Python 使用“应用于多个条件”的自定义方法

我试图根据一些条件更改一列的值，这些条件在下面的函数中使用apply方法进行了详细说明，但是我无法获得下面的预期输出

lst = [np.nan,["google_newyork","google_NewYork","google_LA"],["google_newyork","google_LA"],["google_newyork","google_newyork"]]
  
df = pd.DataFrame(lst, columns =['subcompanies_matches']) 

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然而我得到了这个错误，级数的真值是模糊的。因此，我想知道对于此类问题，apply是否是合适的函数？

试试这个，您不需要数据帧：

def check_subcompanies(subcompanies_concat):
    """ Returns True if subcompanies match else False """
    if  subcompanies_concat != None: 
        subcompany1 = subcompanies_concat[0].lower()
        subcompany2 = [x.lower() for x in subcompanies_concat[1:]]
        if subcompany1 in subcompany2:
            return True
        else:
            return False
    else subcompanies_concat == None:
        return True
df = df.apply(check_subcompanies)

输出：

lst = [["google_newyork","google_newyork"], None, ["google_newyork","google_NewYork","google_LA"], ["google_newyork","google_LA"]]

def check_subcompanies(subcompanies_concat):
    """ Returns True if subcompanies match else False """
    if  subcompanies_concat: 
        subcompany1 = subcompanies_concat[0].lower()
        subcompany2 = [x.lower() for x in subcompanies_concat[1:]]
        if subcompany1 in subcompany2:
            return True
        else:
            return False
    elif(subcompanies_concat == None):
        return True


list(map(check_subcompanies, lst))

---

dataframe的定义不正确，它返回错误，请用更清晰的方法显示您的df，以及您想从check_Subcompanies中得到什么。对不起，现在我做了相应的更正。请尝试回答，您定义的dataframe对LST没有任何好处谢谢您的回答，然而，结果有时会改变，[真，真，真，假]

lst = [["google_newyork","google_newyork"], None, ["google_newyork","google_NewYork","google_LA"], ["google_newyork","google_LA"]]

def check_subcompanies(subcompanies_concat):
    """ Returns True if subcompanies match else False """
    if  subcompanies_concat: 
        subcompany1 = subcompanies_concat[0].lower()
        subcompany2 = [x.lower() for x in subcompanies_concat[1:]]
        if subcompany1 in subcompany2:
            return True
        else:
            return False
    elif(subcompanies_concat == None):
        return True


list(map(check_subcompanies, lst))

[True, True, True, False]