Question

有人可以帮我选择两把钥匙之间的最小值吗？例如，如果我有词典列表：

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score OR alt_score最小的字典。我知道如何找到具有最小score或alt_score的字典：

min(results, key=lambda x:x['alt_score'])

但我不确定如何一次看两个键。我需要类似的东西：

min(results, key=lambda x:x['score', 'alt_score])

或

min(results, key=lambda x:x['score'] or x:x['alt_score'])

结果应该返回：

{
  "model": "short",
  "score": 34,
  "alt_score": 1
}

提前致谢！

Answer 1

min(results, key=lambda x:min(x['score'], x['alt_score']))

Lambdas可以在其中包含几乎任何表达式，包括对min()的内部调用，以获取项目score或alt_score中的较小者。

Answer 2

你可以用这个：

min(results, key=lambda x: min(x['score'], x['alt_score']))

Answer 3

min(results, key=lambda x:min(x['score'], x['alt_score']))

x需要引用内部min（）比较的每个分数。

Answer 4

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
  }, 
]

print(min(results, key=lambda x:min(x['score'],x['alt_score'])))

在python中选择两个最小键

4 个答案: