I'm using Python and Numpy to calculate a best fit polynomial of arbitrary degree. I pass a list of x values, y values, and the degree of the polynomial I want to fit (linear, quadratic, etc.).
This much works, but I also want to calculate r (coefficient of correlation) and r-squared(coefficient of determination). I am comparing my results with Excel's best-fit trendline capability, and the r-squared value it calculates. Using this, I know I am calculating r-squared correctly for linear best-fit (degree equals 1). However, my function does not work for polynomials with degree greater than 1.
Excel is able to do this. How do I calculate r-squared for higher-order polynomials using Numpy?
Here's my function:
import numpy
# Polynomial Regression
def polyfit(x, y, degree):
results = {}
coeffs = numpy.polyfit(x, y, degree)
# Polynomial Coefficients
results['polynomial'] = coeffs.tolist()
correlation = numpy.corrcoef(x, y)[0,1]
# r
results['correlation'] = correlation
# r-squared
results['determination'] = correlation**2
return results
This question is related to
python
math
statistics
numpy
curve-fitting
The answer is
Here's a very simple python function to compute R^2 from the actual and predicted values assuming y and y_hat are pandas series:
def r_squared(y, y_hat):
y_bar = y.mean()
ss_tot = ((y-y_bar)**2).sum()
ss_res = ((y-y_hat)**2).sum()
return 1 - (ss_res/ss_tot)
I originally posted the benchmarks below with the purpose of recommending numpy.corrcoef, foolishly not realizing that the original question already uses corrcoef and was in fact asking about higher order polynomial fits. I've added an actual solution to the polynomial r-squared question using statsmodels, and I've left the original benchmarks, which while off-topic, are potentially useful to someone.
statsmodels has the capability to calculate the r^2 of a polynomial fit directly, here are 2 methods...
import statsmodels.api as sm
import statsmodels.formula.api as smf
# Construct the columns for the different powers of x
def get_r2_statsmodels(x, y, k=1):
xpoly = np.column_stack([x**i for i in range(k+1)])
return sm.OLS(y, xpoly).fit().rsquared
# Use the formula API and construct a formula describing the polynomial
def get_r2_statsmodels_formula(x, y, k=1):
formula = 'y ~ 1 + ' + ' + '.join('I(x**{})'.format(i) for i in range(1, k+1))
data = {'x': x, 'y': y}
return smf.ols(formula, data).fit().rsquared # or rsquared_adj
To further take advantage of statsmodels, one should also look at the fitted model summary, which can be printed or displayed as a rich HTML table in Jupyter/IPython notebook. The results object provides access to many useful statistical metrics in addition to rsquared.
model = sm.OLS(y, xpoly)
results = model.fit()
results.summary()
Below is my original Answer where I benchmarked various linear regression r^2 methods...
The corrcoef function used in the Question calculates the correlation coefficient, r, only for a single linear regression, so it doesn't address the question of r^2 for higher order polynomial fits. However, for what it's worth, I've come to find that for linear regression, it is indeed the fastest and most direct method of calculating r.
def get_r2_numpy_corrcoef(x, y):
return np.corrcoef(x, y)[0, 1]**2
These were my timeit results from comparing a bunch of methods for 1000 random (x, y) points:
- Pure Python (direct
rcalculation)- 1000 loops, best of 3: 1.59 ms per loop
- Numpy polyfit (applicable to n-th degree polynomial fits)
- 1000 loops, best of 3: 326 µs per loop
- Numpy Manual (direct
rcalculation)- 10000 loops, best of 3: 62.1 µs per loop
- Numpy corrcoef (direct
rcalculation)- 10000 loops, best of 3: 56.6 µs per loop
- Scipy (linear regression with
ras an output)- 1000 loops, best of 3: 676 µs per loop
- Statsmodels (can do n-th degree polynomial and many other fits)
- 1000 loops, best of 3: 422 µs per loop
The corrcoef method narrowly beats calculating the r^2 "manually" using numpy methods. It is >5X faster than the polyfit method and ~12X faster than the scipy.linregress. Just to reinforce what numpy is doing for you, it's 28X faster than pure python. I'm not well-versed in things like numba and pypy, so someone else would have to fill those gaps, but I think this is plenty convincing to me that corrcoef is the best tool for calculating r for a simple linear regression.
