from ?predict.lm:

"predict.lm produces a vector of predictions or a matrix of
predictions and bounds with column names fit, lwr, and upr if interval
is set. "

ergo:
predict(model, dfuture, interval = "prediction")[,"fit"]  ## or [,1]
as it's the first column in the returned matrix

is your vector of predicted values that you can plot against
dfuture$date however you would like, e.g. with different colors,
symbols, or whatever. Exactly how you do this depends on what graphics
package you are using. The example in ?predict.lm shows you how to do
it with R's base graphics and overlaying prediction and confidence
intervals.

Cheers,
Bert

On Thu, Oct 26, 2023 at 11:27 AM varin sacha via R-help
<r-help@r-project.org> wrote:
>
> Dear R-Experts,
>
> Here below my R code working but I don't know how to complete/finish my R 
> code to get the final plot with the extrapolation for the10 more years.
>
> Indeed, I try to extrapolate my data with a linear fit over the next 10 
> years. So I create a date sequence for the next 10 years and store as a 
> dataframe to make the prediction possible.
> Now, I am trying to get the plot with the actual data (from year 2004 to 
> 2018) and with the 10 more years extrapolation.
>
> Thanks for your help.
>
> ####################################################
> date <-as.Date(c("2018-12-31", "2017-12-31", "2016-12-31", "2015-12-31", 
> "2014-12-31", "2013-12-31", "2012-12-31", "2011-12-31", "2010-12-31", 
> "2009-12-31", "2008-12-31", "2007-12-31", "2006-12-31", "2005-12-31", 
> "2004-12-31"))
>
> value <-c(15348, 13136, 11733, 10737, 15674, 11098, 13721, 13209, 11099, 
> 10087, 14987, 11098, 13421, 9023, 12098)
>
> model <- lm(value~date)
>
> plot(value~date ,col="grey",pch=20,cex=1.5,main="Plot")
> abline(model,col="darkorange",lwd=2)
>
> dfuture <- data.frame(date=seq(as.Date("2019-12-31"), by="1 year", 
> length.out=10))
>
> predict(model,dfuture,interval="prediction")
> ########################################################
>
> ______________________________________________
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