Python Match Case (Switch) 性能

Question

我期待 Python match/case有平等的时间访问每个案例，但似乎我错了。有什么好的解释为什么吗？
让我们使用以下示例:

def match_case(decimal):
    match decimal:
      case '0':
        return "000"
      case '1':
        return "001"
      case '2':
        return "010"
      case '3':
        return "011"
      case '4':
        return "100"
      case '5':
        return "101"
      case '6':
        return "110"
      case '7':
        return "111"
      case _:
        return "NA"

并定义一个快速工具来测量时间:

import time
def measure_time(funcion):
    def measured_function(*args, **kwargs):
        init = time.time()
        c = funcion(*args, **kwargs)
        print(f"Input: {args[1]} Time: {time.time() - init}")
        return c
    return measured_function

@measure_time
def repeat(function, input):
    return [function(input) for i in range(10000000)]

如果我们运行每个 10000000每种情况下的次数如下:

for i in range(8):
    repeat(match_case, str(i))

# Input: 0 Time: 2.458001136779785
# Input: 1 Time: 2.36093807220459
# Input: 2 Time: 2.6832823753356934
# Input: 3 Time: 2.9995620250701904
# Input: 4 Time: 3.5054492950439453
# Input: 5 Time: 3.815168857574463
# Input: 6 Time: 4.164452791213989
# Input: 7 Time: 4.857251167297363

只是想知道为什么访问时间不同。这不是用查找表优化的吗？。请注意，我对具有相等访问时间的其他方式(即使用字典)不感兴趣。

Answer 1

PEP 622
开发了“match\case”功能来替换这样的代码:

def is_tuple(node):
if isinstance(node, Node) and node.children == [LParen(), RParen()]:
    return True
return (isinstance(node, Node)
        and len(node.children) == 3
        and isinstance(node.children[0], Leaf)
        and isinstance(node.children[1], Node)
        and isinstance(node.children[2], Leaf)
        and node.children[0].value == "("
        and node.children[2].value == ")")

用这样的代码:

def is_tuple(node: Node) -> bool:
match node:
    case Node(children=[LParen(), RParen()]):
        return True
    case Node(children=[Leaf(value="("), Node(), Leaf(value=")")]):
        return True
    case _:
        return False

虽然在最原始的情况下它可能等效于 dict 查找，但通常并非如此。案例模式被设计成看起来像普通的 Python 代码，但实际上它们隐藏了 isinsance和 len当您看到类似 Node() 的代码时，不要执行您期望执行的操作。 .

Essentially this is equivalent to a chain of if ... elif ... else statements. Note that unlike for the previously proposed switch statement, the pre-computed dispatch dictionary semantics does not apply here.