Here's my benchmarking code. I copy-pasted from a Jupyter Notebook (hard not to call it an IPython Notebook...), so I apologize if anything broke on the way. The %timeit magic command requires IPython.
import numpy as np
from scipy import stats
import statsmodels.api as sm
import math
n=1000
x = np.random.rand(1000)*10
x.sort()
y = 10 * x + (5+np.random.randn(1000)*10-5)
x_list = list(x)
y_list = list(y)
def get_r2_numpy(x, y):
slope, intercept = np.polyfit(x, y, 1)
r_squared = 1 - (sum((y - (slope * x + intercept))**2) / ((len(y) - 1) * np.var(y, ddof=1)))
return r_squared
def get_r2_scipy(x, y):
_, _, r_value, _, _ = stats.linregress(x, y)
return r_value**2
def get_r2_statsmodels(x, y):
return sm.OLS(y, sm.add_constant(x)).fit().rsquared
def get_r2_python(x_list, y_list):
n = len(x_list)
x_bar = sum(x_list)/n
y_bar = sum(y_list)/n
x_std = math.sqrt(sum([(xi-x_bar)**2 for xi in x_list])/(n-1))
y_std = math.sqrt(sum([(yi-y_bar)**2 for yi in y_list])/(n-1))
zx = [(xi-x_bar)/x_std for xi in x_list]
zy = [(yi-y_bar)/y_std for yi in y_list]
r = sum(zxi*zyi for zxi, zyi in zip(zx, zy))/(n-1)
return r**2
def get_r2_numpy_manual(x, y):
zx = (x-np.mean(x))/np.std(x, ddof=1)
zy = (y-np.mean(y))/np.std(y, ddof=1)
r = np.sum(zx*zy)/(len(x)-1)
return r**2
def get_r2_numpy_corrcoef(x, y):
return np.corrcoef(x, y)[0, 1]**2
print('Python')
%timeit get_r2_python(x_list, y_list)
print('Numpy polyfit')
%timeit get_r2_numpy(x, y)
print('Numpy Manual')
%timeit get_r2_numpy_manual(x, y)
print('Numpy corrcoef')
%timeit get_r2_numpy_corrcoef(x, y)
print('Scipy')
%timeit get_r2_scipy(x, y)
print('Statsmodels')
%timeit get_r2_statsmodels(x, y)
R-squared is a statistic that only applies to linear regression.
Essentially, it measures how much variation in your data can be explained by the linear regression.
So, you calculate the "Total Sum of Squares", which is the total squared deviation of each of your outcome variables from their mean. . .
where y_bar is the mean of the y's.
Then, you calculate the "regression sum of squares", which is how much your FITTED values differ from the mean
and find the ratio of those two.
Now, all you would have to do for a polynomial fit is plug in the y_hat's from that model, but it's not accurate to call that r-squared.
Here is a link I found that speaks to it a little.
