In [29]:
import pandas as pa
import numpy as np
import matplotlib.pyplot as plt
In [30]:
data = pa.read_csv('food_truck.csv')
data.head()
Out[30]:
Population(10.000) | Profit(10.000 $) | |
---|---|---|
0 | 6.1101 | 17.5920 |
1 | 5.5277 | 9.1302 |
2 | 8.5186 | 13.6620 |
3 | 7.0032 | 11.8540 |
4 | 5.8598 | 6.8233 |
In [31]:
X = data['Population(10.000)']
y = data['Profit(10.000 $)']
Regression lineaire (à la main)¶
In [32]:
x_bar = np.mean(X)
y_bar = np.mean(y)
m = len(X)
t1 = [(X[i] - x_bar) * (y[i] - y_bar) for i in range(m)]
t2 = [(X[i] - x_bar)**2 for i in range(m)]
b1 = np.sum(t1) / np.sum(t2)
b0 = y_bar - b1 * x_bar
y_hat = [b1 * X[i] + b0 for i in range(m)]
In [33]:
plt.grid()
plt.scatter(X, y)
plt.plot(X, y_hat, c='red')
Out[33]:
[<matplotlib.lines.Line2D at 0xffff2f8591d0>]
Regression lineaire (avec Stats
)¶
In [34]:
from scipy import stats
help(stats.linregress)
Help on function linregress in module scipy.stats._stats_mstats_common: linregress(x, y=None, alternative='two-sided') Calculate a linear least-squares regression for two sets of measurements. Parameters ---------- x, y : array_like Two sets of measurements. Both arrays should have the same length. If only `x` is given (and ``y=None``), then it must be a two-dimensional array where one dimension has length 2. The two sets of measurements are then found by splitting the array along the length-2 dimension. In the case where ``y=None`` and `x` is a 2x2 array, ``linregress(x)`` is equivalent to ``linregress(x[0], x[1])``. alternative : {'two-sided', 'less', 'greater'}, optional Defines the alternative hypothesis. Default is 'two-sided'. The following options are available: * 'two-sided': the slope of the regression line is nonzero * 'less': the slope of the regression line is less than zero * 'greater': the slope of the regression line is greater than zero .. versionadded:: 1.7.0 Returns ------- result : ``LinregressResult`` instance The return value is an object with the following attributes: slope : float Slope of the regression line. intercept : float Intercept of the regression line. rvalue : float The Pearson correlation coefficient. The square of ``rvalue`` is equal to the coefficient of determination. pvalue : float The p-value for a hypothesis test whose null hypothesis is that the slope is zero, using Wald Test with t-distribution of the test statistic. See `alternative` above for alternative hypotheses. stderr : float Standard error of the estimated slope (gradient), under the assumption of residual normality. intercept_stderr : float Standard error of the estimated intercept, under the assumption of residual normality. See Also -------- scipy.optimize.curve_fit : Use non-linear least squares to fit a function to data. scipy.optimize.leastsq : Minimize the sum of squares of a set of equations. Notes ----- Missing values are considered pair-wise: if a value is missing in `x`, the corresponding value in `y` is masked. For compatibility with older versions of SciPy, the return value acts like a ``namedtuple`` of length 5, with fields ``slope``, ``intercept``, ``rvalue``, ``pvalue`` and ``stderr``, so one can continue to write:: slope, intercept, r, p, se = linregress(x, y) With that style, however, the standard error of the intercept is not available. To have access to all the computed values, including the standard error of the intercept, use the return value as an object with attributes, e.g.:: result = linregress(x, y) print(result.intercept, result.intercept_stderr) Examples -------- >>> import numpy as np >>> import matplotlib.pyplot as plt >>> from scipy import stats >>> rng = np.random.default_rng() Generate some data: >>> x = rng.random(10) >>> y = 1.6*x + rng.random(10) Perform the linear regression: >>> res = stats.linregress(x, y) Coefficient of determination (R-squared): >>> print(f"R-squared: {res.rvalue**2:.6f}") R-squared: 0.717533 Plot the data along with the fitted line: >>> plt.plot(x, y, 'o', label='original data') >>> plt.plot(x, res.intercept + res.slope*x, 'r', label='fitted line') >>> plt.legend() >>> plt.show() Calculate 95% confidence interval on slope and intercept: >>> # Two-sided inverse Students t-distribution >>> # p - probability, df - degrees of freedom >>> from scipy.stats import t >>> tinv = lambda p, df: abs(t.ppf(p/2, df)) >>> ts = tinv(0.05, len(x)-2) >>> print(f"slope (95%): {res.slope:.6f} +/- {ts*res.stderr:.6f}") slope (95%): 1.453392 +/- 0.743465 >>> print(f"intercept (95%): {res.intercept:.6f}" ... f" +/- {ts*res.intercept_stderr:.6f}") intercept (95%): 0.616950 +/- 0.544475
In [35]:
slope, intercept, r, p, se = stats.linregress(X, y)
In [21]:
y_hat_bis = slope * X + intercept
plt.grid()
plt.scatter(X, y)
plt.plot(X, y_hat, c='red')
plt.plot(X, y_hat_bis, c='green')
Out[21]:
[<matplotlib.lines.Line2D at 0xffff41812c10>]
In [36]:
R2 = r**2
print('R2: {:.2%}'.format(R2))
R2: 70.20%
70% de la variabilité du profit est expliquée par la population.
In [40]:
print('p_value: {:.2%}'.format(p))
print('p_value: ',p)
p_value: 0.00% p_value: 1.0232099778760524e-26
In [ ]: