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返回字典中的三个最大值

dictionary sorting python

Psychotechnopath · 技术社区 · 7 年前

我有以下字典:

'{0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49, 9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408, 16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}'

对于这本字典,我想编写一个函数来返回三具有最高值的键值对(在本例中,键18、19、20)。

我想到了以下几点:

cachedict = nr_of_objects_per_century() #Dictionary mentioned above

def top_3_centuries():
        max_nr_works_list = sorted(cachedict.values())
        top_3_values = []
        for i in range(len(max_nr_works_list)-3, len(max_nr_works_list)):
            top_3_values.append(max_nr_works_list[i])
            print(top_3_values)

这会给我一个要查找的最大值列表。但我该如何从这里开始呢?有没有一种不进行反向查找的方法(这对字典来说很慢,对吗?)我觉得我可以更有效地完成这个任务。

8 回复 | 直到 7 年前

jpp 7 年前

你也可以用 collections.Counter 具有 most_common (内部使用堆队列):

from collections import Counter

dct = {0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49, 
       9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408, 
       16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}

count = Counter(dct)
print(count.most_common(3))  # [(19, 244675), (20, 115878), (18, 111490)]

jpp 7 年前

`heapq.nlargest`

通过使用堆队列,可以避免在此处进行完全排序:

from heapq import nlargest
from operator import itemgetter

dct = {0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49,
       9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408,
       16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}

res = nlargest(3, dct.items(), key=itemgetter(1))

print(res)
# [(19, 244675), (20, 115878), (18, 111490)]

b-fg 7 年前

您可以使用:

a = {0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49,
       9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408,
       16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}

l = sorted(list(a.items()), key=lambda tup: tup[1], reverse=True)[:3]
print(l) # [(19, 244675), (20, 115878), (18, 111490)]

它转换字典 a 放入元组列表中,按 tup[1] ,反转它,得到前3个点击。

Ma0 7 年前

你可以这样做:

dct = {0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49, 9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408, 16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}

res = [next(k for k in dct if dct[k]==v) for v in sorted(dct.values(), reverse=True)[:3]]
print(res)  # -> [19, 20, 18]

分解:

sorted(dct.values(), reverse=True)[:3] ::取字典最大值3。
next(k for k in dct if dct[k]==v) ::返回字典键,其值为上述3个值之一(迭代)。

B. M. 7 年前

通过两个简单步骤:

aux = sorted([(v,k) for (k,v) in dic.items()])
res = [(v,k) for (k,v) in aux[-3:]] 
#[(18, 111490), (20, 115878), (19, 244675)]

比快 nlargest 和 Counter.most_common 在这个例子中。

New2Python 7 年前

这将返回您想要的:

d = {0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49, 9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408, 16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}

print(sorted([(i,j) for i, j in d.items() if j in (sorted(d.values())[-3:])])[-3:])
#[(18, 111490), (19, 244675), (20, 115878)]

Corentin Limier 7 年前

d = {0: 0, 1: 11, 2: 26, 3: 43, 4: 14, 5: 29, 6: 34, 7: 49, 8: 49, 9: 108, 10: 124, 11: 108, 12: 361, 13: 290, 14: 2118, 15: 5408, 16: 43473, 17: 109462, 18: 111490, 19: 244675, 20: 115878, 21: 6960}

d_items_sorted = sorted(d.items(), key=lambda x: x[1], reverse=True)

d_items_sorted[:3]

[(19, 244675), (20, 115878), (18, 111490)]

这是我能得到的最简单的代码,但是对字典进行排序需要花费O(nlogn),并且您应该能够在O(n)中执行相同的操作。

George Bekh-Ivanov 7 年前

你是在寻找最有效的方法还是仅仅是在形式/算法简单性方面的最佳方法?

如果是后者,您应该考虑将字典项作为元组进行排序(您可以使用cachedict.items()获取它们),就像在这个答案中一样。 https://stackoverflow.com/a/613218/10453363

只需按值对元组排序,然后获取最后3个元组(它们是键/值对)