将pandas数据帧转换为具有多个键的字典
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我试图将数据框转换为具有四个键的字典,这四个键都来自列。我还有多个列,我想使用从这四列构建的键返回值。我在循环方式上工作但最终导致内存错误。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
我希望为每个运营商提供一个密钥,如下所示:
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]
答案
您可能不应该在循环时更新rate_keys
。我想你的示例脚本的最后一行应该是读
price[key] = row[carrier]
另一答案
将索引设置为除载波列之外的所有索引,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}
另一答案
IIUC,列表理解如下:
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}
另一答案
如果你这样做
df = df.set_index(['Service', 'Bill','Weight','Zone'])
你基本上有同样的事情
output
print(df.loc[('1DEA', 1, 2, 'N')]['UPS'])
9999.0
另一答案
首先,
temp = df.set_index(['Service', 'Bill', 'Weight', 'Zone']).to_dict()
然后,我们进行字典理解以获得所需的输出,
dict(((k+(i,)), a[i][k]) for i in temp for (k) in temp[i] )
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