【发布时间】：2022-01-17 15:38:20
【问题描述】：
我正在尝试以对数比例绘制我的数据，使用 Theil-Sen 回归作为最佳拟合线。但是，当我在对数尺度 2 上绘制回归线时，它与我的 x=y 线平行，我认为这是不正确的。
X 和 y 的正常比例：
X 和 y 的对数比例：
我发现chaooder 的一个相关解决方案对于半对数尺度上的线性回归会有所帮助。所以目前，我的回归线将从：
y = ax + c 在线性刻度上到 y = 10^^(a log(x)+c) 在我的对数刻度上。但在我的脑海中，我无法理解这是如何解决的，因为我无法计算 a。
这是数据：
index,x,y
0,0.22,0.26
1,0.39,0.1
2,0.4,0.17
3,0.56,0.41
4,0.57,0.12
5,0.62,0.54
6,0.78,0.99
7,0.79,0.35
8,0.8,0.33
9,0.83,0.91
10,0.95,0.81
11,1.08,0.23
12,1.34,0.11
13,1.34,0.44
14,1.35,0.11
15,1.58,0.24
16,1.66,0.71
17,2.11,0.54
18,2.13,0.42
19,2.19,1.72
20,2.25,2.16
21,2.39,0.95
22,2.4,0.16
23,2.73,0.92
24,2.87,1.1
25,2.96,0.27
26,3.12,1.66
27,3.26,0.06
28,3.28,0.68
29,3.34,0.7
30,3.38,1.14
31,3.39,1.81
32,3.41,0.19
33,3.49,1.4
34,3.52,1.57
35,3.6,0.99
36,3.64,1.28
37,3.65,1.68
38,3.89,1.66
39,3.93,1.64
40,4.01,1.04
41,4.07,0.32
42,4.22,0.68
43,4.52,0.57
44,4.53,0.59
45,4.56,0.7
46,4.6,1.15
47,4.62,1.31
48,4.68,1.09
49,5.03,0.48
50,5.06,0.7
51,5.31,0.62
52,5.41,0.21
53,5.45,2.06
54,6.0,0.72
55,6.06,0.36
56,6.64,1.41
57,6.74,0.59
58,6.96,0.95
59,7.01,1.13
60,7.14,1.56
61,7.14,2.82
62,7.19,1.49
63,7.21,0.88
64,7.23,1.31
65,7.55,0.76
66,7.72,0.5
67,7.75,1.65
68,7.77,1.48
69,7.9,1.8
70,7.95,0.68
71,8.03,1.12
72,8.09,2.61
73,8.86,1.71
74,9.31,0.23
75,9.5,2.35
76,9.62,1.84
77,9.91,0.56
78,9.95,1.67
79,10.4,1.15
80,10.8,0.88
81,11.28,1.8
82,11.31,1.58
83,11.43,1.0
84,12.38,2.83
85,13.38,1.45
86,13.9,1.99
87,30.3,1.99
还有我当前的代码：
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.ticker import AutoMinorLocator
from sklearn.linear_model import TheilSenRegressor
for log in [True, False]:
fig,ax = plt.subplots()
data.plot.scatter(ax=ax,
x='x',
y='y',
loglog=log)
vmin = np.amin(data[['x','y']].min().values)*0.8
vmax = np.amax(data[['x','y']].max().values)*1.25
ax.set_xlim(vmin,vmax)
ax.set_ylim(vmin,vmax)
ax.yaxis.set_minor_locator(AutoMinorLocator())
ax.xaxis.set_minor_locator(AutoMinorLocator())
# best fit (ThielSen) line
X = data.x.values[:,np.newaxis]
y = data.y.values
if log:
X = np.log10(X)
y = np.log10(y)
if len(y) > 0:
estimator = TheilSenRegressor(fit_intercept=False) # intentionally set intercept to 0
estimator.fit(X=X,y=y)
y0 = y[0]
x0 = X[0]
y_pred = estimator.predict(np.array([vmin,vmax]).reshape(2,1))
# y_pred = np.power(10,(estimator.predict(X)))
gradient = (y_pred[1] - y_pred[0]) / (vmax - vmin)
intercept = y_pred[1] - gradient * vmax
print(f'gradient: {gradient} \n intercept: {intercept}')
# Theil-Sen regression line
ax.plot([vmin,vmax],y_pred,color='red',lw=1,zorder=1,label='Best fit')
# 1:1 ratio line (black, dashed)
ax.plot([vmin,vmax],[vmin,vmax],lw=1,color='black',ls='--',alpha=0.6,zorder=1,
label='1:1 correlation')
if log:
ax.set_xscale('log');ax.set_yscale('log')
ax.set_title('log-log scale')
fig.savefig('TS_regression_loglog.png')
else:
ax.set_title('normal scale')
fig.savefig('TS_regression_normalscale.png')
【问题讨论】：
标签： python scikit-learn regression