Pandas DataFrame按条件创建分类列：新列全为NA，如何解决？

场景描述 #

现有Pandas DataFrame包含「Vehicle Class」列，需根据自定义分类规则新增「Vehicle Type」列，分类规则如下：

Two_W_Personal = ['MOPED', 'M-CYCLE/SCOOTER', 'M-CYCLE/SCOOTER-WITH SIDE CAR', 'MOTORISED CYCLE (CC > 25CC)']
Two_W_Shared = ['MOTOR CYCLE/SCOOTER-USED FOR HIRE']
Three_W_Personal = ['THREE WHEELER (PERSONAL)']
Three_W_Shared = ['THREE WHEELER (PASSENGER)']
Three_W_Shared_LowSpeed = ['E-RICKSHAW(P)']
Three_W_Goods = ['THREE WHEELER (GOODS)']
Three_W_Goods_LowSpeed = ['E-RICKSHAW WITH CART (G)']
Four_W_Personal = ['MOTOR CAR']
Four_W_Shared = ['MOTOR CAB', 'LUXURY CAB', 'LUXURY CAB']
Bus = ['BUS', 'OMNI BUS', 'OMNI BUS (PRIVATE USE)']
Institution_Bus = ['EDUCATIONAL INSTITUTION BUS']

尝试用以下循环代码赋值，但运行后「Vehicle Type」列全部填充为NA：

for vehicle in master_df['Vehicle Class']:
    if vehicle in Two_W_Personal:
        master_df['Vehicle Type'] = '2W_Personal'
    elif vehicle in Two_W_Shared:
        master_df['Vehicle Type'] = '2W_Shared'
    elif vehicle in Three_W_Personal:
        master_df['Vehicle Type'] = '3W_Personal'
    elif vehicle in Three_W_Shared:
        master_df['Vehicle Type'] = '3W_Shared'
    elif vehicle in Three_W_Shared_LowSpeed:
        master_df['Vehicle Type'] = '3W_Shared_LowSpeed'
    elif vehicle in Three_W_Goods:
        master_df['Vehicle Type'] = '3W_Goods'
    elif vehicle in Three_W_Goods_LowSpeed:
        master_df['Vehicle Type'] = '3W_Goods_LowSpeed'
    elif vehicle in Four_W_Personal:
        master_df['Vehicle Type'] = '4W_Personal'
    elif vehicle in Four_W_Shared:
        master_df['Vehicle Type'] = '4W_Shared'
    elif vehicle in Bus:
        master_df['Vehicle Type'] = 'Bus'
    elif vehicle in Institution_Bus:
        master_df['Vehicle Type'] = 'Institution Bus'
    else: 
        master_df['Vehicle Type'] = 'NA'

错误原因 #

循环逻辑存在核心问题：每次判断后直接给整个「Vehicle Type」列赋值，而非对应行。遍历到最后一个vehicle时，若该值不在任何分类列表中，就会将整列覆盖为NA，导致之前所有赋值结果失效。此外，逐行遍历DataFrame是低效操作，不符合Pandas的向量式设计理念。

解决方法 #

方法1：修复循环逻辑（不推荐，仅作原理演示） #

通过索引定位到具体行赋值，避免覆盖整列：

# 先初始化新列为默认值'NA'
master_df['Vehicle Type'] = 'NA'

# 遍历索引与对应值
for idx, vehicle in enumerate(master_df['Vehicle Class']):
    if vehicle in Two_W_Personal:
        master_df.loc[idx, 'Vehicle Type'] = '2W_Personal'
    elif vehicle in Two_W_Shared:
        master_df.loc[idx, 'Vehicle Type'] = '2W_Shared'
    elif vehicle in Three_W_Personal:
        master_df.loc[idx, 'Vehicle Type'] = '3W_Personal'
    elif vehicle in Three_W_Shared:
        master_df.loc[idx, 'Vehicle Type'] = '3W_Shared'
    elif vehicle in Three_W_Shared_LowSpeed:
        master_df.loc[idx, 'Vehicle Type'] = '3W_Shared_LowSpeed'
    elif vehicle in Three_W_Goods:
        master_df.loc[idx, 'Vehicle Type'] = '3W_Goods'
    elif vehicle in Three_W_Goods_LowSpeed:
        master_df.loc[idx, 'Vehicle Type'] = '3W_Goods_LowSpeed'
    elif vehicle in Four_W_Personal:
        master_df.loc[idx, 'Vehicle Type'] = '4W_Personal'
    elif vehicle in Four_W_Shared:
        master_df.loc[idx, 'Vehicle Type'] = '4W_Shared'
    elif vehicle in Bus:
        master_df.loc[idx, 'Vehicle Type'] = 'Bus'
    elif vehicle in Institution_Bus:
        master_df.loc[idx, 'Vehicle Type'] = 'Institution Bus'

方法2：Pandas向量式映射（推荐，高效简洁） #

构建分类映射字典，用map方法一次性完成赋值：

# 构建映射字典
vehicle_map = {}
# 批量添加各类别映射
vehicle_map.update({v: '2W_Personal' for v in Two_W_Personal})
vehicle_map.update({v: '2W_Shared' for v in Two_W_Shared})
vehicle_map.update({v: '3W_Personal' for v in Three_W_Personal})
vehicle_map.update({v: '3W_Shared' for v in Three_W_Shared})
vehicle_map.update({v: '3W_Shared_LowSpeed' for v in Three_W_Shared_LowSpeed})
vehicle_map.update({v: '3W_Goods' for v in Three_W_Goods})
vehicle_map.update({v: '3W_Goods_LowSpeed' for v in Three_W_Goods_LowSpeed})
vehicle_map.update({v: '4W_Personal' for v in Four_W_Personal})
vehicle_map.update({v: '4W_Shared' for v in Four_W_Shared})
vehicle_map.update({v: 'Bus' for v in Bus})
vehicle_map.update({v: 'Institution Bus' for v in Institution_Bus})

# 生成新列，未匹配项填充为'NA'
master_df['Vehicle Type'] = master_df['Vehicle Class'].map(vehicle_map).fillna('NA')

方法3：使用replace方法实现 #

replace同样支持字典批量替换，效果与map一致：

master_df['Vehicle Type'] = master_df['Vehicle Class'].replace(vehicle_map).fillna('NA')

内容的提问来源于stack exchange，提问作者Ankit Bhatt