使用 R 构建特定的时间序列图

Question

我的问题是我从正态分布生成了一个时间序列并绘制了我的时间序列，但我想将时间序列和轴 X 之间的正区域涂成红色，轴 X 下方的负区域也是如此和我的时间序列。

这是我使用的代码，但它不起作用：

 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

编辑：响应 OP 的附加查询。

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")

原文回复：

这是你的想法吗？

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

这是我不久前为某人编写的一些代码。在这种情况下，正面和负面使用两种不同的颜色。虽然这不是你所追求的，但我想我会分享这个。

# 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
  }

这里是the result！

Answer 3

基础绘图解决方案：

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")

请注意，这不涉及时间序列绘图方法，由于频率值和起始 x 值 1 的缩放比例不同，这种方法很难理解。解决方案如下：

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")