导入数据¶

In [1]:
import pandas as pd
c0m0full = pd.read_csv(r'D:\论文\最后一波一鼓作气\数据\c0m0full.csv')  
c1m1full = pd.read_csv(r'D:\论文\最后一波一鼓作气\数据\c1m1full.csv')  
hsf15full = pd.read_csv(r'D:\论文\最后一波一鼓作气\数据\hsf15full.csv')  
hsf18full = pd.read_csv(r'D:\论文\最后一波一鼓作气\数据\hsf18full.csv')  
city_gender_age_premium_ratio_district15 = pd.read_csv(r'D:\论文\最后一波一鼓作气\数据\city_gender_age_premium_ratio_district15.csv')
In [2]:
#删除城市样本
c1m1full = c1m1full[c1m1full['urban_nbs'] != 'Urban']
#确保15年未整合，18年整合了
c1m1full = c1m1full[(c1m1full['policyintergration2015']==0.0) & (c1m1full['policyintergration2018']==1.0)]
c1m1= c1m1full[['ID', 'c1','m1']]

#删除城市样本
c0m0full = c0m0full[c0m0full['urban_nbs'] != 'Urban']
#确保15年未整合，18年整合了
c0m0full = c0m0full[(c0m0full['policyintergration2015']==0.0) & (c0m0full['policyintergration2018']==1.0)]
c0m0= c0m0full[['ID', 'c0','m0']]

#删除城市样本
hsf15full = hsf15full[hsf15full['urban_nbs'] != 'Urban']
#确保15年未整合，18年整合了
hsf15full = hsf15full[(hsf15full['policyintergration2015']==0.0) & (hsf15full['policyintergration2018']==1.0)]
hsf15= hsf15full[['ID', 'hsf15']]

#删除城市样本
hsf18full = hsf18full[hsf18full['urban_nbs'] != 'Urban']
#确保15年未整合，18年整合了
hsf18full = hsf18full[(hsf18full['policyintergration2015']==0.0) & (hsf18full['policyintergration2018']==1.0)]
hsf18= hsf18full[['ID', 'hsf18']]

最优化方法¶

consumption-based：合并数据¶

In [3]:
data = pd.merge(c0m0, c1m1full,on="ID", how="inner")
data
Out[3]:
ID c0 m0 householdID communityID c1 m1 gender age marriage ... premium2018 r0 r1 r0adjust r1adjust policyintergration2015 policyintergration2018 district GDPgrowthrate urban_nbs
0 64033321002 112.216 60.0 640333210 640333 123.670 0.0 0.0 59.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
1 64033327002 1029.200 6000.0 640333270 640333 1054.764 21000.0 0.0 62.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
2 64033325001 1062.400 3000.0 640333250 640333 78.020 100.0 1.0 66.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
3 64033322001 592.620 2000.0 640333220 640333 859.050 17050.0 1.0 63.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
4 64033330002 2058.400 2000.0 640333300 640333 4025.500 2000.0 1.0 59.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
3882 89676104001 2315.700 840.0 896761040 896761 891.420 8000.0 1.0 61.0 1.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3883 89676114002 1935.145 300.0 896761140 896761 49.800 0.0 0.0 56.0 0.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3884 89676118001 2466.096 1000.0 896761180 896761 661.095 3000.0 0.0 73.0 0.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3885 89676115001 10721.940 800.0 896761150 896761 11638.260 2000.0 1.0 55.0 1.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3886 89676124001 268.422 500.0 896761240 896761 313.242 101.0 1.0 69.0 0.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural

3887 rows × 26 columns

In [4]:
#最优化方法——消费计算
import pandas as pd
import numpy as np

# 计算 E(m0) 和 E(m1)
E_m0 = data['m0'].mean()
E_m1 = data['m1'].mean()

# 计算 E(c0^(-3)) 和 E(c1^(-3))
E_c0_inv2 = (data['c0']**(-3)).mean()
E_c1_inv2 = (data['c1']**(-3)).mean()

# 计算协方差
cov_c0_inv2 = np.cov(data['c0']**(-3) / E_c0_inv2, (data['r0'] - data['r1']) * data['m0'] + data['premium2015'] - data['premium2018'])[0, 1]
cov_c1_inv2 = np.cov(data['c1']**(-3) / E_c1_inv2, (data['r0'] - data['r1']) * data['m1'] + data['premium2015'] - data['premium2018'])[0, 1]

# 计算 gamma
data['gamma112'] = abs(data['premium2015'] - data['premium2018']) + abs(0.5 * (data['r0'] - data['r1']) * (E_m0 + E_m1)) + 0.5 * cov_c0_inv2 + 0.5 * cov_c1_inv2
gamma112= data['gamma112'].mean()
print(gamma112)
data

894.0419995170009
Out[4]:
ID c0 m0 householdID communityID c1 m1 gender age marriage ... r0 r1 r0adjust r1adjust policyintergration2015 policyintergration2018 district GDPgrowthrate urban_nbs gamma112
0 64033321002 112.216 60.0 640333210 640333 123.670 0.0 0.0 59.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 704.829975
1 64033327002 1029.200 6000.0 640333270 640333 1054.764 21000.0 0.0 62.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 704.829975
2 64033325001 1062.400 3000.0 640333250 640333 78.020 100.0 1.0 66.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 704.829975
3 64033322001 592.620 2000.0 640333220 640333 859.050 17050.0 1.0 63.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 704.829975
4 64033330002 2058.400 2000.0 640333300 640333 4025.500 2000.0 1.0 59.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 704.829975
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
3882 89676104001 2315.700 840.0 896761040 896761 891.420 8000.0 1.0 61.0 1.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 753.326538
3883 89676114002 1935.145 300.0 896761140 896761 49.800 0.0 0.0 56.0 0.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 753.326538
3884 89676118001 2466.096 1000.0 896761180 896761 661.095 3000.0 0.0 73.0 0.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 753.326538
3885 89676115001 10721.940 800.0 896761150 896761 11638.260 2000.0 1.0 55.0 1.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 753.326538
3886 89676124001 268.422 500.0 896761240 896761 313.242 101.0 1.0 69.0 0.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 753.326538

