Background
I have a data file which consists of Sea Levels near the Dutch Storm Barrier over time, for various days. The goal is to fit a linear model that describes this evolution of the sea level, and given a certain time frame, make a 5 minute ahead-prediction of the sea level (forecasting).
Approach
Given a certain day (chosen on forehand), I chose a time frame on which I want to fit\train the linear model. After some technical adjustments (see below for actual code), I fitted the model. Then, the linear model object and 5 minutes time range are used in the command 'predict()' for a prediction, and the 'forecast' along with a confidence interval is graphed, just behind the fitted model in the first time frame, all in one plot (see below for an example plot).
Problem
The forecast of the model always over- or under predicts hugely. In terms of magnitude, the forecast is a factor 10^10 (or, equivalently, e+10) off. At the same time, the R^2 and R_adj^2 are 'quite high', (0,972 and 0,9334, respectively) and the model diagnostics (leverages, fitted vs residuals, normal qq) look 'reasonably good'. Hence, my problem/question is: How can the model that fits the data so well, predict/forecast so badly? My only explanation is mistake in the code, but I can't spot it.
The data set
More specifically, the dataset is a data frame, which consists (apart from a index column) of 3 columns: 'date' ( "yyyy-mm-dd" format), 'time' ( "hh:mm:ss" format) and 'water' (integer between approx -150 and 350, sea level in cm). Here's a small slice of the data which already gives rise to the above problem:
> SeaLvlAug30[fitRngAug, ]
date time water
1574161 2010-08-30 04:40:00 253
1574162 2010-08-30 04:40:10 254
1574163 2010-08-30 04:40:20 253
1574164 2010-08-30 04:40:30 250
1574165 2010-08-30 04:40:40 250
1574166 2010-08-30 04:40:50 252
1574167 2010-08-30 04:41:00 250
1574168 2010-08-30 04:41:10 247
1574169 2010-08-30 04:41:20 246
1574170 2010-08-30 04:41:30 245
1574171 2010-08-30 04:41:40 242
1574172 2010-08-30 04:41:50 241
1574173 2010-08-30 04:42:00 242
1574174 2010-08-30 04:42:10 244
1574175 2010-08-30 04:42:20 245
1574176 2010-08-30 04:42:30 247
1574177 2010-08-30 04:42:40 247
1574178 2010-08-30 04:42:50 249
1574179 2010-08-30 04:43:00 250
1574180 2010-08-30 04:43:10 250
Minimal runnable R code
# Construct a time frame of a day with steps of 10 seconds
SeaLvlDayTm <- c(1:8640)*10
# Construct the desired fit Range and prediction Range
ftRng <- c(1:20)
predRng <- c(21:50)
# Construct the desired columns for the data frame
date <- rep("2010-08-30", length(c(ftRng,predRng)))
time <- c("04:40:00", "04:40:10", "04:40:20", "04:40:30", "04:40:40", "04:40:50", "04:41:00",
"04:41:10", "04:41:20", "04:41:30", "04:41:40", "04:41:50", "04:42:00", "04:42:10",
"04:42:20", "04:42:30", "04:42:40", "04:42:50", "04:43:00", "04:43:10", "04:43:20",
"04:43:30", "04:43:40", "04:43:50", "04:44:00", "04:44:10", "04:44:20", "04:44:30",
"04:44:40", "04:44:50", "04:45:00", "04:45:10", "04:45:20", "04:45:30", "04:45:40",
"04:45:50", "04:46:00", "04:46:10", "04:46:20", "04:46:30", "04:46:40", "04:46:50",
"04:47:00", "04:47:10", "04:47:20", "04:47:30", "04:47:40", "04:47:50", "04:48:00",
"04:48:10")
timeSec <- c(1681:1730)*10
water <- c(253, 254, 253, 250, 250, 252, 250, 247, 246, 245, 242, 241, 242, 244, 245, 247,
247, 249, 250, 250, 249, 249, 250, 249, 246, 246, 248, 248, 245, 247, 251, 250,
251, 255, 256, 256, 257, 259, 257, 256, 260, 260, 257, 260, 261, 258, 256, 256,
258, 258)
# Construct the data frame
SeaLvlAugStp2 <- data.frame(date, time, timeSec, water)
# Change the index set of the data frame to correspond that of a year
rownames(SeaLvlAugStp2) <- c(1574161:1574210)
#Use a seperate variable for the time (because of a weird error)
SeaLvlAugFtTm <- SeaLvlAugStp2$timeSec[ftRng]
# Fit the linear model
lmObjAug <- lm(SeaLvlAugStp2$water[ftRng] ~ sin((2*pi)/44700 * SeaLvlAugFtTm)
+ cos((2*pi)/44700 * SeaLvlAugFtTm) + poly(SeaLvlAugFtTm, 3)
+ cos((2*pi)/545 * SeaLvlAugFtTm) + sin((2*pi)/545 * SeaLvlAugFtTm)
+ cos((2*pi)/205 * SeaLvlAugFtTm) + sin((2*pi)/205 * SeaLvlAugFtTm)
+ cos((2*pi)/85 * SeaLvlAugFtTm) + sin((2*pi)/85 * SeaLvlAugFtTm),
data = SeaLvlAug30Stp2[ftRng, ])
# Get information about the linear model fit
summary(lmObjAug)
plot(lmObjAug)
#Compute time range prediction and fit
prdtRngTm <- timeSec[prdtRng]
ftRngTm <- timeSec[ftRng]
#Compute prediction/forecast based on fitted data linear model
prdtAug <- predict(lmObjAug, newdata=data.frame(SeaLvlAugFtTm = prdtRngTm), interval="prediction", level=0.95)
#Calculate lower and upper bound confidence interval prediction
lwrAug <- prdtAug[, 2]
uprAug <- prdtAug[, 3]
#Calculate minimum and maximum y axis plot
yminAug <- min(SeaLvlAug30$water[fitRngAug], SeaLvlAug30$water[prdtRngAug], lwrAug)
ymaxAug <- max(SeaLvlAug30$water[fitRngAug], SeaLvlAug30$water[prdtRngAug], uprAug)
#Make the plot
plot((timeSec/10)[ftRng], SeaLvlAugStp2$water[ftRng], xlim = c(min(timeSec/10), max(prdtRngAug30)), ylim = c(yminAug, ymaxAug), col = 'green', pch = 19, main = "Sea Level high water & prediction August 30 ", xlab = "Time (seconds)", ylab = "Sea Level (cm)")
polygon(c(sort(prdtRngTm/10), rev(sort(prdtRngTm/10))), c(uprAug, rev(lwrAug)), col = "gray", border = "gray")
points(prdtRngTm/10, SeaLvlAug30$water[prdtRngTm/10], col = 'green', pch = 19)
lines(ftRngTm/10, fitted(lmObjAug), col = 'blue', lwd = 2)
lines(prdtRngTm/10, prdtAug[, 1], col = 'blue', lwd = 2)
legend("topleft", legend = c("Observ.", "Predicted", "Conf. Int."), lwd = 2, col=c("green", "blue", "gray"), lty = c(NA, 1, 1), pch = c(19, NA, NA))
Example plot
SeaLvlAug30we can't copy/paste the code to run it. If the code is only forSeaLvlDayTm[fitRngAug], then just edit your example to use that smaller dataset and make your example complete. – MrFlick