本文介绍了如何返回对应于字典中最小值的键列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
找到字典值的最小值 然后返回并提取该值所在的键。 。
让我说一本关于水果总数的字典:
Let say I have a dictionary of total of fruits:
Fruits = {"apple":8, "banana":3, "lemon":5, "pineapple":2,}我希望输出为
["pineapple"]菠萝的价值最小。或者,如果我有这个:
because pineapple has the least value. Or if I have this:
Colour = {"blue":5, "green":2, "purple":6, "red":2}输出将是:
["green","red"]因为绿色和红色都具有
那么如何返回字典中的最小值?
So how do I return the smallest value in dictionary?
推荐答案可以通过两次通过:
>>> colour {'blue': 5, 'purple': 6, 'green': 2, 'red': 2} >>> min_val = min(colour.itervalues()) >>> [k for k, v in colour.iteritems() if v == min_val] ['green', 'red']
另一种方法(需要一些导入,如果需要,您可以选择n个)-该代码只需要第一个(这将是最小值):
An alternative (requires some imports, and means you could take the n many if wanted) - this code just takes the first though (which would be the min value):
from itertools import groupby from operator import itemgetter ordered = sorted(colour.iteritems(), key=itemgetter(1)) bykey = groupby(ordered, key=itemgetter(1)) print map(itemgetter(0), next(bykey)[1]) # ['green', 'red']更多推荐
如何返回对应于字典中最小值的键列表
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