在 y 轴上以对数正态尺度应用线性回归，在 x 轴上应用概率尺度

Question

I am trying to calculate linear regression coefficients, but I keep getting errors related to tuple .

I want to plot a log normal linear regression distribution with Python and calculate the intercept b0 & slope b1 with the following data, and then calculate the y value for x=50 and x=84.1 .

The x axis should be in prob scale and the y axis in log normal scale. I am not sure if the method I wrote is correct for implementing linear regression on a log normal-prob scale and calculating the coefficients.

The code I am using is:

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]

Ideally it would look like something like this, with linear regression straight line of best-fit. ( this is just an example) [1]: https://i.stack.imgur.com/nZG7W.png

Answer 1

you are effectively calculating the covariance over variance to give you the regression coefficient. You can calculate all the values using a list comprehension or numpy, but yes, it's correct.

The one thing I am not sure of, and only you can answer is, is the linear relationship between x and y, or x and log(y)?

Below I use scipy.stats.linregress() to regress it out and plot, you can see the b1 and b0 are the same, hope this answers your question:

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