ggplot的指数拟合，显示回归线和R ^ 2

Question

我正在尝试使用ggplot2和数据包plotly通过数据拟合指数模型，此外，我想显示回归线并获得R ^ 2来检查模型假设< / p>

这是我的数据

SR.irrig<-c(67.39368816,28.7369497,60.18499455,49.32404863,166.393182,222.2902192        ,271.8357323,241.7224707,368.4630364,220.2701789,169.9234274,56.49579274,38.183813,49.337,130.9175233,161.6353594,294.1473982,363.910286,358.3290509,239.8411217,129.6507822           ,32.76462234,30.13952285,52.8365588,67.35426966,132.2303449,366.8785687,247.4012487
            ,273.1931613,278.2790213,123.2425639,45.98362999,83.50199402,240.9945866       
,308.6981358,228.3425602,220.5131914,83.97942185,58.32171185,57.93814837,94.64370151          ,264.7800652,274.258633,245.7294036,155.4177734,77.4523639,70.44223322,104.2283817       ,312.4232116,122.8083088,41.65770103,242.2266084,300.0714687,291.5990173,230.5447786,89.42497778,55.60525466,111.6426307,305.7643166,264.2719213,233.2821407,192.7560296,75.60802862,63.75376269)

temp.pred<-c(2.8,8.1,12.6,7.4,16.1,20.5,20.4,18.4,25.8,14.8,13,5.3,9.4,6.8,15.2,14.3,22.4,23.7,20.8,16.5,7.4,4.61,4.79,8.3,12.1,18.4,22,14.6,15.4,15.5,8.2,10.2,14.8,23.4,20.9,14.5,13,9,2,11.6,13,21,24.7,22.3,10.8,13.2,9.7,15.6,21,10.6,8.3,20.7,24.3,17.9,14.7,5.5,7.,11.7,22.3,17.8,15.5,14.8,2.1,7.3)

temp2 <- data.frame(SR.irrig,temp.pred)

这是我的代码：

gg1 <- ggplot(temp2, aes(x=temp.pred, y=SR.irrig)) + 
  geom_point() + #show points
  stat_smooth(method = 'lm', aes(colour = 'linear'), se = FALSE) +
  stat_smooth(method = 'lm', formula = y ~ poly(x,2), aes(colour = 'polynomial'), se= FALSE)+
  stat_smooth(method = 'nls', formula = y ~ a*exp(b*x), aes(colour = 'Exponential'), se = FALSE, start = list(a=1,b=1))+
  stat_smooth(method = 'nls', formula = y ~ a * log(x) +b, aes(colour = 'logarithmic'), se = FALSE, start = list(a=1,b=1))

对于初始值，我尝试了多种不同的选择，但对于指数模型没有任何作用。

作为输出，我得到以下图形，其中所有模型都包含在内，期望指数

我错过了没有经验的东西。曲线显示？以及如何检查指数拟合的程度如何？

Answer 1

您可以尝试使用nls更好的初始值，并考虑@RichardTelford的建议：

library(tidyverse)
#Data
SR.irrig<-c(67.39368816,28.7369497,60.18499455,49.32404863,166.393182,222.2902192        ,271.8357323,241.7224707,368.4630364,220.2701789,169.9234274,56.49579274,38.183813,49.337,130.9175233,161.6353594,294.1473982,363.910286,358.3290509,239.8411217,129.6507822           ,32.76462234,30.13952285,52.8365588,67.35426966,132.2303449,366.8785687,247.4012487
            ,273.1931613,278.2790213,123.2425639,45.98362999,83.50199402,240.9945866       
            ,308.6981358,228.3425602,220.5131914,83.97942185,58.32171185,57.93814837,94.64370151          ,264.7800652,274.258633,245.7294036,155.4177734,77.4523639,70.44223322,104.2283817       ,312.4232116,122.8083088,41.65770103,242.2266084,300.0714687,291.5990173,230.5447786,89.42497778,55.60525466,111.6426307,305.7643166,264.2719213,233.2821407,192.7560296,75.60802862,63.75376269)

temp.pred<-c(2.8,8.1,12.6,7.4,16.1,20.5,20.4,18.4,25.8,14.8,13,5.3,9.4,6.8,15.2,14.3,22.4,23.7,20.8,16.5,7.4,4.61,4.79,8.3,12.1,18.4,22,14.6,15.4,15.5,8.2,10.2,14.8,23.4,20.9,14.5,13,9,2,11.6,13,21,24.7,22.3,10.8,13.2,9.7,15.6,21,10.6,8.3,20.7,24.3,17.9,14.7,5.5,7.,11.7,22.3,17.8,15.5,14.8,2.1,7.3)

temp2 <- data.frame(SR.irrig,temp.pred)
#Try with better initial vals
fm0 <- nls(log(SR.irrig) ~ log(a*exp(b*temp.pred)), temp2, start = c(a = 1, b = 1))
#Plot
gg1 <- ggplot(temp2, aes(x=temp.pred, y=SR.irrig)) + 
  geom_point() + #show points
  stat_smooth(method = 'lm', aes(colour = 'linear'), se = FALSE) +
  stat_smooth(method = 'lm', formula = y ~ poly(x,2), aes(colour = 'polynomial'), se= FALSE)+
  stat_smooth(method = 'nls', formula = y ~ a*exp(b*x), aes(colour = 'Exponential'), se = FALSE,
              method.args = list(start=coef(fm0)))+
  stat_smooth(method = 'nls', formula = y ~ a * log(x) +b, aes(colour = 'logarithmic'), se = FALSE, start = list(a=1,b=1))
#Display
gg1

输出：

Answer 2

您可以在ggplot中执行此操作，而无需先获取nls模型（尽管最终结果是相同的）。您需要减少minFactor并增加nls控件的最大迭代次数以使模型收敛，但是结果似乎是合理的。请注意如何将参数从stat_smooth传递到nls。

ggplot(temp2, aes(x=temp.pred, y=SR.irrig)) + 
  geom_point() +
  stat_smooth(method = 'lm', 
              formula = y ~ x, 
              mapping = aes(colour = 'linear'), 
              se = FALSE) +
  stat_smooth(method = 'lm', 
              formula = y ~ poly(x,2), 
              mapping = aes(colour = 'polynomial'), 
              se= FALSE)+
  stat_smooth(method = 'nls', 
              formula = y ~ a*exp(b*x), 
              mapping = aes(colour = 'Exponential'), 
              se = FALSE, 
              method.args = list(start   = list(a = 1, b = 1), 
                                 control = list(minFactor = 1/ 8192, 
                                                maxiter = 100))) +
  stat_smooth(method = 'nls', 
              formula = y ~ a * log(x) +b, 
              mapping = aes(colour = 'logarithmic'), 
              se = FALSE, 
              method.args = list(start = list(a=1,b=1)))