3887 rows × 27 columns

In [6]:
#异质性男性 平衡面板
#最优化方法——消费计算 
datamale = data.loc[data['gender'] == 1].copy()
import pandas as pd
import numpy as np

# 计算 E(m0) 和 E(m1)
E_m0 = datamale['m0'].mean()
E_m1 = datamale['m1'].mean()

# 计算 E(c0^(-3)) 和 E(c1^(-3))
E_c0_inv2 = (datamale['c0']**(-3)).mean()
E_c1_inv2 = (datamale['c1']**(-3)).mean()

# 计算协方差
cov_c0_inv2 = np.cov(datamale['c0']**(-3) / E_c0_inv2, (datamale['r0'] - datamale['r1']) * datamale['m0'] + datamale['premium2015'] - datamale['premium2018'])[0, 1]
cov_c1_inv2 = np.cov(datamale['c1']**(-3) / E_c1_inv2, (datamale['r0'] - datamale['r1']) * datamale['m1'] + datamale['premium2015'] - datamale['premium2018'])[0, 1]

# 计算 gamma
datamale['gamma122'] = abs(datamale['premium2015'] - datamale['premium2018']) + abs(0.5 * (datamale['r0'] - datamale['r1']) * (E_m0 + E_m1)) + 0.5 * cov_c0_inv2 + 0.5 * cov_c1_inv2
gamma122= datamale['gamma122'].mean()
print(gamma122)

795.6912224492132
In [7]:
#异质性女性 平衡面板
#最优化方法——消费计算 
datafemale = data.loc[data['gender'] == 0].copy()
import pandas as pd
import numpy as np

# 计算 E(m0) 和 E(m1)
E_m0 = datafemale['m0'].mean()
E_m1 = datafemale['m1'].mean()

# 计算 E(c0^(-3)) 和 E(c1^(-3))
E_c0_inv2 = (datafemale['c0']**(-3)).mean()
E_c1_inv2 = (datafemale['c1']**(-3)).mean()

# 计算协方差
cov_c0_inv2 = np.cov(datafemale['c0']**(-3) / E_c0_inv2, (datafemale['r0'] - datafemale['r1']) * datafemale['m0'] + datafemale['premium2015'] - datafemale['premium2018'])[0, 1]
cov_c1_inv2 = np.cov(datafemale['c1']**(-3) / E_c1_inv2, (datafemale['r0'] - datafemale['r1']) * datafemale['m1'] + datafemale['premium2015'] - datafemale['premium2018'])[0, 1]

# 计算 gamma
datafemale['gamma132'] = abs(datafemale['premium2015'] - datafemale['premium2018']) + abs(0.5 * (datafemale['r0'] - datafemale['r1']) * (E_m0 + E_m1)) + 0.5 * cov_c0_inv2 + 0.5 * cov_c1_inv2
gamma132= datafemale['gamma132'].mean()
print(gamma132)

972.5626036980512
In [8]:
import numpy as np
import pandas as pd

# --------- 异质性条件 ---------
conds = {
    2:  lambda d: d["gender"].eq(1),                                            # 男
    3:  lambda d: d["gender"].eq(0),                                            # 女
    4:  lambda d: d["marriage"].eq(1),                                          # marriage=1
    5:  lambda d: d["marriage"].eq(0),                                          # marriage=0
    6:  lambda d: d["kids15"].eq(1),                                            # kids15=1
    7:  lambda d: d["kids15"].eq(0),                                            # kids15=0
    8:  lambda d: d["age"] < 59,                                                # age<59
    9:  lambda d: d["age"].between(60, 79, inclusive="both"),                   # 60~79
    10: lambda d: d["age"] >= 80,                                               # 80+
    11: lambda d: d["district"].astype(str).str.lower().eq("east"),             # east
    12: lambda d: d["district"].astype(str).str.lower().eq("middle"),           # middle
    13: lambda d: d["district"].astype(str).str.lower().eq("west"),             # west
    14: lambda d: d["hsf15"] > 40,                                              # hsf15>40
    15: lambda d: d["hsf15"].between(25, 40, inclusive="both"),                 # 25~40
    16: lambda d: d["hsf15"] < 25,                                              # <25
    17: lambda d: d["ic15"] > 35000,                                            # ic15>35000
    18: lambda d: d["ic15"].between(5000, 35000, inclusive="both"),             # 5000~35000
    19: lambda d: d["ic15"] < 5000,                                             # <5000
    20: lambda d: d["educationrevised"].isin([6,7,8,9,10,11]),                  # 教育 6-11
    21: lambda d: d["educationrevised"].eq(5),                                  # 教育 5
    22: lambda d: d["educationrevised"].isin([1,2,3,4]),                        # 教育 1-4
}

# --------- 工具函数（更稳健） ---------
def _to_num(s):
    return pd.to_numeric(s, errors="coerce")

def _safe_cov(x, y):
    z = pd.concat([x, y], axis=1).replace([np.inf, -np.inf], np.nan).dropna()
    if len(z) >= 2:
        return float(np.cov(z.iloc[:,0], z.iloc[:,1], ddof=1)[0,1])
    return 0.0

# --------- 批量计算：gamma122 … gamma1222 ---------
_results = {}
for idx, cond_fn in conds.items():
    # 取子样本
    try:
        mask = cond_fn(data)
        sub = data.loc[mask].copy() if isinstance(mask, pd.Series) and len(mask)==len(data) else data.copy()
    except Exception:
        sub = data.copy()

    if sub.empty:
        _results[f"gamma1{idx}2"] = np.nan
        continue

    # 数值化并避免 0 的 -3 次幂产生 inf
    for col in ["c0","c1","r0","r1","m0","m1","premium2015","premium2018"]:
        sub[col] = _to_num(sub[col])

    c0 = sub["c0"].replace(0, np.nan)
    c1 = sub["c1"].replace(0, np.nan)
    r0 = sub["r0"]; r1 = sub["r1"]
    m0 = sub["m0"]; m1 = sub["m1"]
    p15 = sub["premium2015"]; p18 = sub["premium2018"]

