双向插值的替代方法

Question

我写了一些代码来根据两个标准执行插值，即保险金额和免赔额百分比。我一直在努力一次完成插值，所以拆分 filtering.The table hf 包含我用来作为插值结果基础的已知数据 on.Table df 包含新数据需要基于 hf 插值的已开发因子。

现在我的解决方法是首先根据 ded_amount 百分比过滤每个 table，然后执行插值到一个空数据框中并在每个循环后附加。

我觉得这样效率低下，有更好的方法来执行此操作，希望听到一些关于我可以做出的改进的反馈。谢谢

下面提供测试数据。

import pandas as pd
from scipy import interpolate

known_data={'AOI':[80000,100000,150000,200000,300000,80000,100000,150000,200000,300000],'Ded_amount':['2%','2%','2%','2%','2%','3%','3%','3%','3%','3%'],'factor':[0.797,0.774,0.739,0.733,0.719,0.745,0.737,0.715,0.711,0.709]}
new_data={'AOI':[85000,120000,130000,250000,310000,85000,120000,130000,250000,310000],'Ded_amount':['2%','2%','2%','2%','2%','3%','3%','3%','3%','3%']}

hf=pd.DataFrame(known_data)
df=pd.DataFrame(new_data)

deduct_fact=pd.DataFrame()
for deduct in hf['Ded_amount'].unique():
    deduct_table=hf[hf['Ded_amount']==deduct]
    aoi_table=df[df['Ded_amount']==deduct]
    x=deduct_table['AOI']
    y=deduct_table['factor']
    f=interpolate.interp1d(x,y,fill_value="extrapolate")
    xnew=aoi_table[['AOI']]
    ynew=f(xnew)
    append_frame=aoi_table
    append_frame['Factor']=ynew
    deduct_fact=deduct_fact.append(append_frame)

Answer 1

是的，有一种方法可以更有效地执行此操作，而无需制作一堆中间数据帧并附加它们。看看这段代码：

from scipy import interpolate
known_data={'AOI':[80000,100000,150000,200000,300000,80000,100000,150000,200000,300000],'Ded_amount':['2%','2%','2%','2%','2%','3%','3%','3%','3%','3%'],'factor':[0.797,0.774,0.739,0.733,0.719,0.745,0.737,0.715,0.711,0.709]}
new_data={'AOI':[85000,120000,130000,250000,310000,85000,120000,130000,250000,310000],'Ded_amount':['2%','2%','2%','2%','2%','3%','3%','3%','3%','3%']}

hf=pd.DataFrame(known_data)
df=pd.DataFrame(new_data)

# Create this column now
df['Factor'] = None

# I like specifying this explicitly; easier to debug
deduction_amounts = list(hf.Ded_amount.unique())
for deduction_amount in deduction_amounts:
    # You can index a dataframe and call a column in one line
    x, y = hf[hf['Ded_amount']==deduction_amount]['AOI'], hf[hf['Ded_amount']==deduction_amount]['factor']

    f = interpolate.interp1d(x, y, fill_value="extrapolate")

    # This is the most important bit. Lambda function on the dataframe
    df['Factor'] = df.apply(lambda x: f(x['AOI']) if x['Ded_amount']==deduction_amount else x['Factor'], axis=1)

lambda 函数的工作方式是：它逐行遍历 'Factor' 列，并根据其他列的条件为其赋值。

就是returnsdf的AOI列的插值（这就是你说的xnew）如果扣除量匹配，否则就是returns一样的东西回来

双向插值的替代方法

Alternative method for two way interpolation

python

interpolation

scipy