Ert*_*oğa 8 python sorting dictionary list
如何按自定义顺序对其进行排序?
输入:
[
{value: "typeA"},
{value: "typeC"},
{value: "typeB"},
{value: "typeC"},
{value: "typeB"},
{value: "typeA"}
]
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期待结果:
[
{value: "typeB"},
{value: "typeB"},
{value: "typeC"},
{value: "typeC"},
{value: "typeA"},
{value: "typeA"}
]
my_own_order = ['typeB', 'typeC', 'typeA']
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我的python代码现在如下:
result = sorted(input, key=lambda v:v['value'])
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fal*_*tru 17
>>> lst = [
... {'value': "typeA"},
... {'value': "typeC"},
... {'value': "typeB"},
... {'value': "typeC"},
... {'value': "typeB"},
... {'value': "typeA"}
... ]
>>> my_own_order = ['typeB', 'typeC', 'typeA']
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使之间的映射typeB
,typeC
,typeA
0,1,2
>>> order = {key: i for i, key in enumerate(my_own_order)}
>>> order
{'typeA': 2, 'typeC': 1, 'typeB': 0}
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并使用映射进行排序键:
>>> sorted(lst, key=lambda d: order[d['value']])
[{'value': 'typeB'},
{'value': 'typeB'},
{'value': 'typeC'},
{'value': 'typeC'},
{'value': 'typeA'},
{'value': 'typeA'}]
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试试这个:
sorted(input, key=lambda v: my_own_order.index(v['value']))
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