Here is a function to compute the weighted r-squared with Python and Numpy (most of the code comes from sklearn):
from __future__ import division
import numpy as np
def compute_r2_weighted(y_true, y_pred, weight):
sse = (weight * (y_true - y_pred) ** 2).sum(axis=0, dtype=np.float64)
tse = (weight * (y_true - np.average(
y_true, axis=0, weights=weight)) ** 2).sum(axis=0, dtype=np.float64)
r2_score = 1 - (sse / tse)
return r2_score, sse, tse
Example:
from __future__ import print_function, division
import sklearn.metrics
def compute_r2_weighted(y_true, y_pred, weight):
sse = (weight * (y_true - y_pred) ** 2).sum(axis=0, dtype=np.float64)
tse = (weight * (y_true - np.average(
y_true, axis=0, weights=weight)) ** 2).sum(axis=0, dtype=np.float64)
r2_score = 1 - (sse / tse)
return r2_score, sse, tse
def compute_r2(y_true, y_predicted):
sse = sum((y_true - y_predicted)**2)
tse = (len(y_true) - 1) * np.var(y_true, ddof=1)
r2_score = 1 - (sse / tse)
return r2_score, sse, tse
def main():
'''
Demonstrate the use of compute_r2_weighted() and checks the results against sklearn
'''
y_true = [3, -0.5, 2, 7]
y_pred = [2.5, 0.0, 2, 8]
weight = [1, 5, 1, 2]
r2_score = sklearn.metrics.r2_score(y_true, y_pred)
print('r2_score: {0}'.format(r2_score))
r2_score,_,_ = compute_r2(np.array(y_true), np.array(y_pred))
print('r2_score: {0}'.format(r2_score))
r2_score = sklearn.metrics.r2_score(y_true, y_pred,weight)
print('r2_score weighted: {0}'.format(r2_score))
r2_score,_,_ = compute_r2_weighted(np.array(y_true), np.array(y_pred), np.array(weight))
print('r2_score weighted: {0}'.format(r2_score))
if __name__ == "__main__":
main()
#cProfile.run('main()') # if you want to do some profiling
outputs:
r2_score: 0.9486081370449679
r2_score: 0.9486081370449679
r2_score weighted: 0.9573170731707317
r2_score weighted: 0.9573170731707317
This corresponds to the formula (mirror):
with f_i is the predicted value from the fit, y_{av} is the mean of the observed data y_i is the observed data value. w_i is the weighting applied to each data point, usually w_i=1. SSE is the sum of squares due to error and SST is the total sum of squares.
If interested, the code in R: https://gist.github.com/dhimmel/588d64a73fa4fef02c8f (mirror)
You can execute this code directly, this will find you the polynomial, and will find you the R-value you can put a comment down below if you need more explanation.
from scipy.stats import linregress
import numpy as np
x = np.array([1,2,3,4,5,6])
y = np.array([2,3,5,6,7,8])
p3 = np.polyfit(x,y,3) # 3rd degree polynomial, you can change it to any degree you want
xp = np.linspace(1,6,6) # 6 means the length of the line
poly_arr = np.polyval(p3,xp)
poly_list = [round(num, 3) for num in list(poly_arr)]
slope, intercept, r_value, p_value, std_err = linregress(x, poly_list)
print(r_value**2)
I have been using this successfully, where x and y are array-like.
def rsquared(x, y):
""" Return R^2 where x and y are array-like."""
slope, intercept, r_value, p_value, std_err = scipy.stats.linregress(x, y)
return r_value**2
The wikipedia article on r-squareds suggests that it may be used for general model fitting rather than just linear regression.
From scipy.stats.linregress source. They use the average sum of squares method.
import numpy as np
x = np.array(x)
y = np.array(y)
# average sum of squares:
ssxm, ssxym, ssyxm, ssym = np.cov(x, y, bias=1).flat
r_num = ssxym
r_den = np.sqrt(ssxm * ssym)
r = r_num / r_den
if r_den == 0.0:
r = 0.0
else:
r = r_num / r_den
if r > 1.0:
r = 1.0
elif r < -1.0:
r = -1.0
A very late reply, but just in case someone needs a ready function for this:
i.e.
slope, intercept, r_value, p_value, std_err = scipy.stats.linregress(x, y)
as in @Adam Marples's answer.
From yanl (yet-another-library) sklearn.metrics has an r2_score function;
from sklearn.metrics import r2_score
coefficient_of_dermination = r2_score(y, p(x))
Source: Stackoverflow.com