    # 子样本的均值
    E_m0 = m0.mean()
    E_m1 = m1.mean()

    c0_pow = c0 ** (-3)
    c1_pow = c1 ** (-3)
    E_c0_inv2 = c0_pow.replace([np.inf, -np.inf], np.nan).mean()
    E_c1_inv2 = c1_pow.replace([np.inf, -np.inf], np.nan).mean()

    # 协方差（标准化）
    cov0 = cov1 = 0.0
    if pd.notna(E_c0_inv2) and E_c0_inv2 != 0:
        x0 = c0_pow / E_c0_inv2
        y0 = (r0 - r1) * m0 + (p15 - p18)
        cov0 = _safe_cov(x0, y0)
    if pd.notna(E_c1_inv2) and E_c1_inv2 != 0:
        x1 = c1_pow / E_c1_inv2
        y1 = (r0 - r1) * m1 + (p15 - p18)
        cov1 = _safe_cov(x1, y1)

    # 行级 gamma，再取均值（与你示例一致）
    gamma_series = (p15 - p18).abs() + (0.5 * (r0 - r1) * (E_m0 + E_m1)).abs() + 0.5*cov0 + 0.5*cov1
    gamma_val = float(gamma_series.mean())

    name = f"gamma1{idx}2"
    _results[name] = gamma_val
    globals()[name] = gamma_val  # 可选：提升为同名变量

# 打印核对
for idx in range(2, 23):
    key = f"gamma1{idx}2"
    print(f"{key} = {_results.get(key, np.nan)}")

gamma122 = 795.6912224492132
gamma132 = 972.5626036980512
gamma142 = 968.9112657876573
gamma152 = 536.6179933332791
gamma162 = 898.4374643411936
gamma172 = 456.1738551339709
gamma182 = 638.479504002872
gamma192 = 937.0119208117399
gamma1102 = 926.4854623242564
gamma1112 = 1041.4770771710128
gamma1122 = 994.600066804255
gamma1132 = 653.0213728527
gamma1142 = 834.4207863998278
gamma1152 = 974.6068796763933
gamma1162 = 776.7776375893677
gamma1172 = 846.4128232195229
gamma1182 = 584.6022986316312
gamma1192 = 1003.8286917283358
gamma1202 = 1118.927937065343
gamma1212 = 746.2723252894853
gamma1222 = 857.9092498139793

health-based：合并数据¶

In [9]:
dataa = pd.merge(c0m0, c1m1full,on="ID", how="inner")
dataa
Out[9]:
ID c0 m0 householdID communityID c1 m1 gender age marriage ... premium2018 r0 r1 r0adjust r1adjust policyintergration2015 policyintergration2018 district GDPgrowthrate urban_nbs
0 64033321002 112.216 60.0 640333210 640333 123.670 0.0 0.0 59.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
1 64033327002 1029.200 6000.0 640333270 640333 1054.764 21000.0 0.0 62.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
2 64033325001 1062.400 3000.0 640333250 640333 78.020 100.0 1.0 66.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
3 64033322001 592.620 2000.0 640333220 640333 859.050 17050.0 1.0 63.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
4 64033330002 2058.400 2000.0 640333300 640333 4025.500 2000.0 1.0 59.0 1.0 ... 180.0 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
3882 89676104001 2315.700 840.0 896761040 896761 891.420 8000.0 1.0 61.0 1.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3883 89676114002 1935.145 300.0 896761140 896761 49.800 0.0 0.0 56.0 0.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3884 89676118001 2466.096 1000.0 896761180 896761 661.095 3000.0 0.0 73.0 0.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3885 89676115001 10721.940 800.0 896761150 896761 11638.260 2000.0 1.0 55.0 1.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural
3886 89676124001 268.422 500.0 896761240 896761 313.242 101.0 1.0 69.0 0.0 ... 220.0 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural

3887 rows × 26 columns

In [10]:
#计算dh/dm 15
import pandas as pd
import numpy as np
import statsmodels.api as sm
e1 = pd.merge(c0m0,hsf15,on="ID",how="inner")
e1= e1[['m0','hsf15']].copy()
# 删除包含 NaN 或 inf 的行
e1= e1.replace([np.inf, -np.inf], np.nan).dropna()
# 删除包含 0 的行
e1 = e1[(e1['hsf15']!=0) & (e1['m0']!=0)]
# 自变量（X）和因变量（Y）
X = e1['m0']
Y = e1['hsf15']

# 在 X 中添加常数项，以便进行 OLS 回归
X = sm.add_constant(X)

# 拟合 OLS 回归模型
model = sm.OLS(Y, X).fit()

# 输出回归结果
print(model.summary())

# 提取回归系数
coefficients = model.params

# 保存特定自变量的回归系数
h_m15 = coefficients['m0'] 
h_m15

                            OLS Regression Results                            
==============================================================================
Dep. Variable:                  hsf15   R-squared:                       0.000
Model:                            OLS   Adj. R-squared:                  0.000
Method:                 Least Squares   F-statistic:                     1.297
Date:                Mon, 29 Dec 2025   Prob (F-statistic):              0.255
Time:                        18:30:47   Log-Likelihood:                -11963.
No. Observations:                3113   AIC:                         2.393e+04
Df Residuals:                    3111   BIC:                         2.394e+04
Df Model:                           1                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
const         33.6007      0.216    155.587      0.000      33.177      34.024
m0         -1.706e-05    1.5e-05     -1.139      0.255   -4.64e-05    1.23e-05
==============================================================================
Omnibus:                       41.096   Durbin-Watson:                   1.849
Prob(Omnibus):                  0.000   Jarque-Bera (JB):               32.348
Skew:                          -0.167   Prob(JB):                     9.46e-08
Kurtosis:                       2.629   Cond. No.                     1.54e+04
==============================================================================

