将熊猫数据框转换为具有多个键的字典

Posted 2023-03-12

【中文标题】将熊猫数据框转换为具有多个键的字典

【英文标题】：convert pandas dataframe to dictionary with multiple keys

【发布时间】：2019-02-11 00:21:51

【问题描述】：

我正在尝试将数据框转换为具有四个键的字典，这些键都来自列。我还有多个列，我想使用从这四列构建的键返回值。我使用循环的方式工作，但最终出现内存错误。我很好奇有没有更有效的方法呢？

数据框如下所示：

    Service Bill Weight Zone    Resi    UPS FedEx   USPS    DHL
    1DEA           1       2    N      33.02    9999    9999    9999
    1DEA           2       2    N      33.02    9999    9999    9999
    1DEA           3       2    N      33.02    9999    9999    9999

我希望每个运营商都有一个这样的密钥：

    price[('1DEA', '1', '2', 'N', 'UPS')]=33.02
    price[('1DEA', '1', '2', 'N', 'FedEx')]=9999

我试过这个：

    price = 
    carriers = ['UPS', 'FedEx', 'USPS','DHL'] 
    for carrier in carriers:
        for row in rate_keys.to_dict('records'):
              key = (row['Service'], row['Bill Weight'], row['Zone'], 
              row['Resi'], carrier)
              rate_keys[key] = row[carrier]

【参考方案1】：

IIUC，具有这样的列表理解：

carriers = ['UPS', 'FedEx', 'USPS','DHL']
price = (row['Service'], row['Bill Weight'], row['Zone'], row['Resi'], c):row[c]
     for c in carriers for _, row in df.iterrows()

[输出]

('1DEA', 1, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 2, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 3, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 1, 2, 'N', 'FedEx'): 9999,
 ('1DEA', 2, 2, 'N', 'FedEx'): 9999,
 ('1DEA', 3, 2, 'N', 'FedEx'): 9999,
 ('1DEA', 1, 2, 'N', 'USPS'): 9999,
 ('1DEA', 2, 2, 'N', 'USPS'): 9999,
 ('1DEA', 3, 2, 'N', 'USPS'): 9999,
 ('1DEA', 1, 2, 'N', 'DHL'): 9999,
 ('1DEA', 2, 2, 'N', 'DHL'): 9999,
 ('1DEA', 3, 2, 'N', 'DHL'): 9999

【参考方案2】：

如果你这样做

df = df.set_index(['Service', 'Bill','Weight','Zone'])

你基本上有同样的东西

输出

print(df.loc[('1DEA', 1, 2, 'N')]['UPS'])

9999.0

【参考方案3】：

您可能不应该在循环时更新rate_keys。我猜你的示例脚本的最后一行应该是

price[key] = row[carrier]

【参考方案4】：

首先，

temp = df.set_index(['Service', 'Bill', 'Weight', 'Zone']).to_dict()

然后，我们进行字典推导以获得所需的输出，

dict(((k+(i,)), a[i][k]) for i in temp for (k) in temp[i] )

【参考方案5】：

将索引设置为除载体列之外的所有索引，然后堆叠。

df.set_index(['Service', 'Bill Weight', 'Zone', 'Resi']).stack().to_dict()

('1DEA', 1, 2, 'N', 'DHL'): 9999.0,
 ('1DEA', 1, 2, 'N', 'FedEx'): 9999.0,
 ('1DEA', 1, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 1, 2, 'N', 'USPS'): 9999.0,
 ('1DEA', 2, 2, 'N', 'DHL'): 9999.0,
 ('1DEA', 2, 2, 'N', 'FedEx'): 9999.0,
 ('1DEA', 2, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 2, 2, 'N', 'USPS'): 9999.0,
 ('1DEA', 3, 2, 'N', 'DHL'): 9999.0,
 ('1DEA', 3, 2, 'N', 'FedEx'): 9999.0,
 ('1DEA', 3, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 3, 2, 'N', 'USPS'): 9999.0

---

理解

(*r[:4], c): v for r in df.values for c, v in zip(df.columns[4:], r[4:])

('1DEA', 1, 2, 'N', 'DHL'): 9999,
 ('1DEA', 1, 2, 'N', 'FedEx'): 9999,
 ('1DEA', 1, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 1, 2, 'N', 'USPS'): 9999,
 ('1DEA', 2, 2, 'N', 'DHL'): 9999,
 ('1DEA', 2, 2, 'N', 'FedEx'): 9999,
 ('1DEA', 2, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 2, 2, 'N', 'USPS'): 9999,
 ('1DEA', 3, 2, 'N', 'DHL'): 9999,
 ('1DEA', 3, 2, 'N', 'FedEx'): 9999,
 ('1DEA', 3, 2, 'N', 'UPS'): 33.02,
 ('1DEA', 3, 2, 'N', 'USPS'): 9999