问题描述
我正在按如下方式进行回归(df是pandas数据帧):
I am running a regression as follows (df is a pandas dataframe):
import statsmodels.api as sm
est = sm.OLS(df['p'], df[['e', 'varA', 'meanM', 'varM', 'covAM']]).fit()
est.summary()
除其他外，这给了我0.942的R平方.因此，我想绘制原始的y-values和拟合值.为此，我对原始值进行了排序:
Which gave me, among others, an R-squared of 0.942. So then I wanted to plot the original y-values and the fitted values. For this, I sorted the original values:
orig = df['p'].values
fitted = est.fittedvalues.values
args = np.argsort(orig)
import matplotlib.pyplot as plt
plt.plot(orig[args], 'bo')
plt.plot(orig[args]-resid[args], 'ro')
plt.show()
但是，这给了我一个图表，其中的值完全不可用.没有什么可以暗示0.9的R平方.因此，我尝试自己手动进行计算:
This, however, gave me a graph where the values were completely off. Nothing that would suggest an R-squared of 0.9. Therefore, I tried to calculate it manually myself:
yBar = df['p'].mean()
SSTot = df['p'].apply(lambda x: (x-yBar)**2).sum()
SSReg = ((est.fittedvalues - yBar)**2).sum()
1 - SSReg/SSTot
Out[79]: 0.2618159806908984
我做错什么了吗?还是有原因导致我的计算与statsmodels相差甚远? SSTot，SSReg的值分别为48084，35495.
Am I doing something wrong? Or is there a reason why my computation is so far off what statsmodels is getting? SSTot, SSReg have values of 48084, 35495.
推荐答案
如果模型中未包含截距(常量解释变量)，则statsmodels将基于 un-centred 计算R平方总平方和，即.
If you do not include an intercept (constant explanatory variable) in your model, statsmodels computes R-squared based on un-centred total sum of squares, ie.
tss = (ys ** 2).sum() # un-centred total sum of squares
相对于
tss = ((ys - ys.mean())**2).sum() # centred total sum of squares
因此，R平方会更高.
从数学上来说这是正确的 .因为，R平方应指示与简化模型相比，完整模型可以解释多少变化.如果您将模型定义为:
This is mathematically correct. Because, R-squared should indicate how much of the variation is explained by the full-model comparing to the reduced model. If you define your model as:
ys = beta1 . xs + beta0 + noise
则简化后的模型可以是:ys = beta0 + noise，其中beta0的估计值是样本平均值，因此我们具有:noise = ys - ys.mean().这就是去义来自具有拦截功能的模型中的地方.
then the reduced model can be: ys = beta0 + noise, where the estimate for beta0 is the sample average, thus we have: noise = ys - ys.mean(). That is where de-meaning comes from in a model with intercept.
但是通过类似这样的模型:
But from a model like:
ys = beta . xs + noise
您只能减少为:ys = noise.由于noise被假定为零均值，因此您不可以将ys不均值.因此，简化模型中无法解释的变化是不居中的平方总和.
you may only reduce to: ys = noise. Since noise is assumed zero-mean, you may not de-mean ys. Therefore, unexplained variation in the reduced model is the un-centred total sum of squares.
此处在rsquared项目下.将yBar设置为零，我希望您会得到相同的数字.
This is documented here under rsquared item. Set yBar equal to zero, and I would expect you will get the same number.
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