如何让python中的函数在两个列表(每个列表都有自己的x,y)之间进行插值?

#Age Age = ["18","20","25","30","35","40","45","50","55","59"] #Insurance coverage Coverage = ["50000","100000","150000","200000"] #Premiums Data Cover50k_Annual_Premium = [473.05,501.90,588.45,698.05,842.25,1032.65,1298.05,1696.10,2376.85,3380.65] Cover100k_Annual_Premium = [946.10,1003.80,1176.90,1396.10,1684.55,2065.30,2596.05,3392.15,4753.65,6761.25] Cover150k_Annual_Premium = [1419.15,1505.70,1765.30,2094.15,2526.80,3097.95,3894.10,5088.25,7130.50,10141.90] Cover200k_Annual_Premium = [1892.25,2007.60,2353.75,2792.20,3369.10,4130.60,5192.10,6784.35,9507.30,13522.55 ] #Matrix Premiums = np.array([Cover50k_Annual_Premium, Cover100k_Annual_Premium, Cover150k_Annual_Premium, Cover200k_Annual_Premium]) #visualization plt.plot(Cover50k_Annual_Premium, c='Black', ls= '--', marker='o', label='50k') plt.plot(Cover100k_Annual_Premium, c='red', ls= '--', marker='o', label='100k') plt.plot(Cover150k_Annual_Premium, c='blue', ls= '--', marker='o', label='150k') plt.plot(Cover200k_Annual_Premium, c='green', ls= '--', marker='o', label='200k') # interpolate between all premiums # formula: interp = [(i1 + i2)/2.0 for i1, i2 in zip(l1, l2)] interp_premiums = [(Cover50k_Annual_Premium+Cover100k_Annual_Premium+Cover150k_Annual_Premium+Cover200k_Annual_Premium)/4.0 for Cover50k_Annual_Premium, Cover100k_Annual_Premium, Cover150k_Annual_Premium, Cover200k_Annual_Premium in zip(Cover50k_Annual_Premium, Cover100k_Annual_Premium, Cover150k_Annual_Premium, Cover200k_Annual_Premium)] plt.plot(interp_premiums, c='magenta', ls= '--', marker='o', label='interp') plt.legend(loc='upper left',bbox_to_anchor=(1,1)) plt.xticks(list(range(0,10)),Age, rotation='vertical') plt.show() interp_premiums # given Age 27, find the Premiums for 50k, 100k, 150k, 200k # ?????

简而言之,答案是:

from scipy.interpolate import interp1d
AgeNum = np.array(Age, dtype=float)
interps = {Coverage[count]: interp1d(
        AgeNum, param, kind='cubic', fill_value='extrapolate'
        ) for count, param in enumerate(Premiums)}

我们只是通过scipy的三次样条插值1d方法编一个插值函数字典。

不过,我注意到你的绘图有点古怪——我建议不要用记号标签“屏蔽”真正的pot值来覆盖缩放协议。

编辑:

下面是我如何将其全部编码:

import matplotlib.pyplot as plt
import numpy as np

from scipy.interpolate import interp1d

#Age
Age = ["18","20","25","30","35","40","45","50","55","59"]

#Insurance coverage
Coverage = ["50k","100k","150k","200k"]

#Premiums Data
Cover50k_Annual_Premium = [473.05,501.90,588.45,698.05,842.25,1032.65,1298.05,1696.10,2376.85,3380.65]
Cover100k_Annual_Premium = [946.10,1003.80,1176.90,1396.10,1684.55,2065.30,2596.05,3392.15,4753.65,6761.25]
Cover150k_Annual_Premium = [1419.15,1505.70,1765.30,2094.15,2526.80,3097.95,3894.10,5088.25,7130.50,10141.90]
Cover200k_Annual_Premium = [1892.25,2007.60,2353.75,2792.20,3369.10,4130.60,5192.10,6784.35,9507.30,13522.55 ]

#Matrix
Premiums = np.array([Cover50k_Annual_Premium, Cover100k_Annual_Premium, Cover150k_Annual_Premium, Cover200k_Annual_Premium])

#interpolation
AgeNum = np.array(Age, dtype=float)
interps = {Coverage[count]: interp1d(
        AgeNum, param, kind='cubic', fill_value='extrapolate'
        ) for count, param in enumerate(Premiums)}

# 2d interp
coverages = lambda z, y: interp1d(
    x=[50000,100000,150000,200000],
    y=[interps[func_val](z) for func_val in Coverage],
    kind='cubic',
    fill_value='extrapolate'
    )(y)


#visualization
fig = plt.figure()
ax = fig.add_subplot(111)
ax.tick_params(direction='in', pad = 5)

ax.plot(AgeNum, Cover50k_Annual_Premium, c='Black', ls= '--', marker='o', label='50k')
ax.plot(AgeNum, Cover100k_Annual_Premium, c='red', ls= '--', marker='o', label='100k')
ax.plot(AgeNum, Cover150k_Annual_Premium, c='blue', ls= '--', marker='o', label='150k')
ax.plot(AgeNum, Cover200k_Annual_Premium, c='green', ls= '--', marker='o', label='200k')

ages = np.linspace(18,59,100)
ax.plot(ages, interps["50k"](ages))
ax.plot(ages, interps["100k"](ages))
ax.plot(ages, interps["150k"](ages))
ax.plot(ages, interps["200k"](ages))

ax.legend(loc='upper left',bbox_to_anchor=(1,1))

plt.tight_layout()
plt.show()

这样,可以使用字典键调用插值:

In [23]:interps["50k"](27)
Out[23]: array(628.96153198)

In [24]:interps["200k"](27)
Out[24]: array(2515.80282111)

In [40]:coverages(27, 125000)
Out[40]: array(1572.3803509)