安全解包空元组数组 [英] Safe unpack empty tuple array

问题描述

行 import re;print(re.findall("(.*) (.*)", "john smith")) 输出[("john", "smith")]，可以是像 [(first_name, last_name)] = re.findall(...) 一样解压.但是，如果出现不匹配(findall 返回 [])，则此解包会引发 ValueError: not enough values to unpack (expected 1, got 0).

The line import re; print(re.findall("(.*) (.*)", "john smith")) outputs [("john", "smith")], which can be unpacked like [(first_name, last_name)] = re.findall(...). However, in the event of a non-match (findall returning []) this unpacking throws ValueError: not enough values to unpack (expected 1, got 0).

安全解包这个元组数组的正确方法是什么，它可以在匹配 ([("john", "smith")]) 和不匹配 ([]) 场景?

What is the correct way to safely unpack this array of tuples, which would work in both match ([("john", "smith")]) and non-match ([]) scenarios?

推荐答案

一般的答案是在你飞跃之前先看看:

The generic answer is to look before you leap:

if result:
    [(first_name, last_name)] = result

或请求原谅:

try:
    [(first_name, last_name)] = result
except ValueError:
    pass

但是您实际上通过使用 re.findall() 来查找单个结果使事情变得过于复杂.使用 re.seach() 并提取您的匹配组:

but you are actually overcomplicating things by using re.findall() to find a single result. Use re.seach() and extract your matched groups:

match = re.search("(.*) (.*)", value)
if match:
    firstname, lastname = match.groups()

或

try:
    firstname, lastname = re.search("(.*) (.*)", value).groups()
except AttributeError:
    # An attribute error is raised when `re.search()` returned None
    pass

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