Python/Matplotlib：在给定截距和斜率的情况下向图中添加回归线
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Python/Matplotlib: adding regression line to a plot given its intercept and slope
提问by Jonathan
Using the following small dataset:
使用以下小数据集：
bill = [34,108,64,88,99,51]
tip = [5,17,11,8,14,5]
I calculated a best-fit regression line (by hand).
我计算了一个最合适的回归线（手工）。
yi = 0.1462*x - 0.8188 #yi = slope(x) + intercept
I've plotted my original data using Matplotlib like this:
我已经使用 Matplotlib 绘制了我的原始数据，如下所示：
scatter(bill,tip, color="black")
plt.xlim(20,120) #set ranges
plt.ylim(4,18)
#plot centroid point (mean of each variable (74,10))
line1 = plt.plot([74, 74],[0,10], ':', c="red")
line2 = plt.plot([0,74],[10,10],':', c="red")
plt.scatter(74,10, c="red")
#annotate the centroid point
plt.annotate('centroid (74,10)', xy=(74.1,10), xytext=(81,9),
arrowprops=dict(facecolor="black", shrink=0.01),
)
#label axes
xlabel("Bill amount ($)")
ylabel("Tip amount ($)")
#display plot
plt.show()
I am unsure how to get the regression line onto the plot itself. I'm aware that there are plenty of builtin stuff for quickly fitting and displaying best fit lines, but I did this as practice. I know I can start the line at points '0,0.8188' (the intercept), but I don't know how to use the slope value to complete the line (set the lines end points).
我不确定如何将回归线放到图本身上。我知道有很多内置的东西可以快速拟合和显示最佳拟合线，但我这样做是为了练习。我知道我可以在点“0,0.8188”（截距）处开始线，但我不知道如何使用斜率值来完成线（设置线端点）。
Given that for each increase on the x axis, the slope should increase by '0.1462'; for the line coordinates I tried (0,0.8188) for the starting point, and (100,14.62) for the end point. But this line does not pass through my centroid point. It just misses it.
鉴于 x 轴上的每次增加，斜率应增加“0.1462”；对于我尝试的线坐标（0,0.8188）作为起点，（100,14.62）作为终点。但是这条线没有通过我的质心点。它只是想念它。
Cheers, Jon
干杯，乔恩
采纳答案by ImportanceOfBeingErnest
The reasoning in the question partially correct. Having a function f(x) = a*x +b, you may take as first point the interception with the y axis (x=0) as (0, b)(or (0,-0.8188)in this case).
Any other point on that line is given by (x, f(x)), or (x, a*x+b). So looking at the point at x=100 would give you (100, f(100)), plugging in: (100, 0.1462*100-0.8188)= (100,13.8012).
In the case you describe in the question you just forgot to take the binto account.
问题中的推理部分正确。有一个函数f(x) = a*x +b，您可以将与 y 轴 (x=0) 的截距作为第一点(0, b)（或(0,-0.8188)在这种情况下）。
该线上的任何其他点由(x, f(x)), 或给出(x, a*x+b)。因此，查看 x=100 处的点会给您(100, f(100))，插入：(100, 0.1462*100-0.8188)= (100,13.8012)。在您在问题中描述的情况下，您只是忘记考虑了b。
The following shows how to use that function to plot the line in matplotlib:
下面显示了如何使用该函数在 matplotlib 中绘制线条：
import matplotlib.pyplot as plt
import numpy as np
bill = [34,108,64,88,99,51]
tip = [5,17,11,8,14,5]
plt.scatter(bill, tip)
#fit function
f = lambda x: 0.1462*x - 0.8188
# x values of line to plot
x = np.array([0,100])
# plot fit
plt.plot(x,f(x),lw=2.5, c="k",label="fit line between 0 and 100")
#better take min and max of x values
x = np.array([min(bill),max(bill)])
plt.plot(x,f(x), c="orange", label="fit line between min and max")
plt.legend()
plt.show()
Of course the fitting can also be done automatically. You can obtain the slope and intercept from a call to numpy.polyfit:
当然，拟合也可以自动完成。您可以从调用中获取斜率和截距numpy.polyfit：
#fit function
a, b = np.polyfit(np.array(bill), np.array(tip), deg=1)
f = lambda x: a*x + b
The rest in the plot would stay the same.
情节中的其余部分将保持不变。
回答by f5r5e5d
define function fit, get endpoints of data, put in tuples to plot()
定义函数拟合，获取数据端点，将元组放入 plot()
def fit(x):
return 0.1462*x - 0.8188 #yi = slope(x) - intercept
xfit, yfit = (min(bill), max(bill)), (fit(min(bill)), fit(max(bill)))
plt.plot(xfit, yfit,'b')
回答by Enzo Calogero
quick note: I think that the formula for the regression should be
快速说明：我认为回归的公式应该是
return 0.1462*x + 0.8188 #yi = slope(x) + intercept
I mean + instead of -.
我的意思是+而不是-。
