Hello,

I'm glad it helped.
One more thing, I've just noticed that I copied the gam call in lines() but you don't have to refit the model,

b.gam <- predict(fit)

is enough.

Hope this helps,

Rui Barradas

Às 20:35 de 23/04/20, varin sacha escreveu:
David,
Rui,

Many thanks for your response. Thanks Rui it perfectly works.

Best,







Le jeudi 23 avril 2020 à 06:18:46 UTC+2, Rui Barradas <ruipbarra...@sapo.pt> a 
écrit :





Hello,

Inline.

Às 23:29 de 22/04/20, varin sacha via R-help escreveu:
Dear R-experts,

Here below the reproducible example. I can not add the "green" gam curve on the 
plot. The last line of my R code gives me problem. There is a message error that I don't 
understand. Many thanks for your precious help.

#######################################################
a <- as.Date(c("2020-02-25", 
"2020-02-26","2020-02-27","2020-02-28","2020-03-1","2020-03-2","2020-03-3","2020-03-4","2020-03-5","2020-03-6","2020-03-7","2020-03-8","2020-03-9","2020-03-10","2020-03-11","2020-03-12","2020-03-13","2020-03-14","2020-03-15","2020-03-16"))

b<- c(20,28,45,68,89,123,154,190,245,302,460,379,298,300,245,189,165,100,90,78)

df<-data.frame(a,b)

plot(as.Date(a),b,pch=16,type="o",col="blue",ylab="nombre de nouveaux cas 
covid19",xlab="Dates", main="Nombre d’infectés quotidien Covid-19 en Suisse" ,ylim=c(20,460), 
xlim=c(min(as.Date(a)),max(as.Date(a))))

##### fit loess curve with family symmetric
loess.smooth(a,b, span = 2/3, degree = 1,     family = c("symmetric"))
lines(loess.smooth(a,b, span = 2/3, degree = 1,     family = 
"symmetric"),col="purple")

d <-as.numeric(a)

##### Fit GAM
library(mgcv)
fit <- gam(b ~ s(d, bs="cr"),data=df)
summary(fit)
lines(gam(b ~ s(d, bs="cr"),data=df,col="green"))


Instead of this last line try

b.gam <- predict(gam(b ~ s(d, bs = "cr")))
lines(a, b.gam, col = "green")


Hope this helps,

Rui Barradas

############################################################

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