我有两个通过实验确定的变量 ( y1_exp, y2_exp)，它们随着 的变化而变化x，其中随着变化的变化而y2_exp影响（但不影响）。这是数据：y1_expxy1_expy2_exp
library(tidyverse)
library(minpack.lm)
data <- read_csv("x,y1_exp,y2_exp
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105,0.005618042,0.006004894
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1855,0.011587653,0.011161459
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1885,0.011704526,0.011198191
1890,0.011687294,0.011195365
1895,0.011753668,0.011204784
1900,0.011730171,0.01128841
1905,0.01169363,0.011272173
1910,0.011702284,0.011255141
1915,0.011734749,0.011251356
1920,0.011730171,0.011282397
1925,0.011712211,0.011309709
1930,0.011755791,0.011301904
1935,0.011715105,0.011313311
1940,0.011647555,0.011296201
1945,0.011616716,0.01129335
1950,0.011570587,0.011280993
1955,0.011526489,0.011261305
1960,0.011478671,0.011255611
1965,0.011398653,0.011232835
1970,0.011357835,0.011207538
1975,0.011266999,0.01117283
1980,0.011228383,0.011140089
1985,0.011168425,0.011120624
1990,0.011365904,0.011122848
1995,0.011927272,0.01128274
2000,0.012269557,0.011463536
2005,0.012610168,0.011623446
2010,0.012178617,0.011607966
2015,0.012009799,0.011539489
2020,0.011862455,0.011453127
2025,0.011738504,0.011402674
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2365,0.006991019,0.008483845
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使用一阶差分方程（递归计算函数），我计算y2_mod了每个的建模值x。我想拟合的参数是a（参数b是固定的）。我使用 nls.lm 函数进行拟合。这是代码的样子：
fitparms <- c(a = .008)
fn_residuals <- function(fitparms, data) {
b <- .00111
t <- data$x
dT <- t[2] - t[1]
N <- numeric(length(t))
y1_exp <- data$y1_exp
y2_mod <- numeric(length(t))
y2_mod[1] <- data$y2_exp[1]
for (h in seq_len(length(t)-1)) {
y2_mod[h+1] <- (fitparms[1] * (y1_exp[h+1] - y2_mod[h]) + b * (y1_exp[h+1] - y2_mod[h])) * dT + y2_mod[h]
}
data_2 <<- data %>%
add_column(y2_mod = y2_mod)
# residuals
ssqres <- data_2$y2_mod - data_2$y2_exp
# # return predicted vs experimental residual
return(ssqres)
}
# Ploting data
data_2 %>%
gather(y1_exp, y2_exp, y2_mod, key = "y", value = "value") %>%
ggplot(aes(x = x)) +
geom_line(aes(y = value, color = y))
# Fitting
fitval <- nls.lm(par = fitparms,
fn = fn_residuals,
lower = rep(0, 1), #defined lower limit for parameter
control = nls.lm.control(nprint = 1),
data = data)
这是我的输出：
It. 0, RSS = 0.00020291, Par. = 0.008
It. 1, RSS = 8.10975e-05, Par. = 0.00326637
It. 2, RSS = 7.9177e-05, Par. = 0.00349277
It. 3, RSS = 7.90245e-05, Par. = 0.00356081
It. 4, RSS = 7.90137e-05, Par. = 0.00357927
It. 5, RSS = 7.90129e-05, Par. = 0.00358413
It. 6, RSS = 7.90129e-05, Par. = 0.00358539
It. 7, RSS = 7.90129e-05, Par. = 0.00358572
It. 8, RSS = 7.90129e-05, Par. = 0.0035858
# Summary of fit
summary(fitval)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 3.586e-03 6.921e-05 51.81 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.000345 on 664 degrees of freedom
Number of iterations to termination: 8
Reason for termination: Relative error in the sum of squares is at most `ftol'.
# Estimated parameter
parest <- as.list(coef(fitval))
parest
$a
[1] 0.003585804
我是这个领域的新手，所以我的问题是：
- 如果这是解决此类问题的正确（正确）方法（使用 nls.lm 是否正确）？
- 是否有另一种更好的方法来解决为这种曲线拟合找到最佳值的问题？
此外，我没有找到任何可用于为我的问题实现差分方程的包，即我使用y1_exp每个数据x来计算y2_mod. 例如，deSolve 包有一个用于离散时间模型、差分方程（Vignette 第 30 页：在 R 中求解初始值微分方程）的专用求解器（方法 =“迭代”），但是使用此求解器，您只能输入初始值（例如：) y1_exp[1]。
- 那么，有没有更好的方法来为我的类差分方程问题编写代码，或者是否已经存在一些处理此类问题的包？
谢谢！耶内伊