'NoneType' 对象不可订阅 df.apply（lambda 行：（行）[0]，轴=1）

Question

我链接两个table

从 variety_id 到 variety_name

但是当 variety_id 为 null

时我的代码出错

'NoneType' object is not subscriptable

我链接的代码table

        if not df.empty:
   
         df['commodity_name'] = df.apply(lambda row: commodity_name(row)[0], axis=1)
            df['state_name'] = df.apply(lambda row: state_name(row)[0], axis=1)
            df['variety_name'] = df.apply(lambda row: variety_name(row)[0], axis=1)


def commodity_name(self):
    if self.commodity_id:
        return get_commodity_name(self.commodity_id)
    return None


def state_name(self):
    if self.state_id:
        return get_region_name(self.state_id)
    return None


def variety_name(self):
    if self.variety_id:
        return get_variety_name(self.variety_id)
    return None

还编写了映射列的查询

def get_variety_name(variety_id):
    """
    This function is used to return state_name by querying the database based on lgd_state_id
    """
    query = "SELECT commodity_variety_name FROM itrade.commodity_variety_master WHERE commodity_variety_id={}".format(
        variety_id)
    query_result = get_column_value(query)
    if not query_result.empty:
        return list(get_column_value(query)['commodity_variety_name'])
    return None


def get_commodity_name(commodity_id=None):
    """
    This function is used to return commodity_name by querying the database based on commodity_id
    """
    query = "SELECT commodity_id, commodity_name FROM itrade.commodity_master"
    if commodity_id:
        query += " WHERE commodity_id={}".format(commodity_id)
        return list(get_column_value(query)['commodity_name'])
    return get_column_value(query).to_dict(orient='records')

DB table 在最后一行，variety_id 是 none 如果我在其中输入值，那么代码工作正常，就会出现错误

 Modify id  commodity_id    state_id    variety_id  season_id   attribute_value attribute_name  attribute_unit  status  updated_at  created_at
 edit   1   1   1   1   1   10  base_temp   C   1   2021-12-08 05:32:38.565360  2021-11-15 13:53:27.286340
 edit   2   2   2   2   2   32  rainfall    mm  0   2021-11-15 13:53:27.286340  2021-11-15 13:53:27.286340
 edit   3   2   1   5   1   10  temperature Farenhite   1   2021-12-06 08:19:27.754601  2021-12-06 06:58:52.527409
 edit   4   2   1   4   12  12  temperature celcius 1   2021-12-06 07:20:36.682156  2021-12-06 07:20:36.682194
 edit   5   2   35  5   111 111 temperature Farenhite   1   2021-12-06 08:21:01.312057  2021-12-06 08:20:38.318877
 edit   6   2   35  4   2345    1123    temperature (C) celcius 1   2021-12-06 11:36:09.478035  2021-12-06 11:36:01.101701
 edit   7   1   1   2   2   10  base_temp   C   1   2021-12-17 11:08:49.000000  2021-12-17 11:08:49.000000
 edit   8   31  9   60  1   10  base_temp   C   1   2021-12-18 09:26:42.000000  2021-12-18 09:26:42.000000
 edit   9   84  9   1   2   15  base_temp   C   1   2021-12-20 07:48:34.000000  2021-12-20 07:48:34.000000
 edit   10  43  9   NULL    111 1111    temperature (F) celcius 1   2021-12-21 12:14:08.384168  2021-12-21 12:14:08.384215


        if not df.empty:
            df['commodity_name'] = df.apply(lambda row: row.commodity_name()[0], axis=1)
            df['state_name'] = df.apply(lambda row: row.state_name()[0], axis=1)
            df['variety_name'] = df.apply(lambda row: row.variety_name()[0], axis=1)

        response_data = df.to_dict(orient='records')
        print(response_data)
        if column and column in list(df.columns):
            response_data = sorted(response_data, key=lambda i: i[column], reverse=order == 'desc')

        if page:
            return self.get_paginated_response(self.paginate_queryset(response_data))

        return Response(data=response_data, status=status.HTTP_200_OK)


def commodity_name(self, row):
    if row.commodity_id:
        return get_commodity_name(row.commodity_id)
    return None


def state_name(self, row):
    if row.state_id:
        return get_region_name(row.state_id)
    return None


def variety_name(self, row):
    if row.variety_id:
        return get_variety_name(row.variety_id)
    return None

Answer 1

我得到了答案感谢帮助

'NoneType' 对象不可订阅 df.apply（lambda 行：（行）[0]，轴=1）

'NoneType' object is not subscriptable df.apply(lambda row: (row)[0], axis=1)

python

django

integer

nonetype