问

如何在mgcv中设置基准尺寸的最小值？

请叫我姚灬小贱_______ 发布于 2023-02-13 11:27

使用mgcv的惩罚样条,我希望在示例数据中获得10 /年的有效自由度(EDF)(整个期间为60).

library(mgcv)
library(dlnm) 
df <- chicagoNMMAPS

df1<-subset(df, as.Date(date) >= '1995-01-01') 

mod1 <-gam(resp ~ s(time,bs='cr',k=6*15, fx=F)+ s(temp,k=6, bs='cr') + as.factor(dow) 
,family=quasipoisson,na.action=na.omit,data=df1)

在示例数据中,由edf测量的时间的基本维度为56.117,小于每年10.

summary(mod1)


Approximate significance of smooth terms:
           edf Ref.df     F p-value    
s(time) 56.117 67.187 5.369  <2e-16 ***
s(temp)  2.564  3.204 0.998   0.393    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

R-sq.(adj) =  0.277   Deviance explained = 28.2%
GCV score = 1.1297  Scale est. = 1.0959    n = 2192

手动我将通过提供如下的平滑参数来更改edf a

mod1$sp

 s(time)  s(temp) 

23.84809 17.23785

然后我将sp输出插入一个新模型并重新运行它.基本上我会继续改变sp,直到我获得大约60的edf.我将只改变平滑参数的时间.

我将从较低的值开始并检查edf:

mod1a <-gam(resp ~ s(time,bs='cr',k=6*15, fx=F)+ s(temp,k=6, bs='cr') + as.factor(dow) 
,family=quasipoisson,na.action=na.omit,data=df1, sp= c(12.84809,  17.23785 
))
summary(mod1a)
#  edf  62.997

我必须增加平滑参数的时间,以将edf降低到60左右.

mod1b <-gam(resp ~ s(time,bs='cr',k=6*15, fx=F)+ s(temp,k=6, bs='cr') + as.factor(dow) 
,family=quasipoisson,na.action=na.omit,data=df1, sp= c(14.84809,  17.23785 
))
summary(mod1b)
edf  61.393  ## EDF still large, thus I have to increase the sp`

mod1c <-gam(resp ~ s(time,bs='cr',k=6*15, fx=F)+ s(temp,k=6, bs='cr') + as.factor(dow) 
,family=quasipoisson,na.action=na.omit,data=df1, sp=c(16.8190989, 17.23785)) 
summary(mod1c)

edf= 60.005  ## This is what I want to obtain as a final model.

如何通过高效的代码实现最终结果？

1 个回答

我不了解您的模型的细节,但如果您希望最小化(或最大化)edf适合不同的模型sp,optim将完成这项工作.首先,创建一个只返回edf给定不同值的函数sp.

edf.by.sp<-function(sp) {
  model <-gam(resp ~ s(time,bs='cr',k=6*15, fx=F)+ s(temp,k=6, bs='cr') + 
                as.factor(dow),
              family=quasipoisson,
              na.action=na.omit,
              data=df1, 
              sp= c(sp,  17.23785) # Not sure if this quite right.
  )
  abs(summary(model)$s.table['s(time)','edf']-60) # Subtract 60 and flip sign so 60 is lowest.
}

现在,您可以运行optim以最小化edf:

# You could pick any reasonable starting sp value.
# Many optimization methods are available, but in your case
# they work equally well.
best<-optim(12,edf.by.sp,method='BFGS')$par
best
# 16.82708

并且,在插入函数时,你会得到接近0(转换前正好是60):

edf.by.sp(best) # 2.229869e-06

2023-02-13 11:29 回答

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