I am trying to use scikit-learn for polynomial regression. From what I read polynomial regression is a special case of linear regression. I was hopping that maybe one of scikit's generalized linear models can be parameterised to fit higher order polynomials but I see no option for doing that.

I did manage to use a Support Vector Regressor with a poly kernel. That worked well with a subset of my data, but it takes much to long to fit larger data sets so I still need to find something faster (even if trading some precision).

Am I missing something obvious here?


Given data $\mathbf{x}$, a column vector, and $\mathbf{y}$, the target vector, you can perform polynomial regression by appending polynomials of $\mathbf{x}$. For example, consider if

$$ \mathbf{x} = \begin{bmatrix} 2 \\[0.3em] -1 \\[0.3em] \frac{1}{3} \end{bmatrix}$$

Using just this vector in linear regression implies the model:

$$ y = \alpha_1 x $$

We can add columns that are powers of the vector above, which represent adding polynomials to the regression. Below we show this for polynomials up to power 3: $$ X = \begin{bmatrix} 2 & 4 & 8 \\[0.3em] -1 & 1 & -1 \\[0.3em] \frac{1}{3} & \frac{1}{3^2} & \frac{1}{3^3} \end{bmatrix}$$

This is our new data matrix that we use in sklearn's linear regression, and it represents the model:

$$ y = \alpha_1 x + \alpha_2x^2 + \alpha_3x^3$$

Note that I did not add a constant vector of $1$'s, as sklearn will automatically include this.



Polynomial regression is a special case of linear regression. With the main idea of how do you select your features. Looking at the multivariate regression with 2 variables: x1 and x2. Linear regression will look like this: y = a1 * x1 + a2 * x2.

Now you want to have a polynomial regression (let's make 2 degree polynomial). We will create a few additional features: x1*x2, x1^2 and x2^2. So we will get your 'linear regression':

y = a1 * x1 + a2 * x2 + a3 * x1*x2 + a4 * x1^2 + a5 * x2^2

This nicely shows an important concept curse of dimensionality, because the number of new features grows much faster than linearly with the growth of degree of polynomial.

Practice with scikit-learn

You do not need to do all this in scikit. Polynomial regression is already available there (in 0.15 version. Check how to update it here).

from sklearn.preprocessing import PolynomialFeatures
from sklearn import linear_model

X = [[0.44, 0.68], [0.99, 0.23]]
vector = [109.85, 155.72]
predict= [[0.49, 0.18]]
#Edit: added second square bracket above to fix the ValueError problem

poly = PolynomialFeatures(degree=2)
X_ = poly.fit_transform(X)
predict_ = poly.fit_transform(predict)

clf = linear_model.LinearRegression()
clf.fit(X_, vector)
print clf.predict(predict_)
  • 1
    $\begingroup$ What if I do not want to have an interaction terms as x1*x2, do i have to construct X_ manually? there's a parameter "interaction_only" in PolynomialFeatures() constructor, and by default it is False. But setting it to True makes the opposite of what i want: it keeps ONLY the interaction terms, and do not keep x1^2, x2^2, etc. $\endgroup$ – DenisFLASH Dec 1 '14 at 14:59
  • $\begingroup$ The link to youtube claims that the video does not exist anymore. Do you have any other link to that? $\endgroup$ – Markon May 10 '16 at 9:06
  • $\begingroup$ @Markon any video from this list is good enough: youtube.com/results?search_query=curse+of+dimensionality $\endgroup$ – Salvador Dali May 10 '16 at 16:01
  • $\begingroup$ @SalvadorDali that is what dimensionality reduction is for $\endgroup$ – user3916597 Jul 6 '16 at 4:49
  • $\begingroup$ I wonder if we should center the data before or after applying PolynomialFeatures ? $\endgroup$ – renakre May 5 '17 at 20:09

In case you are using a multivariate regression and not just a univariate regression, do not forget the cross terms. For instance if you have two variables $x_1$ and $x_2$, and you want polynomials up to power 2, you should use $y = a_1x_1 + a_2x_2 + a_3x_1^2 + a_4x_2^2 + a_5x_1x_2$ where the last term ($a_5x_1x_2$) is the one I am talking about.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.