How do you calculate Spearman correlation by group in R. I found the following link talking about Pearson correlation by group. But when I tried to replace the type with spearman, it does not work.
https://stats.stackexchange.com/questions/4040/r-compute-correlation-by-group
To calculate Spearman's ρ in R, first, rank the x and y variables. A new data. frame is created to keep the ranked variables. Take the covariance of the variables and divide by the product of the x and y variables' standard deviations to find Spearman's ρ.
The correlation between two numeric variables can be measured with Spearman coefficient. To measure the relationship between numeric variable and categorical variable with > 2 levels you should use eta correlation (square root of the R2 of the multifactorial regression).
The fundamental difference between the two correlation coefficients is that the Pearson coefficient works with a linear relationship between the two variables whereas the Spearman Coefficient works with monotonic relationships as well.
How about this for a base R solution:
df <- data.frame(group = rep(c("G1", "G2"), each = 10),
var1 = rnorm(20),
var2 = rnorm(20))
r <- by(df, df$group, FUN = function(X) cor(X$var1, X$var2, method = "spearman"))
# df$group: G1
# [1] 0.4060606
# ------------------------------------------------------------
# df$group: G2
# [1] 0.1272727
And then, if you want the results in the form of a data.frame:
data.frame(group = dimnames(r)[[1]], corr = as.vector(r))
# group corr
# 1 G1 0.4060606
# 2 G2 0.1272727
EDIT: If you prefer a plyr
-based solution, here is one:
library(plyr)
ddply(df, .(group), summarise, "corr" = cor(var1, var2, method = "spearman"))
very old question, but this tidy
& broom
solution is extremely straightforward. Thus I have to share the approach:
set.seed(123)
df <- data.frame(group = rep(c("G1", "G2"), each = 10),
var1 = rnorm(20),
var2 = rnorm(20))
library(tidyverse)
library(broom)
df %>%
group_by(group) %>%
summarize(correlation = cor(var1, var2,, method = "sp"))
# A tibble: 2 x 2
group correlation
<fct> <dbl>
1 G1 -0.200
2 G2 0.0545
# with pvalues and further stats
df %>%
nest(-group) %>%
mutate(cor=map(data,~cor.test(.x$var1, .x$var2, method = "sp"))) %>%
mutate(tidied = map(cor, tidy)) %>%
unnest(tidied, .drop = T)
# A tibble: 2 x 6
group estimate statistic p.value method alternative
<fct> <dbl> <dbl> <dbl> <chr> <chr>
1 G1 -0.200 198 0.584 Spearman's rank correlation rho two.sided
2 G2 0.0545 156 0.892 Spearman's rank correlation rho two.sided
Since some time/dplyr
version, you need to write this to get results like above and no errors:
df %>%
nest(data = -group) %>%
mutate(cor=map(data,~cor.test(.x$var1, .x$var2, method = "sp"))) %>%
mutate(tidied = map(cor, tidy)) %>%
unnest(tidied) %>%
select(-data, -cor)
Here's another way to do it:
# split the data by group then apply spearman correlation
# to each element of that list
j <- lapply(split(df, df$group), function(x){cor(x[,2], x[,3], method = "spearman")})
# Bring it together
data.frame(group = names(j), corr = unlist(j), row.names = NULL)
Comparing my method, Josh's method, and the plyr solution using rbenchmark:
Dason <- function(){
# split the data by group then apply spearman correlation
# to each element of that list
j <- lapply(split(df, df$group), function(x){cor(x[,2], x[,3], method = "spearman")})
# Bring it together
data.frame(group = names(j), corr = unlist(j), row.names = NULL)
}
Josh <- function(){
r <- by(df, df$group, FUN = function(X) cor(X$var1, X$var2, method = "spearman"))
data.frame(group = attributes(r)$dimnames[[1]], corr = as.vector(r))
}
plyr <- function(){
ddply(df, .(group), summarise, "corr" = cor(var1, var2, method = "spearman"))
}
library(rbenchmark)
benchmark(Dason(), Josh(), plyr())
Which gives the output
> benchmark(Dason(), Josh(), plyr())
test replications elapsed relative user.self sys.self user.child sys.child
1 Dason() 100 0.19 1.000000 0.19 0 NA NA
2 Josh() 100 0.24 1.263158 0.22 0 NA NA
3 plyr() 100 0.51 2.684211 0.52 0 NA NA
So it appears my method is slightly faster but not by much. I think Josh's method is a little more intuitive. The plyr solution is the easiest to code up but it's not the fastest (but it sure is a lot more convenient)!
If you want an efficient solution for large numbers of groups then data.table
is the way to go.
library(data.table)
DT <- as.data.table(df)
setkey(DT, group)
DT[,list(corr = cor(var1,var2,method = 'spearman')), by = group]
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