Python 3.X中多列表比对：如何判断列表是否匹配多个候选列表之一

你编写的判断语句不符合预期，核心是Python的or运算符逻辑规则导致的：array == ["x","2","3"] or ["1","x","3"] or ["1","2","x"]不会判断array是否等于三个列表中的任意一个，实际运算逻辑如下：

• 首先计算array == ["x","2","3"]的结果，结果为真则整个表达式返回真
• 如果上述结果为假，会直接判断["1","x","3"]的布尔值，非空列表的布尔值恒为真，因此整个表达式永远返回真，和array的实际取值无关。

方案一：补全完整相等判断 #

在每个or两侧都编写完整的相等判断逻辑：

array = ["x","x","x"]

if array == ["x","2","3"] or array == ["1","x","3"] or array == ["1","2","x"]:
    print("hello")

方案二：用成员运算简化代码 #

将所有待匹配的列表存入一个容器，用in关键字判断array是否在容器内，代码更易维护：

array = ["x","x","x"]
match_list = (["x","2","3"], ["1","x","3"], ["1","2","x"])

if array in match_list:
    print("hello")

待匹配的选项越多，该方案的简洁性优势越明显，后续调整匹配规则只需要修改match_list即可。

内容的提问来源于stack exchange，提问作者Paul-Jose Sinclair