问题
Here is my sample data:
x = np.array([19.0, 47.0, 34.6, 23.2, 33.5, 28.2,34.8, 15.8, 23.8])
y = np.array([6.12,3.55, 2.67, 2.81, 5.34, 3.75,3.43, 1.44, 0.84])
pl.scatter(x,y, facecolors='b', edgecolors='b', s=24)
x = x[:,np.newaxis]
a, _, _, _ = np.linalg.lstsq(x, y)
pl.plot(x, a*x, 'r-')
pl.xlim(0,50)
pl.ylim(0,7)
You can see in the plot that the linear fit does not reach y=0. How can I find the x-value (i.e. extrapolate the data) at which y=0? And is there a way to get do an error propagation to get the errors for the coefficient?
回答1:
To extrapolate, you just need to pass to plot a longer y array.
Just insert 0 to the array after you fit the line.
y = np.insert(y, 0, 0)
And then pass to plot:
pl.plot(y/a, y, 'r-')
回答2:
the statsmodels package might be easier to use than the relatively low-level lstsq function that's in Numpy. your question is just estimating:
y_i = x_i*a + sigma_i
therefore x=0 will always be at y=0. you might be expecting your code to be estimating:
y_i = a_0 + x_i*a_1 + sigma_i
i.e a_0 is the intercept, and a_1 is the x coefficient.
using statsmodels would require pulling in more packages, but has a much easier interface:
import statsmodels.formula.api as smf
import pandas as pd
df = pd.DataFrame(dict(x=x, y=y))
fit = smf.ols('y ~ x', df).fit()
fit.summary()
would print out:
coef std err t P>|t| [0.025 0.975] Intercept 2.4528 1.960 1.251 0.251 -2.183 7.088 x 0.0303 0.065 0.468 0.654 -0.123 0.183
and you can get x where y=0 via:
-fit.params[0] / fit.params[1]
giving approx -81. if you really wanted to fix the intercept as zero, you would add + 0 to the formula:
fit = smf.ols('y ~ x + 0', df).fit()
this interface goes against the Python "explicit is better than implicit" rule, but copies the "R" language style formulas and (in my experience) most regressions want to estimate the intercept anyway.
来源：https://stackoverflow.com/questions/58057314/how-to-extrapolate-simple-linear-regression-and-get-errors-for-the-coefficients