python - 比较一个列表中的所有项目以相同顺序出现在另一个列表中

Question

在下面的代码中，如何只返回那些值为 True 的值，其中检查列表中的元素顺序在路由列表中遵循？

例如:checklist(["BOMH","DELH","BRGH"]) 应该返回 True ["BOMH","CCUH","DELH","BRGH"] 但不是 ["BOMH","BRGH","CCUH","DELH"]。

def checklhcost2(checklist):
    thcdataselect["shortlist"] = thcdataselect.apply(lambda x:\
                True if all((i in x["routelist"] for i in checklist)) else False)
    return thcdataselect.filter_by(1,"shortlist")

Answer 1

使用映射确定每个值的权重，过滤掉不匹配的那些，并检查其排序是否正确:

master = ["BOMH","DELH","BRGH"]
ordering = {item:idx for idx,item in enumerate(master)}
check1 = ["BOMH","CCUH","DELH","BRGH"]
check2 = ["BOMH","BRGH","CCUH","DELH"]

def check(o, c):
    return all(o[i]==r for i,r in zip((i for i in c if i in o), range(len(o))))

结果:

>>> check(ordering, check1)
True
>>> check(ordering, check2)
False