如何为时间序列局部极值点绘制最佳拟合线或拟合多数点的线条

Hey there! Let's figure out how to add those fitted lines with uncertainty bands for your local minima and maxima in the time series plot. You're already halfway there with finding the extreme points—let's build on that code.

First, a quick note: Time series dates can't be directly used in statistical fitting functions (they need numerical values), so we'll convert datetime indices to matplotlib's numeric date format first. Then we'll compute the fit lines and their confidence intervals, just like the tutorial you mentioned.

Here's the updated code with explanations: #

First, make sure you add the missing import for datetime at the top:

import datetime as dt
from scipy import stats

Then, right after your existing code that finds the minima/maxima and plots the initial scatter points, add this:

# ----------------------
# Process Local Minima
# ----------------------
# Extract valid minima points (drop NaNs)
min_points = index_df.dropna(subset=['min'])[['min']].reset_index()
# Convert dates to numerical values for fitting
min_x_num = plt.date2num(min_points['Date'])
min_y = min_points['min']

# Perform linear regression fit
min_slope, min_intercept, _, _, min_std_err = stats.linregress(min_x_num, min_y)
min_fit_line = min_slope * min_x_num + min_intercept

# Calculate 95% confidence interval for the fit
n_min = len(min_x_num)
dof_min = n_min - 2  # degrees of freedom for linear fit
t_critical = stats.t.ppf(0.975, dof_min)  # 95% confidence level
min_residuals = min_y - min_fit_line
min_mse = np.mean(min_residuals ** 2)
min_ci = t_critical * np.sqrt(min_mse * (1/n_min + (min_x_num - np.mean(min_x_num))**2 / np.sum((min_x_num - np.mean(min_x_num))**2)))

# Convert numerical dates back to datetime for plotting
min_x_dates = plt.num2date(min_x_num)

# ----------------------
# Process Local Maxima
# ----------------------
# Same steps for maxima
max_points = index_df.dropna(subset=['max'])[['max']].reset_index()
max_x_num = plt.date2num(max_points['Date'])
max_y = max_points['max']

max_slope, max_intercept, _, _, max_std_err = stats.linregress(max_x_num, max_y)
max_fit_line = max_slope * max_x_num + max_intercept

n_max = len(max_x_num)
dof_max = n_max - 2
max_residuals = max_y - max_fit_line
max_mse = np.mean(max_residuals ** 2)
max_ci = t_critical * np.sqrt(max_mse * (1/n_max + (max_x_num - np.mean(max_x_num))**2 / np.sum((max_x_num - np.mean(max_x_num))**2)))

max_x_dates = plt.num2date(max_x_num)

# ----------------------
# Final Plotting
# ----------------------
plt.figure(figsize=(12, 6))
# Plot original price data
plt.plot(index_df['Adj Close'].index, index_df['Adj Close'], label='S&P 500 Adj Close', alpha=0.5)
# Plot extreme points
plt.scatter(min_x_dates, min_y, c='r', label='Local Minima')
plt.scatter(max_x_dates, max_y, c='g', label='Local Maxima')
# Plot fit lines and confidence bands
plt.plot(min_x_dates, min_fit_line, 'r--', label='Minima Fit Line')
plt.fill_between(min_x_dates, min_fit_line - min_ci, min_fit_line + min_ci, color='r', alpha=0.2)

plt.plot(max_x_dates, max_fit_line, 'g--', label='Maxima Fit Line')
plt.fill_between(max_x_dates, max_fit_line - max_ci, max_fit_line + max_ci, color='g', alpha=0.2)

# Format plot
plt.title('S&P 500 with Local Extremes & Fitted Lines (95% Confidence)')
plt.xlabel('Date')
plt.ylabel('Adjusted Close Price')
plt.legend()
plt.xticks(rotation=45)
plt.tight_layout()
plt.show()

Key details to note: #

• Date conversion: We use plt.date2num() to turn datetime objects into numbers that fitting functions can work with, then convert back with plt.num2date() for plotting.
• Confidence intervals: The shaded areas represent the 95% confidence range for the fit—this means we're 95% confident the true trend line lies within that band.
• Non-linear fits (optional): If you want a curved fit instead of linear, replace the stats.linregress part with np.polyfit (e.g., np.polyfit(min_x_num, min_y, 3) for a 3rd-degree polynomial). You can calculate confidence bands for polynomials using residual standard deviation as a simpler estimate if needed.

内容的提问来源于stack exchange，提问作者Yi Wen Edwin Ang