Question: I want to implement a decision tree with each leaf being a linear regression, does such a model exist (preferable in sklearn)?

Example case 1:

Mockup data is generated using the formula:

y = int(x) + x * 1.5

Which looks like:

enter image description here

I want to solve this using a decision tree where the final decision results in a linear formula. Something like:

  1. 0 <= x < 1 -> y = 0 + 1.5 * x
  2. 1 <= x < 2 -> y = 1 + 1.5 * x
  3. 2 <= x < 3 -> y = 2 + 1.5 * x
  4. etc.

Which I figured could best be done using a decision tree. I've done some googling and I thought the DecisionTreeRegressor from sklearn.tree could work, but that results in points being assigned a constant value in a range, as shown below: enter image description here


import matplotlib.pyplot as plt
import numpy as np
from sklearn.tree import DecisionTreeRegressor

x = np.linspace(0, 5, 100)
y = np.array([int(i) for i in x]) + x * 1.5

x_train = np.linspace(0, 5, 10)
y_train = np.array([int(i) for i in x_train]) + x_train * 1.5

clf = DecisionTreeRegressor()
clf.fit(x_train.reshape((len(x_train), 1)), y_train.reshape((len(x_train), 1)))

y_result = clf.predict(x.reshape(len(x), 1))
plt.plot(x, y, label='actual results')
plt.plot(x, y_result, label='model predicts')

Example case 2: Instead of one input, there are two inputs: x1 and x2, output is computed by:

  1. x1 = 0 -> y = 1 * x2
  2. x1 = 1 -> y = 3 * x2 + 5
  3. x1 = 6 -> y = -1 * x2 -4
  4. else -> y = x2 * 20 - 100

enter image description here


import matplotlib.pyplot as plt
import random

def get_y(x1, x2):
    if x1 == 0:
        return x2
    if x1 == 1:
        return 3 * x2 + 5
    if x1 == 6:
        return - x2 - 4
    return x2 * 20 - 100

X_0 = [(0, random.random()) for _ in range(100)]
x2_0 = [i[1] for i in X_0]
y_0 = [get_y(i[0], i[1]) for i in X_0]
X_1 = [(1, random.random()) for _ in range(100)]
x2_1 = [i[1] for i in X_1]
y_1 = [get_y(i[0], i[1]) for i in X_1]
X_2 = [(6, random.random()) for _ in range(100)]
x2_2 = [i[1] for i in X_2]
y_2 = [get_y(i[0], i[1]) for i in X_2]
X_3 = [(random.randint(10, 100), random.random()) for _ in range(100)]
x2_3 = [i[1] for i in X_3]
y_3 = [get_y(i[0], i[1]) for i in X_3]
plt.scatter(x2_0, y_0, label='x1 = 0')
plt.scatter(x2_1, y_1, label='x1 = 1')
plt.scatter(x2_2, y_2, label='x1 = 6')
plt.scatter(x2_3, y_3, label='x1 not 0, 1 or 6')

So my question is: does a decision tree with each leaf being a linear regression, exist?

  • $\begingroup$ For anybody interested, I've started working on a project to implement an algorhytm similar to the one described in @BenReiniger answer over at gitlab.com/nathan_vanthof/… $\endgroup$
    – Nathan
    Commented Dec 29, 2019 at 23:19
  • $\begingroup$ Your updated question is more general: modeling a piecewise-linear function where the slopes can change, so they're really n different lines, not just the same line with discontinuities (as in the first example). Also, please add a plot of the updated question, currently the question is a mishmash of both askings, hence hard to understand. And presumably that should be 'x2' on the x-axis of the updated question, not 'x'. $\endgroup$
    – smci
    Commented Dec 30, 2019 at 9:20
  • $\begingroup$ @smci my question is: is there is an existing decision tree with the end leaf being a linear regression? The problem statement is just an example case (minimal replicable problem), changing the example case does not change the question. $\endgroup$
    – Nathan
    Commented Dec 30, 2019 at 10:21
  • $\begingroup$ @smci I've restructured my question, I hope it's clear the question did not change, only the example case I gave :) $\endgroup$
    – Nathan
    Commented Dec 30, 2019 at 10:36
  • 1
    $\begingroup$ I think you are looking for a piecewise linear regression. It does exactly what you need: break the space of y into segments of different lenght, and run a separate linear regression on each of those. $\endgroup$
    – Leevo
    Commented Dec 30, 2019 at 16:57

5 Answers 5


You are looking for Linear Trees.

