I have monthly data from this package: boston, and I did a timeserie
install.packages("fma")
library(fma)
data(package="fma")
View(boston)
timeseries<-ts(boston,start=1)
plot(timeseries)
Now I want to add to this plot two smoothed curves by using local polynomial regression fitting. I was using loess()
, but something must be wrong because it gives me error. This is the code that I am using:
index<-1:nrow(boston)
loess(timeseries ~ index, data = boston, span = 0.8)
Do you wish something like this?
I used {ggplot2}
for the plots. geom_smooth()
allows the use of loess
with span
argument - assuming you want to use loess
.
For the arrangement of the plots, i.e., stacked + one x-axis, I used {grid}
:
# install.packages("fma")
library(fma)
#> Loading required package: forecast
#> Registered S3 method overwritten by 'quantmod':
#> method from
#> as.zoo.data.frame zoo
timeseries <- ts(boston, start = 1)
# plot(timeseries)
library(ggplot2)
df <- as.data.frame(timeseries)
df$index <- 1:35
p1 <- ggplot(data = df, aes(x = index, y = nyase)) +
geom_line() +
geom_smooth(method = 'loess', formula = y~x, span = 0.8) +
theme_minimal() +
theme(axis.title.x = element_blank(), axis.text.x = element_blank())
p2 <- ggplot(data = df, aes(x = index, y = bse)) +
geom_line() +
geom_smooth(method = 'loess', formula = y~x, span = 0.8) +
theme_minimal()
grid::grid.newpage()
grid::grid.draw(rbind(ggplotGrob(p1), ggplotGrob(p2), size = "last"))
gives
Created on 2022-05-19 by the reprex package (v2.0.1)
You can avoid displaying confidence interval around smooth by
adding the argument se = FALSE
to geom_smooth()
.
A short correction on your line loess(...)
, you need to specify the column of your timeseries
object to allow regression, i. e.
loess(timeseries[, 1] ~ index, data = boston, span = 0.8)
# or
loess(timeseries[, 2] ~ index, data = boston, span = 0.8)
Instead of 1 and 2 you could also use "bse"
and "nyase"
. Then, for "bse"
:
index <- 1:35
a <- loess(timeseries[, "bse"] ~ index, data = boston, span = 0.8)
plot(index, timeseries[, "bse"], type = "l")
lines(predict(a), col = "blue")
gives