Question

我有一个主表，其中列出了每种产品的费率以及其包装和风险的各自类别。

class ApplicationController < ActionController::Base
  protected
    def after_sign_in_path_for(resource)
      request.env['omniauth.origin'] || stored_location_for(resource) || root_path
    end
end

在第二张表上，我获得了用户选项，用户将能够根据其他产品费率为这些产品创建任意数量的规则。而且这些规则仅适用于特定的包装或风险等级。

因此，对于下面的示例，产品B的产品A率仅适用于基本包装和良好/中等风险的产品，加上5％。所有包装的产品C的税率均为D加10％，仅是为了避免风险。

df = pd.DataFrame({'package': {0: 'basic', 1: 'medium', 2: 'premium', 3:'basic', 4:'medium', 5:'premium'},
   'risk_bin': {0: 'good/mid', 1: 'good/mid', 2: 'good/mid', 3:'bad', 4:'bad',5:'bad'},
   'A': {0:0.012,1:0.022,2:0.032,3:0.05,4:0.06,5:0.07},
   'B': {0:0.013,1:0.023,2:0.033,3:0.051,4:0.061,5:0.071},
   'C': {0:0.014,1:0.024,2:0.034,3:0.052,4:0.062,5:0.072},
   'D': {0:0.015,1:0.025,2:0.035,3:0.053,4:0.063,5:0.073}})
df = df[df.columns[[4,5,0,1,2,3]]]

因为我可以创建用户想要的规则，所以我需要创建一个循环，然后将值相应地传递给定义的关系。

rules = pd.DataFrame({'rule': {0: '1', 1: '2'},
   'product1': {0: 'B', 1: 'C'},
   'relantioship': {0:'=',1:'='},
   'product2': {0:'A',1:'D'},
   'symbol': {0:'+',1:'-'},
   'value': {0:0.05,1:0.10},
   'package':{0:'basic',1:'all'},
   'risk': {0:'good/mid', 1:'bad'}})
 rules = rules[rules.columns[[5,1,3,2,6,7,0,4]]]

执行此操作时，出现以下错误：

df2 = df.reset_index()

rules_nc = rules['rule'].get_values()
nc_cnt = rules_nc.size     

for i in range(nc_cnt):
    if pd.isnull(rules['rule'][i]):
        break
    product_1 = rules['product1'][i]
    product_2 = rules['product2'][i]
    sym = str(rules['symbol'][i])
    val = rules['value'][i]
    pack= rules['package'][i]
    risk = rules['risk'][i]        

if (df2['risk_bin']==risk) & (df2['package']==pack):
        if sym=='+':
            df2[product_1] = df2[product_2] + val
        if sym=='-':
            df2[product_1] = df2[product_2] - val    
else:
     df2[product_1] =  df2[product_1]

这是我期望这组规则的输出。

 The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

能帮我吗？非常感谢！

Answer 1

这是一种可能的解决方案。不像使用apply那样理想，它比循环更快，但不如矢量解快。我在risk中将risk_bin重命名为rules。

import pandas as pd

df = pd.DataFrame({'package': {0: 'basic', 1: 'medium', 2: 'premium', 3:'basic', 4:'medium', 5:'premium'},
   'risk_bin': {0: 'good/mid', 1: 'good/mid', 2: 'good/mid', 3:'bad', 4:'bad',5:'bad'},
   'A': {0:0.012,1:0.022,2:0.032,3:0.05,4:0.06,5:0.07},
   'B': {0:0.013,1:0.023,2:0.033,3:0.051,4:0.061,5:0.071},
   'C': {0:0.014,1:0.024,2:0.034,3:0.052,4:0.062,5:0.072},
   'D': {0:0.015,1:0.025,2:0.035,3:0.053,4:0.063,5:0.073}})
df = df[df.columns[[4,5,0,1,2,3]]]

rules = pd.DataFrame({'rule': {0: '1', 1: '2'},
   'product1': {0: 'B', 1: 'C'},
   'relantioship': {0:'=',1:'='},
   'product2': {0:'A',1:'D'},
   'symbol': {0:'+',1:'-'},
   'value': {0:0.05,1:0.10},
   'package':{0:'basic',1:'all'},
   'risk_bin': {0:'good/mid', 1:'bad'}})
rules = rules[rules.columns[[5,1,3,2,6,7,0,4]]]

def fun(row):
    if row["symbol"] == "+":
        row[row["product1"]] = row[row["product2"]] + row["value"]
    else:
        row[row["product1"]] = row[row["product2"]] - row["value"]
    return row

# here you look for all the rows where rules match with the given columns
df1 = pd.merge(df.reset_index(), rules, on=["package", "risk_bin"])
# here you what a rule for `all` package
df2 = pd.merge(df.reset_index(),
               rules[rules["package"]=='all'].loc[:, rules.columns != "package"],
               on=["risk_bin"])
# now you apply the function to both df
df1 = df1.apply(lambda x: fun(x), axis=1)
df2 = df2.apply(lambda x: fun(x), axis=1)

#select the indices in df1 and df2
bad_idx = df.index.isin(df1["index"].tolist()+df2["index"].tolist())

#concat all together
res = pd.concat([df1[df.columns], df2[df.columns], df[~bad_idx]],ignore_index=True)

使用if迭代规则

1 个答案: