У меня есть две экспериментально определенные переменные (y1_exp, y2_exp), которые изменяются с x, где y2_exp влияет на y1_exp с изменением x (однако y1_exp не влияет на y2_exp),Вот данные:
library(tidyverse)
library(minpack.lm)
data <- read_csv("x,y1_exp,y2_exp
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1625,0.011895883,0.011159118
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1765,0.011540083,0.011148072
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1785,0.011728344,0.01111432
1790,0.011871734,0.011236811
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1800,0.011876553,0.011194875
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1810,0.011800381,0.01116778
1815,0.011834656,0.011162121
1820,0.011859992,0.01115675
1825,0.011904479,0.01120268
1830,0.011887852,0.01121683
1835,0.011930542,0.01120834
1840,0.011897012,0.01121683
1845,0.011772847,0.011230774
1850,0.011648206,0.01120268
1855,0.011587653,0.011161459
1860,0.011609475,0.011155129
1865,0.011606612,0.011167355
1870,0.01163715,0.011165474
1875,0.011702345,0.011181461
1880,0.011630658,0.011184283
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
2030,0.011623752,0.011331657
2035,0.011528515,0.011290242
2040,0.011422442,0.011249917
2045,0.011297187,0.011192985
2050,0.011224108,0.011134442
2055,0.01113363,0.01109932
2060,0.011045227,0.011065237
2065,0.011351941,0.011120562
2070,0.011576883,0.01120359
2075,0.011567037,0.011223171
2080,0.011478605,0.011215587
2085,0.011391364,0.011166053
2090,0.011298781,0.01111384
2095,0.011181323,0.011080428
2100,0.011093248,0.011002934
2105,0.010982501,0.010963566
2110,0.010883381,0.01090522
2115,0.010814714,0.010859742
2120,0.010721912,0.010796415
2125,0.010683917,0.010766847
2130,0.010580852,0.010725531
2135,0.010116913,0.010564735
2140,0.009553532,0.010407921
2145,0.009338262,0.010417143
2150,0.009155087,0.010448563
2155,0.008953537,0.010404576
2160,0.008783894,0.010359066
2165,0.008633234,0.010342206
2170,0.008503224,0.010270291
2175,0.008354554,0.010227032
2180,0.00827382,0.010135737
2185,0.008141687,0.010084595
2190,0.008032464,0.010089726
2195,0.007964806,0.010009439
2200,0.007898842,0.009945284
2205,0.008107711,0.009910075
2210,0.007992595,0.009844644
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2245,0.007153126,0.009055356
2250,0.006969728,0.008921778
2255,0.006750942,0.008829258
2260,0.006683758,0.008575119
2265,0.006760147,0.008532829
2270,0.006586153,0.008526863
2275,0.006708381,0.008433861
2280,0.006529335,0.008342521
2285,0.00642451,0.008397442
2290,0.006863021,0.008618997
2295,0.006855758,0.008662227
2300,0.006786828,0.008711552
2305,0.006929332,0.008715441
2310,0.006950265,0.008713845
2315,0.006885391,0.008718283
2320,0.006809618,0.008671087
2325,0.007080174,0.008741003
2330,0.007041501,0.008670396
2335,0.006937773,0.008681892
2340,0.006842371,0.00864445
2345,0.007110533,0.008602165
2350,0.006981684,0.008549515
2355,0.006870447,0.008522203
2360,0.007093723,0.00856686
2365,0.006991019,0.008483845
2370,0.006858281,0.008487377
2375,0.007004398,0.008393025
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3320,0.006360987,0.00659698")
Используя разностное уравнение первого порядка (рекурсивно вычисленная функция), я рассчитал смоделированное значение y2_mod для каждого x. Параметр, который я хотел установить, равен a (параметр b фиксирован). Я использовал функцию nls.lm для примерки. Вот как выглядит код:
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)
Это мой вывод:
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
Я новичок в этой области, поэтому мои вопросы:
- , если это правильный (правильный) способ решения таких проблем (правильно ли использовать nls.lm)?
- Есть ли еще лучший способ решения проблемы нахождения оптимальногозначение для такой подгонки кривой?
Кроме того, я не нашел ни одного пакета, который можно было бы использовать для реализации разностных уравнений для моей задачи, т.е. я использовал y1_exp данных на каждый x для расчета y2_mod. Например, пакет deSolve имеет специализированный решатель (method = "iteration") для моделей с дискретным временем, разностных уравнений (стр. 30 Vignette: решение дифференциальных уравнений начальных значений в R), однако с помощью этого решателя вы можете вводить только начальные значения (например: y1_exp[1]).
- Итак, есть ли лучший способ написания кода для моей задачи по уравнению разностей доброго вида или уже существует какой-то пакет, который занимается такими проблемами?
Спасибо! Jernej