有人能帮我选择两个键之间的最小值吗?例如,如果我有字典列表:
results = [
{
"model": "short",
"score": 34,
"alt_score": 1
},
{
"model": "med",
"score": 22,
"alt_score": 11
},
{
"model": "tall",
"score": 42,
"alt_score": 90
},
{
"model": "xtall",
"score": 83,
"alt_score": 15
},
]
我想选择具有最小的score
或{score
或{
min(results, key=lambda x:x['alt_score'])
但我不知道怎么同时看两把钥匙。我需要这样的东西:
min(results, key=lambda x:x['score', 'alt_score])
或者
min(results, key=lambda x:x['score'] or x:x['alt_score'])
结果应返回:
^{pr2}$提前谢谢!在
min(results, key=lambda x:min(x['score'], x['alt_score']))
x需要为内部min()比较引用每个分数。在
Lambdas中几乎可以有任何表达式,包括对}。在
min()
的内部调用,以获得该项的较小值,score
或{你可以用这个:
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