How draw a loess line in ts plot - r

Question

I tried hours to figure out how I can make my loess line work. The problem is I do not know much (lets say near nothing). I only have to use R for one course in university.
I created a fake table the real table is for download here
I have to make a timeline plot that worked surprisingly well. But now I have to add two loess lines with different spans. My Problem is I don't know how the command really works. I mean I know it should be something like loess(..~.., data=..). The step where I'm stuck is marked with "WHAT BELONGS HERE" in the given code below.
table <- structure(list(
Months = c("1980-06", "1980-07", "1980-08", "1980-09",
"1980-10", "1980-11", "1980-12", "1981-01"),
Total = c(75000, 70000, 60000, 73000, 72000, 71000, 76000, 71000)),
.Names = c("Monts", "Total of Killed Pigs"),
row.names = c(NA, 4L), class = "data.frame")
ts.obj <- ts(table$`Total of Killed Pigs`, start = c(1980, 1), frequency = 2)
plot(ts.obj)
trend1 <- loess(# **WHAT BELONGS HERE?**, data = table, span =1)
predict1 <- predict(trend1)
lines(predict1, col ="blue")
That is my original code:
obj <- read.csv(file="PATH/monthly-total-number-of-pigs-sla.csv", header=TRUE, sep=",")
ts.obj <- ts(obj$Monthly.total.number.of.pigs.slaughtered.in.Victoria..Jan.1980...August.1995, start = c(1980, 1), frequency = 12)
plot(ts.obj)
trend1 <- loess (WHAT BELONGS HERE?, data = obj, span =1)
predict1 <- predict (trend1)
lines(predict1, col="blue")

Answer 1

We can do away with the data argument as the time series is univariate (just one variable).
The formula ts.obj ~ index(ts.obj) can be read as
value as a function of time
as ts.obj will give you the values, and index(ts.obj) will give you the time index for those values, and the tilde ~ specifies that the first is a function of, or dependent on, the other.
library(zoo) # for index()
plot(ts.obj)
trend1 <- loess(ts.obj ~ index(ts.obj), span=1)
trend2 <- loess(ts.obj ~ index(ts.obj), span=2)
trend3 <- loess(ts.obj ~ index(ts.obj), span=3)
pred <- sapply(list(trend1, trend2, trend3), predict)
matlines(index(ts.obj), pred, lty=1, col=c("blue", "red", "orange"))
zoo isn't strictly required. If you replace index(ts.obj) with as.numeric(time(ts.obj)) you should be fine, I think.

Answer 2

In case you were wanting to go with ggplot2:
library(ggplot2)
library(dplyr)
table <- structure(list(
Months = c("1980-06", "1980-07", "1980-08", "1980-09",
"1980-10", "1980-11", "1980-12", "1981-01"),
Total = c(75000, 70000, 60000, 73000, 72000, 71000, 76000, 71000)),
.Names = c("Months", "Total"),
row.names = c(NA, 8L), class = "data.frame")
Change to proper dates:
table <- table %>% mutate(Months = as.Date(paste0(Months,"-01")))
Plot:
ggplot(table, aes(x=Months, y=Total)) +
geom_line() +
geom_smooth(span=1, se= FALSE, color ="red") +
geom_smooth(span=2, se= FALSE, color ="green") +
geom_smooth(span=3, se= FALSE) +
theme_minimal()