Re: [R] Plot for 10 years extrapolation

2023-10-28 Thread varin sacha via R-help 
Dear Rui,

I really thank you a lot for your precious R help. It is exactly what I was 
trying to do! Once more, many thanks!

Best,
Sacha






Le vendredi 27 octobre 2023 à 09:36:18 UTC+2, Rui Barradas 
 a écrit : 





Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:
> 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")
> 
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
Hello,

Here is a way with base R graphics. Explained in the code comments.




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)

dfuture <- data.frame(date = seq(as.Date("2019-12-31"), by="1 year", 
length.out=10))



predfuture <- predict(model, dfuture, interval="prediction")
dfuture <- cbind(dfuture, predfuture)

# start the plot with the required x and y limits
xlim <- range(c(date, dfuture$date))
ylim <- range(c(value, dfuture$fit))

plot(value ~ date, col="grey", pch=20, cex=1.5, main="Plot"
      , xlim = xlim, ylim = ylim)

# abline extends the fitted line past the x value (date)
# limit making the next ten years line ugly and not even
# completely overplotting the abline drawn line
abline(model, col="darkorange", lwd=2)
lines(fit ~ date, dfuture
      # , lty = "dashed"
      , lwd=2
      , col = "black")

# if lines() is used for both the interpolated and extrapolated
# values you will have a gap between both fitted and predicted lines
# but it is closer to what you want

# get the fitted values first (interpolated values)
ypred <- predict(model)


plot(value ~ date, col="grey", pch=20, cex=1.5, main="Plot"

      , xlim = xlim, ylim = ylim)

# plot the interpolated values
lines(ypred ~ date, col="darkorange", lwd = 2)
# and now the extrapolated values
# I use normal orange to make the difference more obvious
lines(fit ~ date, dfuture, lty = "dashed", lwd=2, col = "orange")



Hope this helps,

Rui Barradas


-- 
Este e-mail foi analisado pelo software antivírus AVG para verificar a presença 
de vírus.
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Re: [R] Plot for 10 years extrapolation

2023-10-27 Thread Rui Barradas 

Às 19:23 de 26/10/2023, varin sacha via R-help escreveu:

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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and provide commented, minimal, self-contained, reproducible code.

Hello,

Here is a way with base R graphics. Explained in the code comments.




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)

dfuture <- data.frame(date = seq(as.Date("2019-12-31"), by="1 year", 
length.out=10))




predfuture <- predict(model, dfuture, interval="prediction")
dfuture <- cbind(dfuture, predfuture)

# start the plot with the required x and y limits
xlim <- range(c(date, dfuture$date))
ylim <- range(c(value, dfuture$fit))

plot(value ~ date, col="grey", pch=20, cex=1.5, main="Plot"
 , xlim = xlim, ylim = ylim)

# abline extends the fitted line past the x value (date)
# limit making the next ten years line ugly and not even
# completely overplotting the abline drawn line
abline(model, col="darkorange", lwd=2)
lines(fit ~ date, dfuture
  # , lty = "dashed"
  , lwd=2
  , col = "black")

# if lines() is used for both the interpolated and extrapolated
# values you will have a gap between both fitted and predicted lines
# but it is closer to what you want

# get the fitted values first (interpolated values)
ypred <- predict(model)

plot(value ~ date, col="grey", pch=20, cex=1.5, main="Plot"
 , xlim = xlim, ylim = ylim)

# plot the interpolated values
lines(ypred ~ date, col="darkorange", lwd = 2)
# and now the extrapolated values
# I use normal orange to make the difference more obvious
lines(fit ~ date, dfuture, lty = "dashed", lwd=2, col = "orange")



Hope this helps,

Rui Barradas


--
Este e-mail foi analisado pelo software antivírus AVG para verificar a presença 
de vírus.
www.avg.com

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Re: [R] Plot for 10 years extrapolation

2023-10-26 Thread Bert Gunter 
Incidentally, if all you wanted to do was plot fitted values, the
predict method is kinda overkill, as it's just the fitted line from
the model. But I assume you wanted to plot CI's/PI's also, as the
example illustrated.

-- Bert

On Thu, Oct 26, 2023 at 1:56 PM Bert Gunter  wrote:
>
> 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
>  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")
> > 
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.

__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Plot for 10 years extrapolation

2023-10-26 Thread Bert Gunter 
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
 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")
> 
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.