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 1.54e+04. This might indicate that there are
strong multicollinearity or other numerical problems.
Out[10]:
np.float64(-1.7062284276591698e-05)
In [11]:
#计算dh/dm 18
import pandas as pd
import statsmodels.api as sm
f1 = pd.merge(c1m1,hsf18,on="ID",how="inner")
f1= f1[['m1','hsf18']].copy()
# 删除包含 NaN 或 inf 的行
f1= f1.replace([np.inf, -np.inf], np.nan).dropna()
# 删除包含 0 的行
f1 = f1[(f1['hsf18']!=0) & (f1['m1']!=0)]
# 自变量（X）和因变量（Y）
X = f1['m1']
Y = f1['hsf18']

# 在 X 中添加常数项，以便进行 OLS 回归
X = sm.add_constant(X)

# 拟合 OLS 回归模型
model = sm.OLS(Y, X).fit()

# 输出回归结果
print(model.summary())

# 提取回归系数
coefficients = model.params

# 保存特定自变量的回归系数
h_m18 = coefficients['m1'] 
h_m18

                            OLS Regression Results                            
==============================================================================
Dep. Variable:                  hsf18   R-squared:                       0.000
Model:                            OLS   Adj. R-squared:                  0.000
Method:                 Least Squares   F-statistic:                     1.579
Date:                Mon, 29 Dec 2025   Prob (F-statistic):              0.209
Time:                        18:30:48   Log-Likelihood:                -26371.
No. Observations:                5630   AIC:                         5.275e+04
Df Residuals:                    5628   BIC:                         5.276e+04
Df Model:                           1                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
const         51.5258      0.359    143.352      0.000      50.821      52.230
m1         -1.359e-05   1.08e-05     -1.257      0.209   -3.48e-05    7.61e-06
==============================================================================
Omnibus:                    14923.644   Durbin-Watson:                   1.748
Prob(Omnibus):                  0.000   Jarque-Bera (JB):              441.401
Skew:                          -0.235   Prob(JB):                     1.42e-96
Kurtosis:                       1.711   Cond. No.                     3.42e+04
==============================================================================

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 3.42e+04. This might indicate that there are
strong multicollinearity or other numerical problems.
Out[11]:
np.float64(-1.3590936773055398e-05)
In [12]:
import pandas as pd
import numpy as np

# 计算 E(m0) 和 E(m1)
E_m0 = dataa['m0'].mean()
E_m1 = dataa['m1'].mean()

# 计算 E(c0^(-3)) 和 E(c1^(-3))
E_c0_inv2 = (dataa['c0']**(-3)).mean()
E_c1_inv2 = (dataa['c1']**(-3)).mean()

# 计算协方差
cov_c0_inv2 = np.cov((0.019743 * h_m15)/ (E_c0_inv2 * dataa['r0']), (dataa['r0'] - dataa['r1']) * dataa['m0'] + dataa['premium2015'] - dataa['premium2018'])[0, 1]
cov_c1_inv2 = np.cov((0.019743 * h_m18)/ (E_c1_inv2 * dataa['r1']), (dataa['r0'] - dataa['r1']) * dataa['m1'] + dataa['premium2015'] - dataa['premium2018'])[0, 1]

# 计算 gamma
dataa['gamma113'] = abs(dataa['premium2015'] - dataa['premium2018']) + abs(0.5 * (dataa['r0'] - dataa['r1']) * (E_m0 + E_m1)) + 0.5 * cov_c0_inv2 + 0.5 * cov_c1_inv2
gamma113 = dataa['gamma113'].mean()
print(gamma113)
dataa

689.4542339864197
Out[12]:
ID c0 m0 householdID communityID c1 m1 gender age marriage ... r0 r1 r0adjust r1adjust policyintergration2015 policyintergration2018 district GDPgrowthrate urban_nbs gamma113
0 64033321002 112.216 60.0 640333210 640333 123.670 0.0 0.0 59.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 500.242209
1 64033327002 1029.200 6000.0 640333270 640333 1054.764 21000.0 0.0 62.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 500.242209
2 64033325001 1062.400 3000.0 640333250 640333 78.020 100.0 1.0 66.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 500.242209
3 64033322001 592.620 2000.0 640333220 640333 859.050 17050.0 1.0 63.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 500.242209
4 64033330002 2058.400 2000.0 640333300 640333 4025.500 2000.0 1.0 59.0 1.0 ... 0.6167 0.700 0.544192 0.660903 0.0 1.0 east 0.118005 Rural 500.242209
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
3882 89676104001 2315.700 840.0 896761040 896761 891.420 8000.0 1.0 61.0 1.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 548.738772
3883 89676114002 1935.145 300.0 896761140 896761 49.800 0.0 0.0 56.0 0.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 548.738772
3884 89676118001 2466.096 1000.0 896761180 896761 661.095 3000.0 0.0 73.0 0.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 548.738772
3885 89676115001 10721.940 800.0 896761150 896761 11638.260 2000.0 1.0 55.0 1.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 548.738772
3886 89676124001 268.422 500.0 896761240 896761 313.242 101.0 1.0 69.0 0.0 ... 0.6500 0.725 0.624462 0.685108 0.0 1.0 west 0.284050 Rural 548.738772

3887 rows × 27 columns

In [13]:
#健康based异质性探索——男性
dataamale = dataa.loc[dataa['gender'] == 1].copy()
import pandas as pd
import numpy as np

