如何用arctan函数拟合近似给定数据集的曲线？

Hey there! Let's work through fitting your dataset with an arctangent function—since your data follows that classic sigmoidal, saturating shape of arctan(x), this is a great choice. I'll use R for all examples, matching your existing code.

Step 1: Organize Your Data #

First, let's wrap your data into a data frame for easier handling:

xs <- seq(1000,50000,by=1000)
data <- c(0.101, 0.169, 0.209, 0.256, 0.289, 0.373, 0.391, 0.418, 0.477, 0.528, 
          0.579, 0.570, 0.602, 0.657, 0.690, 0.720, 0.747, 0.764, 0.781, 0.811, 
          0.842, 0.847, 0.864, 0.889, 0.871, 0.894, 0.897, 0.915, 0.926, 0.931, 
          0.932, 0.940, 0.950, 0.963, 0.956, 0.967, 0.967, 0.963, 0.971, 0.980, 
          0.986, 0.988, 0.986, 0.985, 0.993, 0.992, 0.994, 0.991, 0.995, 0.993)
df <- data.frame(x = xs, y = data)

Step 2: Choose an Arctangent Model #

The standard arctan(x) ranges from -π/2 to π/2, but your data starts around 0.1 and saturates near 1. We need a scaled, shifted version of the function to match this. A flexible model for your data is:

y = k + m * atan(n * x)

• k: The baseline (y-value when x approaches 0)
• m: Scales the arctan output to match your data's range
• n: Controls how quickly the curve rises (smaller values = slower rise)

Based on your data's trend, we can guess initial parameters:

• k ≈ -0.03 (since x=1000 gives y≈0.1, and atan(n*1000) will be small)
• m ≈ 0.65 (to scale the arctan's max output (~π/2 ≈1.57) to reach near 1)
• n ≈ 0.0002 (so n*50000 ≈10, where atan(10) is close to π/2)

Step 3: Fit the Model with Nonlinear Least Squares #

We'll use R's nls() function for nonlinear fitting. If you run into convergence issues, the nlsLM() function from the minpack.lm package is more robust:

Option 1: Basic nls()

# Define the model
arctan_model <- y ~ k + m * atan(n * x)

# Fit with initial guesses
fit <- nls(arctan_model, data = df, start = list(k = -0.03, m = 0.65, n = 0.0002))

# View fit details
summary(fit)

Option 2: Robust Fit with nlsLM()

# Install the package if you haven't already
# install.packages("minpack.lm")
library(minpack.lm)

fit_lm <- nlsLM(arctan_model, data = df, start = list(k = -0.03, m = 0.65, n = 0.0002))
summary(fit_lm)

The summary will give you parameter estimates, standard errors, and goodness-of-fit metrics like the residual standard error.

Step 4: Visualize the Fit #

Let's plot your original data alongside the fitted curve to check how well it matches:

# Generate predicted values from the fit
df$y_pred <- predict(fit, newdata = df)

# Create the plot
plot(df$x, df$y, pch = 16, col = "steelblue", 
     main = "Arctangent Fit to Your Data",
     xlab = "x", ylab = "y")
lines(df$x, df$y_pred, col = "firebrick", lwd = 2)
legend("bottomright", 
       legend = c("Original Data", "Fitted Arctan Curve"),
       col = c("steelblue", "firebrick"),
       pch = c(16, NA), lty = c(NA, 1))

Step 5: Evaluate the Fit #

To make sure the fit is good, check the residuals (differences between observed and predicted values):

residuals <- df$y - df$y_pred
plot(df$x, residuals, pch = 16, col = "gray",
     main = "Residuals of Arctan Fit",
     xlab = "x", ylab = "Residual")
abline(h = 0, col = "firebrick", lwd = 2)

If residuals are randomly scattered around 0 (no obvious patterns), the model is a good fit.

Bonus: Try a Shifted Arctan Model #

If you want to account for a horizontal shift (moving the "inflection point" of the curve), use this model instead:

y = a * atan(b*(x - c)) + d

You can guess initial parameters (e.g., a=0.5, b=0.0001, c=15000, d=0.5) and fit it the same way as above. Compare AIC values between models—lower AIC means a better fit.

内容的提问来源于stack exchange，提问作者Andrew Davidson