将曲线拟合到特定数据 [英] Fitting a curve to specific data

问题描述

我的论文中包含以下数据：

  28 45 
 91 14 
 102 11 
 393 5 
 4492 1.77 
  

我需要对此进行拟合。如果我绘制它，那么这就是我得到的。

我认为某种指数曲线应该适合此数据。我正在使用GNUplot。有人可以告诉我哪种曲线适合该曲线以及可以使用哪些初始参数吗？

解决方案

以防万一 R 是一个选项，下面是您可能使用的两种方法的示意图。

第一种方法：评估一组候选模型的拟合优度

这可能是最好的方法，因为它利用了您可能已经知道或期望的变量之间的关系。

 ＃读入数据
 dat<-read.table（text = xy 
 28 45 
 91 14 
 102 11 
 393 5 
 4492 1.77，标头= TRUE）
 
＃快速目测
图（dat ）; lines（dat）
  

 ＃可能的模型有很多……只是在弥补
＃需要付出更多的努力，应该增加一些更好的候选者。
＃可能的模型种类繁多... 
模型<-list（lm（y〜x，data = dat）， 
 lm（y〜I（1 / x），数据= dat），
 lm（y〜log（x），数据= dat），
 nls（y〜I（1 / / x * a）+ b * x，data = dat，start = list（a = 1，b = 1）），
 nls（y〜（a + b * log（x）），data = dat， start = setNames（coef（lm（y〜log（x），data = dat）），c（ a， b）））），
 nls（y〜I（exp（1）^（ a + b * x）），data = dat，start = list（a = 0，b = 0）），
 nls（y〜I（1 / x * a）+ b，data = dat，start = list（a = 1，b = 1））
）
 
＃快速查看这些模型的视觉拟合
库（ggplot2）
 ggplot （dat，aes（x，y））+ geom_point（大小= 5）+ 
 stat_smooth（方法= lm，公式= as.formula（models [[1]]），大小= 1，se =假，颜色=黑色）+ 
 stat_smooth（方法= lm，公式= as.formula（models [[2]]），大小= 1，se =假，颜色= blue）+ 
 stat_smooth（方法= lm，公式= as.formula（模型[ [3]]），大小= 1，se =否，颜色=黄色）+ 
 stat_smooth（方法= nls，公式= as.formula（models [[4]]），数据= dat，方法.args = list（开始= list（a = 0，b = 0）），大小= 1，se = FALSE，颜色= red，线型= 2）+ 
 stat_smooth（method = nls，公式= as.formula（models [[5]]），data = dat，method.args = list（start = setNames（coef（lm（y〜log（x），data = dat））），c（ a， b））），大小= 1，se =否，颜色=绿色，线型= 2）+ 
 stat_smooth（方法= nls，公式= as.formula（models [[6]]），数据= dat，method.args = list（start = list（a = 0，b = 0）），大小= 1，se = FALSE，color = violet）+ 
 stat_smooth（方法= nls，公式= as.formula（models [[7]]），data = dat，m ethod.args = list（start = list（a = 0，b = 0）），大小= 1，se = FALSE，color = orange，线型= 2）
  

橙色曲线看起来不错。让我们看看当我们测量这些模型的相对优度时，它的排名是...

 ＃计算AIC和AICc （对于小样本）对于每个
＃模型，以查看哪种模型最好，即具有最低的AIC 
库（AICcmodavg）；库（plyr）; library（stringr）
 ldply（models，function（mod）{data.frame（AICc = AICc（mod），AIC = AIC（mod），model = deparse（formula（mod）））}）
 
 AICc AIC模型
 1 70.23024 46.23024 y〜x 
 2 44.37075 20.37075 y〜I（1 / x）
 3 67.00075 43.00075 y〜log（x）
 4 43.82083 19.82083 y〜I（1 / x * a）+ b * x 
 5 67.00075 43.00075 y〜（a + b * log（x））
 6 52.75748 28.75748 y〜I（exp（1 ）^（a + b * x））
 7 44.37075 20.37075 y〜I（1 / x * a）+ b 
 
＃y〜I（1 / x * a）+ b * x是在此曲线中尝试过的最佳模型
＃它非常适合该图并且具有拟合统计量的最佳优势
＃毫无疑问，对nls的理解更好，数据也更好可以找到合适的
＃函数。也许这里的优化方法可能是
＃有用的：http://stats.stackexchange.com/a/21098/7744 
  

第二种方法：使用遗传编程来搜索大量模型

这似乎是一种黑暗的方法进行曲线拟合。您不必在开始时指定太多，尽管也许我做错了...

 ＃符号回归使用遗传编程
＃http://rsymbolic.org/projects/rgp/wiki/Symbolic_Regression 
库（rgp）
＃这可能会花费一些时间并抛出
＃a很多警告... 
 result1<-symbolicRegression（y〜x，
 data = dat，functionSet = mathFunctionSet，
 stopCondition = makeStepsStopCondition（2000）
＃检查结果，它们每次都会不同... 
（symbreg< -result1 $ population [[which.min（sapply（result1 $ population，result1 $ fitnessFunction））]]）
 
函数（x）
 tan（（x-x + tan（x））* x）
＃非常奇怪... 
 
＃检查外观是否合适
 ggplot（）+ geom_point（data = dat，aes（x，y），size = 3）+ 
 geom_line（data = data.frame（symbx = dat $ x，symby = sapply（dat $ x，symbreg） ），aes（symbx，symby），颜色=红色）
  

实际上是非常差的视觉效果。也许需要更多的努力才能从基因编程中获得高质量的结果...

