[英]Apply linear regression in a Log-Normal scale in the y axis and a Prob scale in the x axis
我正在尝试计算线性回归系数,但我不断收到与tuple
相关的错误。
我想用 Python plot 对数正态线性回归分布,并用以下数据计算截距b0
和斜率b1
,然后计算x=50
和x=84.1
的y
值。
x axis
应为概率比例, y axis
应为对数正常比例。 我不确定我写的方法对于在对数正态概率尺度上实现线性回归和计算系数是否正确。
我正在使用的代码是:
from matplotlib import pyplot as plt
import seaborn
import probscale
from pylab import *
import numpy as np
# Permeability values (mD)
y = [283, 650, 565, 407, 714, 500, 730, 900, 420, 591, 381, 430, 324, 440, 1212, 315, 450]
# Permeability values in descending order (y, mD)
y.sort(reverse = True)
print('Permeability values in Descending Order :', y)
# Percentage of samples with larger permeability (x, %)
x = tuple([round(n/len(y)*100, 1) for n in range(len(y))])
print('Percentage of samples with larger permeability :', x)
# Plot
fig, ax = plt.subplots(figsize=(10, 8))
ax.set_xlim(0.01, 99)
ax.set_xscale('prob')
ax.set_ylim(1e0, 1e4)
ax.set_yscale('log')
seaborn.despine(fig=fig)
plt.plot(x, y, 'go')
plt.title('Permeability Variation')
plt.ylabel('Permebility, md')
plt.xlabel('Percent of Samples with Larger Permeability, %')
plt.grid(True)
plt.show()
# Mean for x and y
mean_x = np.mean(x)
mean_y = np.mean(y)
# Total number of values
m = len(x)
# Calculate b1 and b0
numer = 0
denom = 0
for i in range(m):
numer += (x[i] - mean_x) * (y[i] - mean_y)
denom += (x[i] - mean_x) ** 2
b1 = numer / denom
b0 = mean_y - (b1 * mean_x)
# Print coefficients
print('b1 = ', b1, 'b0 = ', b0)
# Calculate permeability at 84.1% and 50% probability (Percentiles)
# Calculate variance for permeability distribution (VDP)
k1 = b0 + b1 * 50
k2 = b0 + b1 * 84.1
# Dykstra Parsons Formula 'VDP' (k1=@50% Percentile and k2=@84.1% Percentile)
vdp = (k1 - k2) / k1
print('vdp = ', vdp)
# Calculate r^2 score (Coefficient of Correlation)
sumofsquares = 0
sumofresiduals = 0
for i in range(m):
y_pred = b0 + b1 * x[i]
sumofsquares += (y[i] - mean_y) ** 2
sumofresiduals += (y[i] - y_pred) ** 2
score = 1 - (sumofresiduals / sumofsquares)
print('R^2 score = ', score)[![enter image description here][1]][1]
理想情况下,它看起来像这样,具有最佳拟合的线性回归直线。 (这只是一个例子)[1]: https://i.stack.imgur.com/nZG7W.png
您正在有效地计算方差的协方差,从而为您提供回归系数。 您可以使用列表推导或 numpy 计算所有值,但是是的,它是正确的。
我不确定,只有你能回答的一件事是,是 x 和 y 之间的线性关系,还是 x 和 log(y) 之间的线性关系?
下面我使用scipy.stats.linregress()
来回归它和 plot,你可以看到 b1 和 b0 是相同的,希望这能回答你的问题:
from matplotlib import pyplot as plt
import seaborn
import probscale
import numpy as np
from scipy import stats
y = [283, 650, 565, 407, 714, 500, 730, 900, 420, 591, 381, 430, 324, 440, 1212, 315, 450]
y.sort(reverse = True)
x = tuple([round(n/len(y)*100, 1) for n in range(len(y))])
slope, intercept, r_value, p_value, std_err = stats.linregress(x,y)
fig, ax = plt.subplots(figsize=(10, 8))
ax.set_xlim(0.01, 99)
ax.set_xscale('prob')
ax.set_ylim(1e0, 1e4)
ax.set_yscale('log')
seaborn.despine(fig=fig)
plt.plot(x, y, 'go')
plt.plot(x,list(map(lambda i:intercept+slope*i,x)), '--k')
print(slope,intercept)
-7.2679081487585195 889.7839128827538
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