<p>正如@mujjiga所指出的,扩展是该过程的重要组成部分</p>
<p>我想提请大家注意另外两个关键点:</p>
<ul>
<li>模型选择决定您解决某类问题的能力</李>
<li>新的<code>scklearn</code>API,帮助您标准化解决方案开发</李>
</ul>
<p>让我们从数据集开始:</p>
<pre><code>import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
x = np.arange(14)
df = pd.DataFrame({'x': x, 'p': 15-x})
df['y'] = 1e4/df['p']
</code></pre>
<p>然后我们导入一些感兴趣的<code>sklearn</code>API对象:</p>
<pre><code>from sklearn.svm import SVR
from sklearn.linear_model import LinearRegression
from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import StandardScaler, RobustScaler, FunctionTransformer
</code></pre>
<p>首先,我们为目标值创建一个scaler函数:</p>
<pre><code>ysc = StandardScaler()
</code></pre>
<p>注意,我们可以使用不同的<a href="https://scikit-learn.org/stable/modules/classes.html#module-sklearn.preprocessing" rel="nofollow noreferrer">scalers</a>,或者构建一个<a href="https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.FunctionTransformer.html" rel="nofollow noreferrer">custom transformation</a></p>
<pre><code># Scaler robust against outliers:
ysc = RobustScaler()
# Logarithmic Transformation:
ysc = FunctionTransformer(func=np.log, inverse_func=np.exp, check_inverse=True)
</code></pre>
<p>我们使用我们选择的缩放器缩放目标:</p>
<pre><code>ysc.fit(df[['y']])
df['yn'] = ysc.transform(df[['y']])
</code></pre>
<p>我们还构建了一个<a href="https://scikit-learn.org/stable/modules/generated/sklearn.pipeline.make_pipeline.html#sklearn.pipeline.make_pipeline" rel="nofollow noreferrer">pipeline</a>,其中包含特征标准化器和<strong>所选模型</strong>(我们调整了参数以提高拟合度)。我们使用管道将其适配到您的数据集:</p>
<pre><code>reg = make_pipeline(StandardScaler(), SVR(kernel='rbf', C=1e3, epsilon=1e-3))
reg.fit(df[['p']], df['yn'])
</code></pre>
<p>此时,我们可以预测值并将其转换回原始比例:</p>
<pre><code>df['ynhat'] = reg.predict(df[['p']])
df['yhat'] = ysc.inverse_transform(df[['ynhat']])
</code></pre>
<p>我们检查适合度分数:</p>
<pre><code>reg.score(df[['p']], df['yn']) # 0.9999646718755011
</code></pre>
<p>我们还可以计算每个点的绝对和相对误差:</p>
<pre><code>df['yaerr'] = df['yhat'] - df['y']
df['yrerr'] = df['yaerr']/df['y']
</code></pre>
<p>最终结果是:</p>
<pre><code> x p y yn ynhat yhat yaerr yrerr
0 0 15 666.666667 -0.834823 -0.833633 668.077018 1.410352 0.002116
1 1 14 714.285714 -0.794636 -0.795247 713.562403 -0.723312 -0.001013
2 2 13 769.230769 -0.748267 -0.749627 767.619013 -1.611756 -0.002095
3 3 12 833.333333 -0.694169 -0.693498 834.128425 0.795091 0.000954
4 4 11 909.090909 -0.630235 -0.629048 910.497550 1.406641 0.001547
5 5 10 1000.000000 -0.553514 -0.555029 998.204445 -1.795555 -0.001796
6 6 9 1111.111111 -0.459744 -0.460002 1110.805275 -0.305836 -0.000275
7 7 8 1250.000000 -0.342532 -0.341099 1251.697707 1.697707 0.001358
8 8 7 1428.571429 -0.191830 -0.193295 1426.835676 -1.735753 -0.001215
9 9 6 1666.666667 0.009105 0.010458 1668.269984 1.603317 0.000962
10 10 5 2000.000000 0.290414 0.291060 2000.764717 0.764717 0.000382
11 11 4 2500.000000 0.712379 0.690511 2474.088446 -25.911554 -0.010365
12 12 3 3333.333333 1.415652 1.416874 3334.780642 1.447309 0.000434
13 13 2 5000.000000 2.822199 2.821420 4999.076799 -0.923201 -0.000185
</code></pre>
<p>从图形上看,它导致:</p>
<pre><code>fig, axe = plt.subplots()
axe.plot(df['p'], df['y'], label='$y(p)$')
axe.plot(df['p'], df['yhat'], 'o', label='$\hat{y}(p)$')
axe.set_title(r"SVR Fit for $y(x) = \frac{k}{x-a}$")
axe.set_xlabel('$p = x-a$')
axe.set_ylabel('$y, \hat{y}$')
axe.legend()
axe.grid()
</code></pre>
<p><a href="https://i.stack.imgur.com/X3IhE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X3IhE.png" alt="Fit Result"/></a></p>
<h2>线性化</h2>
<p>在上面的示例中,我们不能使用<code>poly</code>内核,而是必须使用<code>rbf</code>内核。这是因为,如果我们打算用多项式拟合有理函数,我们最好在拟合之前首先用<code>p = x/(x-b)</code>代换变换数据。在这种情况下,它将仅仅归结为执行线性回归。下面的示例显示了它的工作原理:</p>
<p>缩放器和转换也可以组合成管道。我们定义了一条线性化和缩放问题的管道:</p>
<pre><code># Rational Fraction Substitution with consecutive Standardization
ysc = make_pipeline(
FunctionTransformer(func=lambda x: x/(x+1),
inverse_func=lambda x: x/(1-x),
check_inverse=True),
StandardScaler()
)
</code></pre>
<p>然后我们可以使用经典的OLS回归数据:</p>
<pre><code>reg = make_pipeline(StandardScaler(), LinearRegression())
reg.fit(df[['p']], df['yn'])
</code></pre>
<p>这提供了正确的结果:</p>
<pre><code>reg.score(df[['p']], df['yn']) # 0.9999998722172933
</code></pre>
<p>第二种解决方案利用了已知的线性化,因此无需对模型进行参数化</p>