python - 结构模式匹配和无穷大

Question

我正在计算非负p的Lp距离函数。对于除 p = 0 和 p = ∞ 之外的所有情况，内置的 pow() 函数都可以很好地发挥作用。在了解结构模式匹配之前，我使用过字典和异常处理:

from math import sqrt, inf
distance_function = {   0.0: lambda x, y: int(x != 0.0) + int(y != 0.0),
                        1.0: lambda x, y: abs(x) + abs(y), # Likely a tad faster than 'pow()'       
                        inf: lambda x, y: max(abs(x), abs(y))}                  
def lp_distance(x, y, p): 
    try:                return distance_function[p](x, y)                   
    except KeyError:    return pow(pow(abs(x), p) + pow(abs(y), p), 1.0/p)

有些人不希望这里有异常(exception)。所以我将代码片段重写为以下代码:

def lp_distance(x, y, p):
    match p:
        case 0.0:           return int(x != 0.0) + int(y != 0.0)
        case 1.0:           return abs(x) + abs(y)      
        # The line below triggers "SyntaxError: name capture 'inf' makes remaining patterns unreachable"
        case inf:           return max(abs(x), abs(y))
        # But the following works:
        case p if p == inf: return max(abs(x), abs(y))
        case _:             return pow(pow(abs(x), p) + pow(abs(y), p), 1.0/p)

为什么 case inf: 不正确(Python v3.10.2)？

Answer 1

在 case声明，a simple name is a pattern that captures (assigns) to that name 。相比之下，a dotted name is a patterns that refers to the value of that name .

In simple terms NAME will always succeed and it will set NAME = <subject>.

In simple terms NAME1.NAME2 will succeed only if <subject> == NAME1.NAME2

仅使用case inf:表示要匹配的值无条件分配给名称 inf – 名称是否先前已绑定(bind)并不重要。
你想要的是 case math.inf: ，表示与该值进行比较。

import math

def lp_distance(x, y, p):
    match p:
        case 0.0:
            return int(x != 0.0) + int(y != 0.0)
        case 1.0:
            return abs(x) + abs(y)      
        # compare against a value by using its dotted name
        case math.inf:
            return max(abs(x), abs(y))
        case _:
            return pow(pow(abs(x), p) + pow(abs(y), p), 1.0/p)