Question

我有一个字典的python列表：

mylist = [
{'id':0, 'weight':10, 'factor':1, 'meta':'ABC'},
{'id':1, 'weight':5, 'factor':1, 'meta':'ABC'},
{'id':2, 'weight':5, 'factor':2, 'meta':'ABC'},
{'id':3, 'weight':1, 'factor':1, 'meta':'ABC'}
]

按重量排序该列表的最有效/最干净的方法是什么（数值）。结果列表应如下所示：

mylist = [
{'id':3, 'weight':1, 'factor':1, 'meta':'ABC'},
{'id':1, 'weight':5, 'factor':1, 'meta':'ABC'},
{'id':2, 'weight':5, 'factor':2, 'meta':'ABC'},
{'id':0, 'weight':10, 'factor':1, 'meta':'ABC'},
]

Answer 1

mylist.sort(key=lambda d: (d['weight'], d['factor']))

或

import operator
mylist.sort(key=operator.itemgetter('weight', 'factor'))

Answer 2

以下内容应该有效：

def cmp_dict(x, y):
    weight_diff = y['weight'] - x['weight']
    if weight_diff == 0:
        return y['factor'] - x['factor']
    else:
        return weight_diff

myList.sort(cmp_dict)

Answer 3

我接受了dF对灵感的回答，但这是我最终确定的方案：

@staticmethod
def ordered_list(mylist):
    def sort_func(d):
        return (d['weight'], d['factor'])

    mylist.sort(key=sort_func)

Answer 4

decoratedlist = [(item[weight], item) for item in mylist]
decoratedlist.sort()
results = [item for (key, item) in decoratedlist]

在python中排序字典列表

4 个答案: