Question

我已将数据标准化并应用回归分析来预测收益率（y）。但我的预测输出也给出了标准化（0到1）我想用我正确的数据编号预测答案，而不是0到1。

数据：

Total_yield(y)    Rain(x)  
      64799.30   720.1  
      77232.40   382.9  
      88487.70  1198.2  
      77338.20   341.4  
      145602.05   406.4 
      67680.50   325.8 
      84536.20   791.8 
      99854.00   748.6 
      65939.90  1552.6 
      61622.80  1357.7
      66439.60   344.3

接下来，我使用以下代码对数据进行了规范化：

from sklearn.preprocessing import Normalizer
import pandas
import numpy
dataframe = pandas.read_csv('/home/desktop/yield.csv')
array = dataframe.values
X = array[:,0:2]
scaler = Normalizer().fit(X)
normalizedX = scaler.transform(X)
print(normalizedX)

     Total_yield      Rain
0       0.999904  0.013858
1       0.999782  0.020872
2       0.999960  0.008924
3       0.999967  0.008092
4       0.999966  0.008199
5       0.999972  0.007481
6       0.999915  0.013026
7       0.999942  0.010758
8       0.999946  0.010414
9       0.999984  0.005627
10      0.999967  0.008167

接下来，我使用此标准化值使用以下代码计算R-sqaure：

array=normalizedX
data = pandas.DataFrame(array,columns=['Total_yield','Rain'])
import statsmodels.formula.api as smf
lm = smf.ols(formula='Total_yield ~ Rain', data=data).fit()
lm.summary()

输出：

<class 'statsmodels.iolib.summary.Summary'>
"""
                            OLS Regression Results                            
==============================================================================
Dep. Variable:            Total_yield   R-squared:                       0.752
Model:                            OLS   Adj. R-squared:                  0.752
Method:                 Least Squares   F-statistic:                     1066.
Date:                Thu, 09 Feb 2017   Prob (F-statistic):          2.16e-108
Time:                        14:21:21   Log-Likelihood:                 941.53
No. Observations:                 353   AIC:                            -1879.
Df Residuals:                     351   BIC:                            -1871.
Df Model:                           1                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
Intercept      1.0116      0.001    948.719      0.000         1.009     1.014
Rain          -0.3013      0.009    -32.647      0.000        -0.319    -0.283
==============================================================================
Omnibus:                      408.798   Durbin-Watson:                   1.741
Prob(Omnibus):                  0.000   Jarque-Bera (JB):            40636.533
Skew:                          -4.955   Prob(JB):                         0.00
Kurtosis:                      54.620   Cond. No.                         10.3
==============================================================================

现在，R-square = 0.75，

regression model : y =  b0 + b1  *x

Yield  =  b0 + b1 * Rain

Yield  =  intercept + coefficient for Rain * Rain

Now when I use my data value for Rain data then it will gives this answer :
Yield  =  1.0116    + ( -0.3013 * 720.1(mm)) = -215.95

-215.95yield is wrong, 

 And when I use normalize value for rain data then predicted yield comes in normalize value in between 0 to 1.

 I want predict if rainfall will be 720.1 mm then how many yield will be there? 

If anyone help me how to get predicted yield ? I want to compare  Predicted yield vs given yield.

Answer 1

首先，在这种情况下不应使用Normalizer。它不会跨功能进行规范化。它沿着行进行。你可能不想要它。

使用MinMaxScaler或RobustScaler缩放每个功能。有关详细信息，请参阅preprocessing docs。

其次，这些类具有x = np.asarray([720.1,382.9,1198.2,341.4,406.4,325.8, 791.8,748.6,1552.6,1357.7,344.3]).reshape(-1,1) y = np.asarray([64799.30,77232.40,88487.70,77338.20,145602.05,67680.50, 84536.20,99854.00,65939.90,61622.80,66439.60]).reshape(-1,1) scalerx = RobustScaler() x_scaled = scalerx.fit_transform(x) scalery = RobustScaler() y_scaled = scalery.fit_transform(y)函数，可以将预测的y值转换回原始单位。

statsmodel.OLS

在这些缩放数据上调用x_scaled_test = scalerx.transform([720.1])。在预测时，首先转换测试数据：

Yield_scaled  =  b0 + b1 * x_scaled_test

在此值上应用回归模型并获得结果。 y的结果将根据缩放数据。

Yield_original = scalery.inverse_transform(Yield_scaled)

因此，对其进行逆变换以获取原始单位的数据。

@Entity
@Table(name = "organization")
@Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class Organization implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(name = "orgname")
    private String orgname;

    @Column(name = "orgaddress")
    private String orgaddress;

    @OneToMany(mappedBy = "organization")
    @JsonIgnore
    @Cache(usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
    private Set<Branch> branches = new HashSet<>();

//geter()
//setter()

}

但在我看来，这种线性模型不会给出太多准确性，因为当我绘制数据时，这就是结果。

此数据不适用于线性模型。使用其他技术，或获取更多数据。

规范化数据后，使用回归分析如何预测y？

1 个答案: