list))

I believe it is far from optimal, but you can tweak it as much as you want to get the k top numbers and have them returned in their original order.我相信它远非最佳,但您可以尽可能多地调整它以获得前 k 个数字并让它们按原始顺序返回。 Output: Output:

[3.5, 5.6, 8.9]

How about this?这个怎么样?

[m for m in my_list if m in sorted(my_list)[-3:]]

You're building a new list of 'm' items, from the top 3 items.您正在从前 3 个项目中构建一个新的“m”项目列表。 The list comprehension keeps your items in order.列表理解使您的项目井井有条。

The order of your example is such that you could just sort it and take the 3 top items, but I think you mean that you might have the top 3 items NOT in order, like:您的示例的顺序是这样的,您可以对其进行排序并获取前 3 个项目,但我认为您的意思是您可能没有按顺序排列前 3 个项目,例如:

my_list = [3.5, 1.2, 0.3, 7.8, 3.3]

which results in这导致

[3.5,7.8,3.3]

Here's my own answer, it works for the problem.这是我自己的答案,它可以解决问题。

my_list = [3.5, 1.6, 2.4, 8.9, 5.6]
top_k = sorted(my_list)[-3:]
result = list(filter(lambda x : x in top_k,my_list))
print(result)

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