# 计算 E(m0) 和 E(m1)
E_m0 = dataamale['m0'].mean()
E_m1 = dataamale['m1'].mean()

# 计算 E(c0^(-3)) 和 E(c1^(-3))
E_c0_inv2 = (dataamale['c0']**(-3)).mean()
E_c1_inv2 = (dataamale['c1']**(-3)).mean()

# 计算协方差
cov_c0_inv2 = np.cov((0.019743 * h_m15)/ (E_c0_inv2 * dataamale['r0']), (dataamale['r0'] - dataamale['r1']) * dataamale['m0'] + dataamale['premium2015'] - dataamale['premium2018'])[0, 1]
cov_c1_inv2 = np.cov((0.019743 * h_m18)/ (E_c1_inv2 * dataamale['r1']), (dataamale['r0'] - dataamale['r1']) * dataamale['m1'] + dataamale['premium2015'] - dataamale['premium2018'])[0, 1]

# 计算 gamma
dataamale['gamma123'] = abs(dataamale['premium2015'] - dataamale['premium2018']) + abs(0.5 * (dataamale['r0'] - dataamale['r1']) * (E_m0 + E_m1)) + 0.5 * cov_c0_inv2 + 0.5 * cov_c1_inv2
gamma123 = dataamale['gamma123'].mean()
print(gamma123)

701.6096874269684
In [14]:
#健康based异质性探索——女性
dataafemale = dataa.loc[dataa['gender'] == 0].copy()
import pandas as pd
import numpy as np

# 计算 E(m0) 和 E(m1)
E_m0 = dataafemale['m0'].mean()
E_m1 = dataafemale['m1'].mean()

# 计算 E(c0^(-3)) 和 E(c1^(-3))
E_c0_inv2 = (dataafemale['c0']**(-3)).mean()
E_c1_inv2 = (dataafemale['c1']**(-3)).mean()

# 计算协方差
cov_c0_inv2 = np.cov((0.019743 * h_m15)/ (E_c0_inv2 * dataafemale['r0']), (dataafemale['r0'] - dataafemale['r1']) * dataafemale['m0'] + dataafemale['premium2015'] - dataafemale['premium2018'])[0, 1]
cov_c1_inv2 = np.cov((0.019743 * h_m18)/ (E_c1_inv2 * dataafemale['r1']), (dataafemale['r0'] - dataafemale['r1']) * dataafemale['m1'] + dataafemale['premium2015'] - dataafemale['premium2018'])[0, 1]

# 计算 gamma
dataafemale['gamma133'] = abs(dataafemale['premium2015'] - dataafemale['premium2018']) + abs(0.5 * (dataafemale['r0'] - dataafemale['r1']) * (E_m0 + E_m1)) + 0.5 * cov_c0_inv2 + 0.5 * cov_c1_inv2
gamma133 = dataafemale['gamma133'].mean()
print(gamma133)

676.297559935399
In [15]:
import numpy as np
import pandas as pd

PHI = 0.019743  # 你的 phi_tilde

# ========== 工具函数 ==========
def safe_filter(df, cond_fn):
    """对 df 应用条件；若缺列/异常则返回原 df（不筛选）"""
    try:
        m = cond_fn(df)
        if isinstance(m, pd.Series) and len(m) == len(df):
            return df.loc[m].copy()
    except Exception:
        pass
    return df.copy()

def _to_num(s):
    return pd.to_numeric(s, errors="coerce")

def _safe_cov(x, y):
    z = pd.concat([x, y], axis=1).replace([np.inf, -np.inf], np.nan).dropna()
    if len(z) >= 2:
        return float(np.cov(z.iloc[:,0], z.iloc[:,1], ddof=1)[0,1])
    return 0.0

def _align_health_for_subset(dataa_full, data_sub, h_full, id_col="ID"):
    """
    将全样本 h_m15/h_m18 与子样本对齐（索引或按 ID）。
    支持：Series（索引=dataa.index 或 =ID）、array-like。
    """
    if isinstance(h_full, pd.Series):
        if h_full.index.equals(dataa_full.index):
            return h_full.loc[data_sub.index]
        if id_col in data_sub.columns and h_full.index.isin(data_sub[id_col]).any():
            s = h_full.reindex(data_sub[id_col])
            s.index = data_sub.index
            return s
        try:
            return h_full.loc[data_sub.index]
        except Exception:
            return pd.Series(np.nan, index=data_sub.index)
    # array-like
    try:
        base = pd.Series(h_full, index=dataa_full.index)
        return base.loc[data_sub.index]
    except Exception:
        return pd.Series(np.nan, index=data_sub.index)

def compute_gamma_health_explore(dataa_full, data_sub, h_m15, h_m18):
    """
    你的“健康 based 探索”口径（与示例一致）：
      1) 子样本内均值：E_m0, E_m1；Ec0 = E[c0^(-2)], Ec1 = E[c1^(-2)]
      2) 协方差：
         cov0 = Cov( (PHI*h_m15)/(Ec0*r0), (r0-r1)*m0 + p15 - p18 )
         cov1 = Cov( (PHI*h_m18)/(Ec1*r1), (r0-r1)*m1 + p15 - p18 )
      3) 行级 gamma_i = |p15 - p18| + |0.5*(r0-r1)*(E_m0+E_m1)| + 0.5*cov0 + 0.5*cov1
         返回子样本内 gamma_i 的均值
    """
    # 数值化
    for col in ["c0","c1","r0","r1","m0","m1","premium2015","premium2018"]:
        data_sub[col] = _to_num(data_sub[col])

    # 子样本均值
    E_m0 = data_sub["m0"].mean()
    E_m1 = data_sub["m1"].mean()

    # 处理 c0/c1 的 0，避免 **(-2) 出现 inf
    c0 = data_sub["c0"].replace(0, np.nan)
    c1 = data_sub["c1"].replace(0, np.nan)
    Ec0 = (c0 ** (-3)).replace([np.inf, -np.inf], np.nan).mean()
    Ec1 = (c1 ** (-3)).replace([np.inf, -np.inf], np.nan).mean()

