Adding label at the end of the line by plot function in R

Question

I create a line chart by plot() in R and I don't know how to add label at each line end. All I can find online is to by using ggplot() instead. Can anyone do that by plot()? My table is "case_study1", and I assign each row to a vector as "V20xx" and plot it as a line chart by using plot(). (Since I just beginning using R so this is what I can do for plotting a line chart.)

V2014 <- as.numeric(case_study1[2,])
V2013 <- as.numeric(case_study1[3,])
V2012 <- as.numeric(case_study1[4,])
V2011 <- as.numeric(case_study1[5,])
V2010 <- as.numeric(case_study1[6,])
V2009 <- as.numeric(case_study1[7,])
V2008 <- as.numeric(case_study1[8,])
V2007 <- as.numeric(case_study1[9,])
plot(V2015, type = "o", col = "red", xlab = "Month", ylab = "Temperature", main = "Temperature chart")
lines(V2014, type = "o", col = "blue")
lines(V2013, type = "o", col = "yellow")
lines(V2012, type = "o", col = "chartreuse")
lines(V2011, type = "o", col = "chocolate")
lines(V2010, type = "o", col = "brown")
lines(V2009, type = "o", col = "burlywood")
lines(V2008, type = "o", col = "coral")
lines(V2007, type = "o", col = "cyan")

The plot I got

Welcome to SO, StevenCui! This is a good first question, but realize that since we don't have your data, any answers will either be guesses or with data that is not yours. It tends to help if you can make your question more reproducible to include sample data, see https://stackoverflow.com/q/5963269, [mcve], and https://stackoverflow.com/tags/r/info for discussions of `dput`, `data.frame`, and/or `read.table`. Further, by "label" do you mean a way to identify the year to each line/color? You could try `legend(...)` for that. — r2evans, Sep 27 '22 at 17:12
the endpoints of your lines are so close together I would use a legend instead as well — rawr, Sep 27 '22 at 17:17

r2evans · Answer 1 · 2022-09-27T19:00:50.897

Sample data,

case_study1 <- structure(c(18.1265311110765, 18.1445806138217, 20.29444403667, 19.3943562610075, 18.4467029515654, 19.1106318701059, 18.913005082868, 20.1609526062384, 19.4320843210444, 19.3427841216326, 17.5637686550617, 21.8841708265245, 18.4573832275346, 16.7219386268407, 17.2237817412242, 18.5469768662006, 18.4829838024452, 20.7631137836725, 31.0647214353085, 29.5135084576905, 32.6102900002152, 29.3641985105351, 25.1434186547995, 27.994190084748, 27.0048756664619, 27.5333320870996, 31.3106615673751, 41.3828093148768,  40.9396879393607, 40.9130710875615, 37.9195794519037, 35.6223422279581, 39.8471474340186, 35.4306885786355, 38.9832761064172, 39.5527733126655, 50.0248330328614, 49.7703500119969, 51.4038161402568, 47.6908210013062, 44.733499577269, 49.0990697136149, 47.0381244253367, 50.6246182974428, 50.8727697608992, 48.1883749496192, 49.9094971902668, 51.4024006286636, 47.858812619932, 43.8532895939425, 51.0264792805538, 47.2120952093974, 50.0444536274299, 52.3428046917543, 59.7227923031896, 58.1244931928813,  60.2706275274977, 59.0147362891585, 52.6608879743144, 62.2188456263393, 57.7123434329405, 58.3746858304366, 62.6367503125221, 60.9077196735889, 56.8138210857287, 59.812376066111, 60.5437361020595, 51.2970413453877, 60.3391999211162, 57.3215152909979, 56.4301307220012, 62.8979858038947, 50.1909973882139, 45.9235293632373, 49.24725244008, 52.3605857389048, 42.8149003852159, 51.3142714761198, 47.4335271269083, 47.0058694463223, 52.0639638807625, 41.0728860516101, 36.0847078282386, 39.904410911724, 42.5762362144887,  32.67773448769, 42.9504046197981, 35.6693080635741, 37.4426508890465, 40.1027930378914), .Dim = 9:10, .Dimnames = list(Year = c("2007", "2008", "2009", "2010", "2011", "2012", "2013", "2014", "2015"), Month = c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10")))
case_study1
#       Month
# Year          1        2        3        4        5        6        7        8        9       10
#   2007 18.12653 19.34278 31.06472 41.38281 50.02483 48.18837 59.72279 60.90772 50.19100 41.07289
#   2008 18.14458 17.56377 29.51351 40.93969 49.77035 49.90950 58.12449 56.81382 45.92353 36.08471
#   2009 20.29444 21.88417 32.61029 40.91307 51.40382 51.40240 60.27063 59.81238 49.24725 39.90441
#   2010 19.39436 18.45738 29.36420 37.91958 47.69082 47.85881 59.01474 60.54374 52.36059 42.57624
#   2011 18.44670 16.72194 25.14342 35.62234 44.73350 43.85329 52.66089 51.29704 42.81490 32.67773
#   2012 19.11063 17.22378 27.99419 39.84715 49.09907 51.02648 62.21885 60.33920 51.31427 42.95040
#   2013 18.91301 18.54698 27.00488 35.43069 47.03812 47.21210 57.71234 57.32152 47.43353 35.66931
#   2014 20.16095 18.48298 27.53333 38.98328 50.62462 50.04445 58.37469 56.43013 47.00587 37.44265
#   2015 19.43208 20.76311 31.31066 39.55277 50.87277 52.34280 62.63675 62.89799 52.06396 40.10279

Similar plot, using matplot (to reduce your iteration):

cols <- c("red", "blue", "yellow", "chartreuse", "chocolate", "brown", "burlywood", "coral", "cyan")
matplot(
  t(case_study1), type='b', pch=1, lty=1, col=cols,
  xlim = c(1, ncol(case_study1) + 3),
  xlab = "Month", ylab = "Temperature",
  main = "Temperature chart")
legend("bottomright", legend = 2007:2015, col = cols, pch = 1)

To adapt this to your code using multiple calls to lines, add xlim= to your first plot call, then add the legend(..) call.

To do this in ggplot2 is not cosmic, but it should be melted into a long format.

melted <- reshape2::melt(case_study1)
ggplot(melted, aes(Month, value, color = factor(Year), group = Year)) +
  geom_line() +
  labs(title = "Temperature chart", x = "Month", y = "Temperature") +
  scale_color_discrete(name = "Year") +
  scale_x_continuous(breaks = seq(1, 10, by = 2))

Adding label at the end of the line by plot function in R

1 Answers1