信用：曲线拟合答案1 ，通过曲线拟合答案2 /stackoverflow.com/users/516548/g-grothendieck\">G。格洛腾迪克。

I have the following data in my thesis:

28 45
91 14
102 11
393 5
4492 1.77

I need to fit a curve into this. If I plot it, then this is what I get.

I think some kind of exponential curve should fit this data. I am using GNUplot. Can someone tell me what kind of curve will fit this and what initial parameters I can use?

解决方案

Just in case R is an option, here's a sketch of two methods you might use.

First method: evaluate the goodness of fit of a set of candidate models

This is probably the best way as it takes advantage of what you might already know or expect about the relationship between the variables.

# read in the data
dat <- read.table(text= "x y 
28 45
91 14
102 11
393 5
4492 1.77", header = TRUE)

# quick visual inspection
plot(dat); lines(dat)

# a smattering of possible models... just made up on the spot
# with more effort some better candidates should be added
# a smattering of possible models...
models <- list(lm(y ~ x, data = dat), 
               lm(y ~ I(1 / x), data = dat),
               lm(y ~ log(x), data = dat),
               nls(y ~ I(1 / x * a) + b * x, data = dat, start = list(a = 1, b = 1)), 
               nls(y ~ (a + b * log(x)), data = dat, start = setNames(coef(lm(y ~ log(x), data = dat)), c("a", "b"))),
               nls(y ~ I(exp(1) ^ (a + b * x)), data = dat, start = list(a = 0,b = 0)),
               nls(y ~ I(1 / x * a) + b, data = dat, start = list(a = 1,b = 1))
)

# have a quick look at the visual fit of these models
library(ggplot2)
ggplot(dat, aes(x, y)) + geom_point(size = 5) +
    stat_smooth(method = lm, formula = as.formula(models[[1]]), size = 1, se = FALSE, color = "black") + 
    stat_smooth(method = lm, formula = as.formula(models[[2]]), size = 1, se = FALSE, color = "blue") + 
    stat_smooth(method = lm, formula = as.formula(models[[3]]), size = 1, se = FALSE, color = "yellow") + 
    stat_smooth(method = nls, formula = as.formula(models[[4]]), data = dat, method.args = list(start = list(a = 0,b = 0)), size = 1, se = FALSE, color = "red", linetype = 2) + 
    stat_smooth(method = nls, formula = as.formula(models[[5]]), data = dat, method.args = list(start = setNames(coef(lm(y ~ log(x), data = dat)), c("a", "b"))), size = 1, se = FALSE, color = "green", linetype = 2) +
    stat_smooth(method = nls, formula = as.formula(models[[6]]), data = dat, method.args = list(start = list(a = 0,b = 0)), size = 1, se = FALSE, color = "violet") +
    stat_smooth(method = nls, formula = as.formula(models[[7]]), data = dat, method.args = list(start = list(a = 0,b = 0)), size = 1, se = FALSE, color = "orange", linetype = 2)

The orange curve looks pretty good. Let's see how it ranks when we measure the relative goodness of fit of these models are...

# calculate the AIC and AICc (for small samples) for each 
# model to see which one is best, ie has the lowest AIC
library(AICcmodavg); library(plyr); library(stringr)
ldply(models, function(mod){ data.frame(AICc = AICc(mod), AIC = AIC(mod), model = deparse(formula(mod))) })

      AICc      AIC                     model
1 70.23024 46.23024                     y ~ x
2 44.37075 20.37075                y ~ I(1/x)
3 67.00075 43.00075                y ~ log(x)
4 43.82083 19.82083    y ~ I(1/x * a) + b * x
5 67.00075 43.00075      y ~ (a + b * log(x))
6 52.75748 28.75748 y ~ I(exp(1)^(a + b * x))
7 44.37075 20.37075        y ~ I(1/x * a) + b

# y ~ I(1/x * a) + b * x is the best model of those tried here for this curve
# it fits nicely on the plot and has the best goodness of fit statistic
# no doubt with a better understanding of nls and the data a better fitting
# function could be found. Perhaps the optimisation method here might be
# useful also: http://stats.stackexchange.com/a/21098/7744

Second method: use genetic programming to search a vast amount of models

This seems to be a kind of wild shot in the dark approach to curve-fitting. You don't have to specify much at the start, though perhaps I'm doing it wrong...

# symbolic regression using Genetic Programming
# http://rsymbolic.org/projects/rgp/wiki/Symbolic_Regression
library(rgp)
# this will probably take some time and throw
# a lot of warnings...
result1 <- symbolicRegression(y ~ x, 
             data=dat, functionSet=mathFunctionSet,
             stopCondition=makeStepsStopCondition(2000))
# inspect results, they'll be different every time...
(symbreg <- result1$population[[which.min(sapply(result1$population, result1$fitnessFunction))]])

function (x) 
tan((x - x + tan(x)) * x) 
# quite bizarre...

# inspect visual fit
ggplot() + geom_point(data=dat, aes(x,y), size = 3) +
  geom_line(data=data.frame(symbx=dat$x, symby=sapply(dat$x, symbreg)), aes(symbx, symby), colour = "red")

Actually a very poor visual fit. Perhaps there's a bit more effort required to get quality results from genetic programming...

Credits: Curve fitting answer 1, curve fitting answer 2 by G. Grothendieck.

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