    # 对齐 h_m15 / h_m18
    h15 = _align_health_for_subset(dataa_full, data_sub, h_m15)
    h18 = _align_health_for_subset(dataa_full, data_sub, h_m18)

    r0 = data_sub["r0"].replace(0, np.nan)
    r1 = data_sub["r1"].replace(0, np.nan)
    m0 = data_sub["m0"]; m1 = data_sub["m1"]
    p15 = data_sub["premium2015"]; p18 = data_sub["premium2018"]

    # 协方差（Ec0/Ec1 为 0 或 NaN 时相应项置 0）
    cov0 = cov1 = 0.0
    if pd.notna(Ec0) and Ec0 != 0:
        a0 = (PHI * h15) / (Ec0 * r0)
        y0 = (r0 - r1) * m0 + (p15 - p18)
        cov0 = _safe_cov(a0, y0)
    if pd.notna(Ec1) and Ec1 != 0:
        a1 = (PHI * h18) / (Ec1 * r1)
        y1 = (r0 - r1) * m1 + (p15 - p18)
        cov1 = _safe_cov(a1, y1)

    # 行级 gamma，再取均值
    gamma_series = (p15 - p18).abs() + (0.5 * (data_sub["r0"] - data_sub["r1"]) * (E_m0 + E_m1)).abs() + 0.5*cov0 + 0.5*cov1
    return float(gamma_series.mean())

# ========== 异质性条件 ==========
conds = {
    2:  lambda d: d["gender"].eq(1),                                            # 男
    3:  lambda d: d["gender"].eq(0),                                            # 女
    4:  lambda d: d["marriage"].eq(1),                                          # marriage=1
    5:  lambda d: d["marriage"].eq(0),                                          # marriage=0
    6:  lambda d: d["kids15"].eq(1),                                            # kids15=1
    7:  lambda d: d["kids15"].eq(0),                                            # kids15=0
    8:  lambda d: d["age"] < 59,                                                # age<59
    9:  lambda d: d["age"].between(60, 79, inclusive="both"),                   # 60~79
    10: lambda d: d["age"] >= 80,                                               # 80+
    11: lambda d: d["district"].astype(str).str.lower().eq("east"),             # east
    12: lambda d: d["district"].astype(str).str.lower().eq("middle"),           # middle
    13: lambda d: d["district"].astype(str).str.lower().eq("west"),             # west
    14: lambda d: d["hsf15"] > 40,                                              # hsf15>40
    15: lambda d: d["hsf15"].between(25, 40, inclusive="both"),                 # 25~40
    16: lambda d: d["hsf15"] < 25,                                              # <25
    17: lambda d: d["ic15"] > 35000,                                            # ic15>35000
    18: lambda d: d["ic15"].between(5000, 35000, inclusive="both"),             # 5000~35000
    19: lambda d: d["ic15"] < 5000,                                             # <5000
    20: lambda d: d["educationrevised"].isin([6,7,8,9,10,11]),                  # 教育 6-11
    21: lambda d: d["educationrevised"].eq(5),                                  # 教育 5
    22: lambda d: d["educationrevised"].isin([1,2,3,4]),                        # 教育 1-4
}

# ========== 批量计算：gamma123 … gamma1223 ==========
_results = {}
for idx, cond_fn in conds.items():
    sub = safe_filter(dataa, cond_fn)  # 按你的示例，health-based 用 dataa 作为主样本
    if sub.empty:
        val = np.nan
    else:
        val = compute_gamma_health_explore(dataa_full=dataa, data_sub=sub, h_m15=h_m15, h_m18=h_m18)
    name = f"gamma1{idx}3"
    _results[name] = val
    globals()[name] = val  # 可选：注册为同名变量

# 打印核对
for idx in range(2, 23):
    key = f"gamma1{idx}3"
    print(f"{key} = {_results.get(key, np.nan)}")

gamma123 = 701.6096874269684
gamma133 = 676.297559935399
gamma143 = 742.1281289436823
gamma153 = 493.6350274056347
gamma163 = 694.3024694656491
gamma173 = 356.45453692191603
gamma183 = 671.0503320289147
gamma193 = 693.9424656194997
gamma1103 = 571.7774025800886
gamma1113 = 600.9391876768528
gamma1123 = 763.7401963175519
gamma1133 = 601.8149627531205
gamma1143 = 621.8740168169526
gamma1153 = 795.3126953827027
gamma1163 = 586.7692087079621
gamma1173 = 694.1307975452706
gamma1183 = 555.6822076078612
gamma1193 = 739.0296290542765
gamma1203 = 760.9427888447271
gamma1213 = 813.6137702363884
gamma1223 = 656.7930492144557

用完全信息法求解¶

In [16]:
d1 = pd.merge(c0m0, c1m1, on="ID", how="inner")
d2 = pd.merge(d1, hsf15, on="ID", how="inner")
d3 = pd.merge(d2, hsf18, on="ID", how="inner")
d3
Out[16]:
ID c0 m0 c1 m1 hsf15 hsf18
0 64033321002 112.216 60.0 123.670 0.0 54.436016 10.551840
1 64033327002 1029.200 6000.0 1054.764 21000.0 28.096293 39.754467
2 64033322001 592.620 2000.0 859.050 17050.0 21.303569 59.803845
3 64033330002 2058.400 2000.0 4025.500 2000.0 34.523055 68.626626
4 64033341001 5436.500 500.0 1806.578 2500.0 50.384267 78.657775
... ... ... ... ... ... ... ...
3750 89676104001 2315.700 840.0 891.420 8000.0 27.522919 42.058152
3751 89676114002 1935.145 300.0 49.800 0.0 35.235805 15.917436
3752 89676118001 2466.096 1000.0 661.095 3000.0 32.251424 54.653869
3753 89676115001 10721.940 800.0 11638.260 2000.0 32.757347 73.821948
3754 89676124001 268.422 500.0 313.242 101.0 34.706036 95.825352

3755 rows × 7 columns

In [17]:
import numpy as np
import pandas as pd
# 参数
sigma = 3.0
phi_tilde = 0.019743

d3["B_bar"] = (d3["c0"]**(1 - sigma)) + (1 - sigma) * phi_tilde * (d3["hsf15"] - d3["hsf18"])
d3["gamma111"] = d3["c1"] - d3["B_bar"]**(1 / (1 - sigma))

gamma111= d3["gamma111"].mean()
print(gamma111)
d3

1996.3867642406108
Out[17]:
ID c0 m0 c1 m1 hsf15 hsf18 B_bar gamma111
0 64033321002 112.216 60.0 123.670 0.0 54.436016 10.551840 -1.732731 NaN
1 64033327002 1029.200 6000.0 1054.764 21000.0 28.096293 39.754467 0.460336 1053.290118
2 64033322001 592.620 2000.0 859.050 17050.0 21.303569 59.803845 1.520225 858.238953
3 64033330002 2058.400 2000.0 4025.500 2000.0 34.523055 68.626626 1.346614 4024.638256
4 64033341001 5436.500 500.0 1806.578 2500.0 50.384267 78.657775 1.116408 1805.631570
... ... ... ... ... ... ... ... ... ...
3750 89676104001 2315.700 840.0 891.420 8000.0 27.522919 42.058152 0.573938 890.100020
3751 89676114002 1935.145 300.0 49.800 0.0 35.235805 15.917436 -0.762805 NaN
3752 89676118001 2466.096 1000.0 661.095 3000.0 32.251424 54.653869 0.884583 660.031762
3753 89676115001 10721.940 800.0 11638.260 2000.0 32.757347 73.821948 1.621477 11637.474684
3754 89676124001 268.422 500.0 313.242 101.0 34.706036 95.825352 2.413371 312.598293

3755 rows × 9 columns

In [18]:
#异质性男性
import numpy as np
import pandas as pd
d4 = pd.merge(d1, hsf18full, on="ID", how="inner")
dmale = d4.loc[d4['gender'] == 1].copy()

# 参数
sigma = 3.0
phi_tilde = 0.019743

dmale["B_bar"] = (dmale["c0"]**(1 - sigma)) + (1 - sigma) * phi_tilde * (dmale["hsf15"] - dmale["hsf18"])
dmale["gamma121"] = dmale["c1"] - dmale["B_bar"]**(1 / (1 - sigma))

gamma121= dmale["gamma121"].mean()
print(gamma121)

1968.4893177925992
In [19]:
#异质性女性
import numpy as np
import pandas as pd
d4 = pd.merge(d1, hsf18full, on="ID", how="inner")
dfemale = d4.loc[d4['gender'] == 0].copy()

# 参数
sigma = 3.0
phi_tilde = 0.019743

dfemale["B_bar"] = (dfemale["c0"]**(1 - sigma)) + (1 - sigma) * phi_tilde * (dfemale["hsf15"] - dfemale["hsf18"])
dfemale["gamma131"] = dfemale["c1"] - dfemale["B_bar"]**(1 / (1 - sigma))

gamma131= dfemale["gamma131"].mean()
print(gamma131)

2030.5047297378765
In [20]:
import numpy as np
import pandas as pd

# ===== 1) 构造 d4 =====
# 这里假定 d1 已经包含了 hsf15、c0、c1、gender 等列
d4 = pd.merge(d1, hsf18full, on="ID", how="inner")

# 若某些分组变量是字符串（比如 district），确保为字符串
if "district" in d4.columns:
    d4["district"] = d4["district"].astype(str)

# ===== 2) 异质性条件 =====
conds = {
    2:  lambda d: d["gender"].eq(1),                                            # 男
    3:  lambda d: d["gender"].eq(0),                                            # 女
    4:  lambda d: d["marriage"].eq(1),                                          # marriage=1
    5:  lambda d: d["marriage"].eq(0),                                          # marriage=0
    6:  lambda d: d["kids15"].eq(1),                                            # kids15=1
    7:  lambda d: d["kids15"].eq(0),                                            # kids15=0
    8:  lambda d: d["age"] < 59,                                                # age<59
    9:  lambda d: d["age"].between(60, 79, inclusive="both"),                   # 60~79
    10: lambda d: d["age"] >= 80,                                               # 80+
    11: lambda d: d["district"].str.lower().eq("east"),                         # east
    12: lambda d: d["district"].str.lower().eq("middle"),                       # middle
    13: lambda d: d["district"].str.lower().eq("west"),                         # west
    14: lambda d: d["hsf15"] > 40,                                              # hsf15>40
    15: lambda d: d["hsf15"].between(25, 40, inclusive="both"),                 # 25~40
    16: lambda d: d["hsf15"] < 25,                                              # <25
    17: lambda d: d["ic15"] > 35000,                                            # ic15>35000
    18: lambda d: d["ic15"].between(5000, 35000, inclusive="both"),             # 5000~35000
    19: lambda d: d["ic15"] < 5000,                                             # <5000
    20: lambda d: d["educationrevised"].isin([6,7,8,9,10,11]),                  # 教育 6-11
    21: lambda d: d["educationrevised"].eq(5),                                  # 教育 5
    22: lambda d: d["educationrevised"].isin([1,2,3,4]),                        # 教育 1-4
}

# ===== 3) 核心计算（逐行算，再对子样本取均值） =====
sigma = 3.0
phi_tilde = 0.019743
power = 1.0 - sigma             # = -2
inv_power = 1.0 / power         # = -0.5

def compute_gamma_subset_like_yours(df_sub: pd.DataFrame) -> float:
    if df_sub.empty:
        return float("nan")
    # 不做 base>0 保护
    B_bar = (df_sub["c0"] ** power) + power * phi_tilde * (df_sub["hsf15"] - df_sub["hsf18"])
    gamma_i = df_sub["c1"] - (B_bar ** inv_power)
    return float(gamma_i.mean())  # mean 默认会跳过 NaN

