Python 3.x 如何在一个函数中将名称拆分为名字、中间名和姓氏，并同时分配到列中？

我喜欢使用函数将

df

中的

name

列拆分为名字、中间名和姓氏。但这给了我一个错误：

ValueError: too many values to unpack (expected 3)

我的代码：

df['FIRST_NAME'], df['MIDDLE_NAME'], df['LAST_NAME'] = \
    df.apply(split_name, var='NAME_V2', axis=1)


def split_name(df, var):
    first_name = ''
    middle_name = ''
    last_name = ''
    full_name = df[var]
    name_entity = full_name.split()
    name_entity_length = len(name_entity)

    if name_entity_length == 1:
        last_name = full_name

    elif name_entity_length == 2:
        first_name = name_entity[0]
        last_name = name_entity[-1]

    elif name_entity_length >= 3:
        first_name = name_entity[0]
        middle_name = name_entity[1:-1]
        last_name = name_entity[-1]     

    return (first_name, middle_name, last_name)

数据帧：

NAME_V2      FIRST_NAME    MIDDLE_NAME     LAST_NAME
John Smith     John                          Smith
Smith                                        Smith
J O I Smith     J             O I            Smith

以下是我的方法：

def split_name(df, var):
    sub_df = df[var].str.split('\\s+', expand=True)
    result = []

    for _, row in sub_df.iterrows():
        info = {'FirstName': '', 'MiddleName': '', 'LastName': ''}
        n = row.count()

        if n == 0:
            pass
        elif n == 1:
            info['LastName'] = row.iloc[0]
        elif n == 2:
            info['FirstName'], info['LastName'] = row.iloc[:2]
        else:
            info['FirstName'] = row.iloc[0]
            info['LastName'] = row.iloc[-1]
            info['MiddleName'] = ' '.join([(string or '') for string in row.iloc[1:-1]])
        result.append(info)
    return pd.DataFrame(result, index=df.index)

split_name(df, 'NAME_V2')

结果:

  FirstName MiddleName LastName
0      John               Smith
1                         Smith
2         J        O I    Smith

您可以将其连接到原始数据帧。

以下是我的方法：

def split_name(df, var):
    sub_df = df[var].str.split('\\s+', expand=True)
    result = []

    for _, row in sub_df.iterrows():
        info = {'FirstName': '', 'MiddleName': '', 'LastName': ''}
        n = row.count()

        if n == 0:
            pass
        elif n == 1:
            info['LastName'] = row.iloc[0]
        elif n == 2:
            info['FirstName'], info['LastName'] = row.iloc[:2]
        else:
            info['FirstName'] = row.iloc[0]
            info['LastName'] = row.iloc[-1]
            info['MiddleName'] = ' '.join([(string or '') for string in row.iloc[1:-1]])
        result.append(info)
    return pd.DataFrame(result, index=df.index)

split_name(df, 'NAME_V2')

df.NAME_V2.str.extractall(r"\b(\w*)\s*(.*)\s*\b(\w+$)").fillna("").rename({0:"First_Name",1:"Middle_Name",2:"Last_Name"},axis=1)
Out[17]: 
        First_Name Middle_Name Last_Name
  match                                 
0 0           John                 Smith
1 0                                Smith
2 0              J         O I     Smith

结果:

  FirstName MiddleName LastName
0      John               Smith
1                         Smith
2         J        O I    Smith

---

您可以将其连接到原始数据帧。

您可以发布演示数据帧和预期输出吗？

apply

返回一个包含3个元素的元组序列；它在LHS上需要一件事：删除元组。仅以a、b、c的形式返回，看看它是否有效。您可以发布演示数据帧和预期输出吗？

apply

返回一个包含3个元素的元组序列；它在LHS上需要一件事：删除元组。只将其作为a、b、c返回，看看是否有效

df.NAME_V2.str.extractall(r"\b(\w*)\s*(.*)\s*\b(\w+$)").fillna("").rename({0:"First_Name",1:"Middle_Name",2:"Last_Name"},axis=1)
Out[17]: 
        First_Name Middle_Name Last_Name
  match                                 
0 0           John                 Smith
1 0                                Smith
2 0              J         O I     Smith