cor(airquality[airquality$Month == 1, c("Temp", "Humidity")])
gives you a 2 * 2 covariance matrix rather than a number. I bet you want a single number for each Month, so use
## cor(Temp, Humidity | Month)
with(airquality, mapply(cor, split(Temp, Month), split(Humidity, Month)) )
and you will obtain a vector.
Have a read around ?split and ?mapply; they are very useful for "by group" operations, although they are not the only option. Also read around ?cor, and compare the difference between
a <- rnorm(10)
b <- rnorm(10)
cor(a, b)
cor(cbind(a, b))
The answer you linked in your question is doing something similar to cor(cbind(a, b)).
Reproducible example
The airquality dataset in R does not have Humidity column, so I will use Wind for testing:
## cor(Temp, Wind | Month)
x <- with(airquality, mapply(cor, split(Temp, Month), split(Wind, Month)) )
# 5 6 7 8 9
#-0.3732760 -0.1210353 -0.3052355 -0.5076146 -0.5704701
We get a named vector, where names(x) gives Month, and unname(x) gives correlation.
Thank you very much! It worked just perfectly! I was trying to figure out how to obtain a vector with the R^2 for each correlation too, but I can't... Any ideas?
cor(x, y) is like fitting a standardised linear regression model:
coef(lm(scale(y) ~ scale(x) - 1)) ## remember to drop intercept
The R-squared in this simple linear regression is just the square of the slope. Previously we have x storing correlation per group, now R-squared is just x ^ 2.