# ===== 4) 批量生成 gamma121 … gamma1221 =====
results = {}
for idx, cond_fn in conds.items():
    # 如果条件里用到的列不存在，会抛错；这里捕获后把该组记为 NaN（避免改变其它组结果）
    try:
        mask = cond_fn(d4)
        sub = d4.loc[mask].copy() if isinstance(mask, pd.Series) and len(mask) == len(d4) else d4.copy()
    except Exception:
        sub = pd.DataFrame(columns=d4.columns)  # 列缺失时该组置空
    name = f"gamma1{idx}1"
    results[name] = compute_gamma_subset_like_yours(sub)
    globals()[name] = results[name]  # 可选：注册为同名变量

# 打印核对
for idx in range(2, 23):
    key = f"gamma1{idx}1"
    print(f"{key} = {results.get(key, np.nan)}")

gamma121 = 1968.4893177925992
gamma131 = 2030.5047297378765
gamma141 = 2091.756836228619
gamma151 = 1516.9958081205616
gamma161 = 2009.6788446394628
gamma171 = 752.4025142905097
gamma181 = 2513.485763337119
gamma191 = 1492.2099449302036
gamma1101 = 963.7132207091026
gamma1111 = 2318.3816169376114
gamma1121 = 2056.9974605815864
gamma1131 = 1614.5882274330595
gamma1141 = 2312.716443873066
gamma1151 = 1938.9117698957555
gamma1161 = 1559.7312849130187
gamma1171 = 2566.5778502883627
gamma1181 = 2138.1528481056553
gamma1191 = 1882.5176488438392
gamma1201 = 2677.6344623042633
gamma1211 = 2276.608300019149
gamma1221 = 1740.557558427637
In [21]:
# -*- coding: utf-8 -*-
import pandas as pd

# 1) 行索引与数据
rows = [
    "全样本",
    "男性","女性",
    "有配偶","无配偶",
    "有子女","无子女",
    "小于59 岁","60 岁—79 岁","80 岁及以上",
    "东部","中部","西部",
    "健康状况较好","健康状况中等","健康状况较差",
    "较高收入","中等收入","较低收入",
    "教育程度较高","教育程度中等","教育程度较低",
]

data = [
[gamma111, gamma112, gamma113],
[gamma121, gamma122, gamma123],
[gamma131, gamma132, gamma133],
[gamma141, gamma142, gamma143],
[gamma151, gamma152, gamma153],
[gamma161, gamma162, gamma163],
[gamma171, gamma172, gamma173],
[gamma181, gamma182, gamma183],
[gamma191, gamma192, gamma193],
[gamma1101, gamma1102, gamma1103],
[gamma1111, gamma1112, gamma1113],
[gamma1121, gamma1122, gamma1123],
[gamma1131, gamma1132, gamma1133],
[gamma1141, gamma1142, gamma1143],
[gamma1151, gamma1152, gamma1153],
[gamma1161, gamma1162, gamma1163],
[gamma1171, gamma1172, gamma1173],
[gamma1181, gamma1182, gamma1183],
[gamma1191, gamma1192, gamma1193],
[gamma1201, gamma1202, gamma1203],
[gamma1211, gamma1212, gamma1213],
[gamma1221, gamma1222, gamma1223],
]

# 2) 多级列索引
cols = pd.MultiIndex.from_tuples([
    ("完全信息方法",""),
    ("最优化方法","仅假设效用函数\n的消费部分"),
    ("最优化方法","仅假设效用函数\n的健康部分"),
])

df = pd.DataFrame(data, index=rows, columns=cols)

# 3) 分组起始行（加粗横线）
group_starts = {
    "男性",           # 性别组
    "有配偶",         # 婚姻组
    "有子女",         # 子女组
    "45 岁—59 岁",    # 年龄组
    "东部",           # 地区组
    "健康状况较好",   # 健康组
    "较高收入",       # 收入组
    "教育程度较高"    # 教育组
}

def row_borders(row):
    label = row.name
    if label in group_starts:
        return ['border-top: 2px solid #4a4a4a'] * len(row)
    return [''] * len(row)

# 4) 样式与展示
styler = (
    df.style
      .set_table_styles([
          {'selector': 'th.col_heading.level0',
           'props': [('font-weight', '700'),
                     ('border-bottom','1px solid #4a4a4a')]},
          {'selector': 'th.col_heading.level1',
           'props': [('font-weight', '700')]},
          {'selector': 'th.row_heading',
           'props': [('font-weight', '700')]},
          {'selector': 'table',
           'props': [('border-collapse','collapse'),
                     ('font-family','-apple-system,BlinkMacSystemFont,Segoe UI,Roboto,PingFang SC,Helvetica,Arial')]}
      ])
      .format(precision=0)
      .set_properties(**{
          'text-align': 'center',
          'padding': '6px',
          'border':'1px solid #a0a0a0'
      })
      .apply(row_borders, axis=1)
)

# 在 Jupyter 中显示
styler
Out[21]:
  完全信息方法 最优化方法
  仅假设效用函数 的消费部分 仅假设效用函数 的健康部分
全样本 1996 894 689
男性 1968 796 702
女性 2031 973 676
有配偶 2092 969 742
无配偶 1517 537 494
有子女 2010 898 694
无子女 752 456 356
小于59 岁 2513 638 671
60 岁—79 岁 1492 937 694
80 岁及以上 964 926 572
东部 2318 1041 601
中部 2057 995 764
西部 1615 653 602
健康状况较好 2313 834 622
健康状况中等 1939 975 795
健康状况较差 1560 777 587
较高收入 2567 846 694
中等收入 2138 585 556
较低收入 1883 1004 739
教育程度较高 2678 1119 761
教育程度中等 2277 746 814
教育程度较低 1741 858 657