Question

I have two plots displaying supply and demand, and one plot in which I have subtracted the demand from the supply to show the resulting asymmetry. I would like to shade the area between the x-axis and the negative part of the asymmetry, to show the extent of the deficit.

I currently use the following code:

plot.asymmetry <- ggplot(data=df.overview.month, 
                         aes(x=Date.Time, y=Asymmetry)) +    
      geom_area(data=subset(df.overview.month, Asymmetry < 0),     
                         aes(x=Date.Time, y=Asymmetry)) 

However – as could be expected – this does not shade the area between geom_line and the x-axis, but only between negative values of the asymmetry data, which is something else entirely, as shown in the resulting graph:

Is there any way to overcome this problem?

/Edit: some example data:

time.initial <- as.POSIXct("2010-12-31 23:00:00", tz="GMT")
Date.Time<-vector()
for(i in 1:24) {
Date.Time[i] <- time.initial + i*3600
}

Demand<-vector()
for(i in 0:23) {
Demand[i+1] <- 155 + 20*sin((pi/12)*i - (pi/2)) + 10*sin((pi/4380)*i + (pi/2))
}

Supply<-vector()
for(i in 0:23) {
Supply[i+1] <- 165 + 5*sin((pi/4380)*i - (pi/2)) + rnorm(1, mean=0, sd=0.20*165)
}

df.overview.month <- data.frame(Date.Time, Demand, Supply, Asymmetry=Supply-Demand)

Answer 1

What about this as inspiration. Now you only need to add additional data points where the asymmetry is equal to zero (like @baptiste suggested). I create a new column which is NA when the asymmetry is above zero, in this way no geom_ribbon will be drawn there. Just subsetting the data will not lead to the required plot.

df.overview.month$Assym_ribbon = ifelse(df.overview.month$Asymmetry > 0, 
                                        NA, 
                                        df.overview.month$Asymmetry)
ggplot(aes(x = Date.Time, y = Asymmetry), 
         data = df.overview.month) + 
   geom_line() + 
   geom_ribbon(aes(ymin = 0, ymax = Assym_ribbon), 
         data = , fill = "red")

Some additional notes about the way you constructed your example. The most important one is that R is vectorized. For example:

set.seed(1)
Supply<-vector()
for(i in 0:23) {
  Supply[i+1] <- 165 + 
           5*sin((pi/4380)*i - 
           (pi/2)) + 
           rnorm(1, mean=0, sd=0.20*165)
}

is equivalent to:

set.seed(1)
i = 0:23
Supply_vec <- 165 + 5*sin((pi/4380)*i - 
               (pi/2)) + 
               rnorm(length(i), mean=0, sd=0.20*165)

> all.equal(Supply_vec, Supply)
[1] TRUE

In this case the reduction in code is modest, but in other (more realistic) settings using vectorization will save you dozens of lines of code.