Linear Trees differ from Decision Trees because they compute linear approximation (instead of constant ones) fitting simple Linear Models in the leaves.

For a project of mine, I developed linear-tree: a python library to build Model Trees with Linear Models at the leaves.

enter image description here

linear-tree is developed to be fully integrable with scikit-learn.

from sklearn.linear_model import *
from lineartree import LinearTreeRegressor, LinearTreeClassifier

regr = LinearTreeRegressor(base_estimator=LinearRegression())
regr.fit(X, y)

clf = LinearTreeClassifier(base_estimator=RidgeClassifier())
clf.fit(X, y)

LinearTreeRegressor and LinearTreeClassifier are provided as scikit-learn BaseEstimator. They are wrappers that build a decision tree on the data fitting a linear estimator from sklearn.linear_model. All the models available in sklearn.linear_model can be used as linear estimators.

Compare Decision Tree with Linear Tree:

enter image description here

enter image description here

  • $\begingroup$ does this work on big big data/tree ? e.g. 1M leafs ? each leaf 80 feature ? $\endgroup$
    – user702846
    Commented May 20, 2021 at 12:47
  • $\begingroup$ it can reach a max_depth of 20, but consider that linear trees usually converge faster than decision trees, i.e. less deep structures... the more you want to go deep or be precise in the splits, the higher the training time $\endgroup$ Commented May 20, 2021 at 12:56

I think the easiest way to do this would be to have a decision tree where the final decision results in a linear formula.

Setting aside whether this is actually easiest/best, this type of model does exist, usually called "model-based recursive partitioning". See e.g. https://stats.stackexchange.com/q/78563/232706
There are several packages in R for this: partyfit (and the older simpler party), mob, Cubist; unfortunately, there don't seem to be any in Python yet. Here's a discussion on including it in sklearn, from mid-2018.


I would suggest using spline regression. or some polynomial regression.

Why? what you are basically approximating is the (almost) step-wise function. See here

More info in here and a great background introduction in here.

  • $\begingroup$ Thank you for your answer. It does answer my question but I now realize I simplified it too much. I've added some further explanation on what I exactly need. $\endgroup$
    – Nathan
    Commented Dec 29, 2019 at 14:36

Even after your update, I think Noah's hint to spline regression is the best way to approach the problem. Here is a brief example in R:

# Generate data
x <- -50:100
y <- 0.001*x^3
df = data.frame(y,x)

# Linear regression
pred_ols = predict(reg_ols, newdata=df)

# GAM with regression splines
reg_gam = gam(y~s(x,5), data=df)
pred_gam = predict(reg_gam, newdata=df)

# Plot prediction and actual data
df2 = data.frame(x,y,pred_ols, pred_gam)
ggplot(df2, aes(x)) +                    
  geom_line(aes(y=y),size=1, colour="red") +  
  geom_line(aes(y=pred_ols),size=1, colour="blue") +
  geom_line(aes(y=pred_gam),size=1, colour="black", linetype = "dashed")  

So I have some function which is the data generating process (red line in the plot) and I want to get a good fit on this function. OLS (linear regression) will not work well (blue line in plot) but a GAM with splines will fit very well (black dashed).

enter image description here

This model looks like $y_i=\beta_0 + \beta_1 x_{1,i} + u_i$ (so 2D-like), but of course you can expand the model to $y_i=\beta_0 + \beta_1 x_{1,i} + ... + \beta_k x_{k,i} + u_i$, where $k$ is the number of "variables" in the model (so kD-like).

Since you seem to be on Python, you would need to look for a Py implementation of GAMs, such as Statsmodels of PyGAM.

Chapter 7 of "Introduction to Statistical Learning" covers splines regression. You may have a look at the Python Lab for the book.


Create a new column based on your formula and train your decision tree. Then, based on the outcome class of the tree, pass the data to different regression.

This would be a sort of Stacking.enter image description here

Please treat this just an engineering approach, it might not be a good solution based on your data and need.


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