我正在尝试使用线性回归估算熊猫数据框中的缺失值
`
for index in [missing_data_df.horsepower.index]:
i = 0
if pd.isnull(missing_data_df.horsepower[index[i]]):
#linear regression equation
a = 0.25743277 * missing_data_df.displacement[index[i]] + 0.00958711 *
missing_data_df.weight[index[i]] + 25.874947903262651
# replacing "nan" values in dataframe using .set_value
missing_data_df.set_value(index[i],"horsepower",a)
i+=1
`
它正在执行。但数据框中缺少的值（nan）不会通过变量'a'中的线性回归被预测值替代。有什么建议吗？
以下是包含缺失数据的数据框
`
>>> missing_data_df:
mpg cylinders displacement horsepower weight acceleration \
10 NaN 4.0 133.0 115.0 3090.0 17.5
11 NaN 8.0 350.0 165.0 4142.0 11.5
12 NaN 8.0 351.0 153.0 4034.0 11.0
13 NaN 8.0 383.0 175.0 4166.0 10.5
14 NaN 8.0 360.0 175.0 3850.0 11.0
17 NaN 8.0 302.0 140.0 3353.0 8.0
38 25.0 4.0 98.0 NaN 2046.0 19.0
39 NaN 4.0 97.0 48.0 1978.0 20.0
133 21.0 6.0 200.0 NaN 2875.0 17.0
337 40.9 4.0 85.0 NaN 1835.0 17.3
343 23.6 4.0 140.0 NaN 2905.0 14.3
361 34.5 4.0 100.0 NaN 2320.0 15.8
367 NaN 4.0 121.0 110.0 2800.0 15.4
382 23.0 4.0 151.0 NaN 3035.0 20.5
model_year origin car_name
10 70.0 2.0 citroen ds-21 pallas
11 70.0 1.0 chevrolet chevelle concours (sw)
12 70.0 1.0 ford torino (sw)
13 70.0 1.0 plymouth satellite (sw)
14 70.0 1.0 amc rebel sst (sw)
17 70.0 1.0 ford mustang boss 302
38 71.0 1.0 ford pinto
39 71.0 2.0 volkswagen super beetle 117
133 74.0 1.0 ford maverick
337 80.0 2.0 renault lecar deluxe
343 80.0 1.0 ford mustang cobra
361 81.0 2.0 renault 18i
367 81.0 2.0 saab 900s
382 82.0 1.0 amc concord dl
`
最佳答案
您可以为此使用apply和lambda：
missing_data_df['horsepower']= missing_data_df.apply(
lambda row:
0.25743277 * row.displacement + 0.00958711 * row.weight + 25.874947903262651
if np.isnan(row.horsepower) else row.horsepower, axis=1)
关于python - 在python中使用线性回归估算缺失值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/44097633/