首页 > 解决方案 > 对对象字典进行排名的最佳方法是什么?

问题描述

给定以下类:

class Comparison():

    def __init__(self, value):
        self.value = value
        self.rank = 0

class Comparisons(dict):

    def __init__(self):
        super(Comparisons, self).__init__()

    def rank(self):
        # Method to examine each Comparison instance and
        # assign Comparison.rank based on Comparison.value

rank()该方法检查对象并分配等级的有效方法是什么?例如:

comparisons = Comparisons()

# store some Comparison instances
comparisons['one'] = Comparison(10)
comparisons['two'] = Comparison(5)
comparisons['three'] = Comparison(1)

# function to rank the comparisons
comparisons.rank()

print(comparisons['one'].rank)
print(comparisons['two'].rank)
print(comparisons['three'].rank)

回报:

3
2
1

如果该rank()方法可以处理关系,那将更加有益。

标签: pythonsortingrank

解决方案


这看起来像最直观(天真?)的方式:

def rank(self):
    sorted_comparisons = sorted(self.values(), key=lambda c: c.value)
    for rank, comparison in enumerate(sorted_comparisons, 1):
        comparison.rank = rank

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