ggplot to create correlation graph

Question

I am trying to create linear correlation graph between to variables using ggplot2:

dput(sum)
structure(list(Date = structure(c(15218, 15248, 15279, 15309, 
15340, 15371, 15400, 15431, 15461, 15492, 15522, 15553), class = "Date"), 
    Teams = c(87, 142, 173, 85, 76, 76, 93, 49, 169, 139, 60, 
    120), Scores = c("67101651", "62214988", "63183320", "66750198", 
    "61483322", "67546775", "75290893", "60713372", "77879142", 
    "70290302", "83201853", "83837301")), .Names = c("Date", 
"Teams", "Scores"), row.names = c(NA, 12L), class = "data.frame")

this is my command:

ggplot(sum, aes(x = Scores, y = Teams, group=1)) + 
    geom_smooth(method = "lm", se=TRUE, size=2, formula = lm(Teams ~ Scores))

I get this error:

Error in eval(expr, envir, enclos) : object 'Teams' not found

any ideas?

For any future reader of this thread - to call a data frame like one of the most used base R functions is not a good idea. On this occasion, may I remind that df is also a base R function (although less often used than sum) — tjebo
– tjebo, Commented Jan 24, 2021 at 11:54
Does this answer your question? Adding a regression line on a ggplot — tjebo
– tjebo, Commented Jan 24, 2021 at 12:02

Sven Hohenstein · Accepted Answer · 2012-09-14 18:23:36Z

1

If you want to specify the formula for, e.g., linear model, use y ~ poly(x, 1). You don't need to change the formula parameter as long as you want a simple linear regression (it's the default for method = "lm"):

ggplot(sum, aes(x = Scores, y = Teams, group = 1)) +
  geom_smooth(method = "lm", formula = y ~ poly(x, 1), se = TRUE, size = 2)

I also would recommend using Scores as numeric values (as.numeric(Scores)) if you don't want this variable to be categorial. This would change the regression line.

Score as categorial variable:

categorial

Score as numeric variable:

numeric

edited Sep 14, 2012 at 18:23

answered Sep 14, 2012 at 18:15

Sven Hohenstein

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2 Comments

user1471980 Over a year ago

is there an easy way to print the r square value as a legend?

Sven Hohenstein Over a year ago

An easy way would be to print it in the the title. Just add opts(title = bquote(R^2 ~ ":" ~ .(summary(lm(Teams ~ as.numeric(Scores), sum))$r.squared))) to the plot.

Johnny5ish · Accepted Answer · 2021-01-25 21:04:40Z

1

Here's another option using stat_cor from the ggpubr package. This code will plot your points and display the correlation and p value. You can change "pearson" to "spearman" if you have non-normal data.

ggplot(sum, aes(x = Scores, y = Teams, group = 1)) +
  geom_point(aes()) +
  geom_smooth(method = "lm", se = TRUE, size = 2) +
  stat_cor(method = "pearson", cor.coef.name = "r", vjust = 1, size = 4)

edited Jan 25, 2021 at 21:04

answered Jan 23, 2021 at 21:51

Johnny5ish

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