[R] fitting a sinus curve

[Hans W Borchers]

maaariiianne <marianne.zeyringer <at> ec.europa.eu> writes:

> Dear R community! 
> I am new to R and would be very grateful for any kind of help. I am a PhD
> student and need to fit a model to an electricity load profile of a
> household (curve with two peaks). I was thinking of looking if a polynomial
> of 4th order,  a sinus/cosinus combination or a combination of 3 parabels
> fits the data best. I have problems with the sinus/cosinus regression: 

time <- c(
0.00, 0.15,  0.30,  0.45, 1.00, 1.15, 1.30, 1.45, 2.00, 2.15, 2.30, 2.45, 
3.00, 3.15, 3.30, 3.45, 4.00, 4.15, 4.30, 4.45, 5.00, 5.15, 5.30, 5.45, 6.00, 
6.15, 6.30, 6.45, 7.00, 7.15, 7.30, 7.45, 8.00, 8.15, 8.30, 8.45, 9.00, 9.15, 
9.30, 9.45, 10.00, 10.15, 10.30, 10.45, 11.00, 11.15, 11.30, 11.45, 12.00, 
12.15, 12.30, 12.45, 13.00, 13.15, 13.30, 13.45, 14.00, 14.15, 14.30, 14.45, 
15.00, 15.15, 15.30, 15.45, 16.00, 16.15, 16.30, 16.45, 17.00, 17.15, 17.30, 
17.45, 18.00, 18.15, 18.30, 18.45, 19.00, 19.15, 19.30, 19.45, 20.00, 20.15, 
20.30, 20.45, 21.00, 21.15, 21.30, 21.45, 22.00, 22.15, 22.30, 22.45, 23.00, 
23.15, 23.30, 23.45) 
watt <- c(
94.1, 70.8, 68.2, 65.9, 63.3, 59.5, 55, 50.5, 46.6, 43.9, 42.3, 41.4, 40.8, 
40.3, 39.9, 39.5, 39.1, 38.8, 38.5, 38.3, 38.3, 38.5, 39.1, 40.3, 42.4, 45.6, 
49.9, 55.3, 61.6, 68.9, 77.1, 86.1, 95.7, 105.8, 115.8, 124.9, 132.3, 137.6, 
141.1, 143.3, 144.8, 146, 147.2, 148.4, 149.8, 151.5, 153.5, 156, 159, 162.4, 
165.8, 168.4, 169.8, 169.4, 167.6, 164.8, 161.5, 158.1, 154.9, 151.8, 149, 
146.5, 144.4, 142.7, 141.5, 140.9, 141.7, 144.9, 151.5, 161.9, 174.6, 187.4, 
198.1, 205.2, 209.1, 211.1, 212.2, 213.2, 213, 210.4, 203.9, 192.9, 179, 
164.4, 151.5, 141.9, 135.3, 131, 128.2, 126.1, 124.1, 121.6, 118.2, 113.4, 
107.4, 100.8)

> df<-data.frame(time,  watt) 
> lmfit <- lm(time ~ watt + cos(time) + sin(time),  data = df)

Your regression formula does not make sense to me.
You seem to expect a periodic function within 24 hours, and if not it would
still be possible to subtract the trend and then look at a periodic solution.
Applying a trigonometric regression results in the following approximations:

    library(pracma)
    plot(2*pi*time/24, watt, col="red")
    ts  <- seq(0, 2*pi, len = 100)
    xs6 <- trigApprox(ts, watt, 6)
    xs8 <- trigApprox(ts, watt, 8)
    lines(ts, xs6, col="blue", lwd=2)
    lines(ts, xs8, col="green", lwd=2)
    grid()

where as examples the trigonometric fits of degree 6 and 8 are used.
I would not advise to use higher orders, even if the fit is not perfect.

Hans Werner

> Thanks a lot,  
> Marianne