Construct a specific plot of time series using R - r

Question

My problem is that I generate a time series from normal distribution and I plot my time series but I want to color in red the positive area between the time series and the axe X, the same for the negative area below the axe X and my time series.
This is the code I use but it does not work :
x1<-rnorm(250,0.4,0.9)
x <- as.matrix(x1)
t <- ts(x[,1], start=c(1,1), frequency=30)
plot(t,main="Daily closing price of Walterenergie",ylab="Adjusted close Returns",xlab="Times",col="blue")
plot(t,xlim=c(2,4),main="Daily closing price of Walterenergie",ylab="Adjusted close Returns",xlab="Times",col="blue")
abline(0,0)
z1<-seq(2,4,0.001)
cord.x <- c(2,z1,4)
cord.y <- c(0,t(z1),0)
polygon(cord.x,cord.y,col='red')

Answer 1

Edit: In response to OP's additional query.
library(ggplot2)
df <- data.frame(t=1:nrow(x),y=x)
df$fill <- ifelse(x>0,"Above","Below")
ggplot(df)+geom_line(aes(t,y),color="grey")+
geom_ribbon(aes(x=t,ymin=0,ymax=ifelse(y>0,y,0)),fill="red")+
geom_ribbon(aes(x=t,ymin=0,ymax=ifelse(y<0,y,0)),fill="blue")+
labs(title="Daily closing price of Walterenergie",
y="Adjusted close Returns",
x="Times")
Original response:
Is this what you had in mind?
library(ggplot2)
df <- data.frame(t=1:nrow(x),y=x)
ggplot(df)+geom_line(aes(t,y),color="grey")+
geom_ribbon(aes(x=t,ymin=0,ymax=y),fill="red")+
labs(title="Daily closing price of Walterenergie",
y="Adjusted close Returns",
x="Times")

Answer 2

This is some code I had written a while ago for someone. In this case two different colors are used for positive and negative. Although this is not exactly what you're after, I thought I'll share this.
# Set a seed to get a reproducible example
set.seed(12345)
num.points <- 100
# Create some data
x.vals <- 1:num.points
values <- rnorm(n=num.points, mean=0, sd=10)
# Plot the graph
plot(x.vals, values, t="o", pch=20, xlab="", ylab="", las=1)
abline(h=0, col="darkgray", lwd=2)
# We need to find the intersections of the curve with the x axis
# Those lie between positive and negative points
# When the sign changes the product between subsequent elements
# will be negative
crossings <- values[-length(values)] * values[-1]
crossings <- which(crossings < 0)
# You can draw the points to check (uncomment following line)
# points(x.vals[crossings], values[crossings], col="red", pch="X")
# We now find the exact intersections using a proportion
# See? Those high school geometry problems finally come in handy
intersections <- NULL
for (cr in crossings)
{
new.int <- cr + abs(values[cr])/(abs(values[cr])+abs(values[cr+1]))
intersections <- c(intersections, new.int)
}
# Again, let's check the intersections
# points(intersections, rep(0, length(intersections)), pch=20, col="red", cex=0.7)
last.intersection <- 0
for (i in intersections)
{
ids <- which(x.vals<=i & x.vals>last.intersection)
poly.x <- c(last.intersection, x.vals[ids], i)
poly.y <- c(0, values[ids], 0)
if (max(poly.y) > 0)
{
col="green"
}
else
{
col="red"
}
polygon(x=poly.x, y=poly.y, col=col)
last.intersection <- i
}
And here's the result!

Answer 3

Base plotting solution:
x1<-rnorm(250,0.4,0.9)
x <- as.matrix(x1)
# t <- ts(x[,1], start=c(1,1), frequency=30)
plot(x1,main="Daily closing price of Walterenergie",ylab="Adjusted close Returns",xlab="Times",col="blue", type="l")
polygon( c(0,1:250,251), c(0, x1, 0) , col="red")
Note this doesn't deal with the time-series plotting method which is rather difficult to understand because of differences in scaling by the frequency value and a starting x value of 1. The solution to that is below:
plot(t,main="Daily closing price of Walterenergie",
ylab="Adjusted close Returns",xlab="Times",col="blue", type="l")
polygon( c(1,1+(0:250)/30), c(0, t, 0) , col="red")