I have a data frame in R (taken from the dplyr's site here):
library(dplyr)
gdf <-
tibble(g = c(1, 1, 2, 3), v1 = 10:13, v2 = 20:23) %>%
group_by(g)
gdf
Resulting to:
# A tibble: 4 × 3
# Groups: g [3]
g v1 v2
<dbl> <int> <int>
1 1 10 20
2 1 11 21
3 2 12 22
4 3 13 23
Now I have a vector :
y <- rnorm(4);y
I want to measure the correlation of y with v1 and the correlation of y with v2 simultaneously.
The across()
function might do the job
gdf %>% mutate(across(v1:v2, ~ cor(.x,y)))
but R reports me an error :
Error: Problem with `mutate()` input `..1`.
ℹ `..1 = across(v1:v2, ~cor(.x, y))`.
x incompatible dimensions
ℹ The error occurred in group 1: g = 1.
Run `rlang::last_error()` to see where the error occurred.
CodePudding user response:
Since cor()
requires same dimension for both x
and y
, you cannot group rows together, otherwise, they will not have 4 elements to match with 4 values in y
.
Prepare data and library
library(dplyr)
gdf <-
tibble(g = c(1, 1, 2, 3), v1 = 10:13, v2 = 20:23)
y <- rnorm(4)
[1] 0.59390132 0.91897737 0.78213630 0.07456498
mutate()
If you want to keep v1
and v2
in the output, use the .names
argument to indicate the names of the new columns. {.col}
refers to the column name that across
is acting on.
gdf %>% mutate(across(v1:v2, ~ cor(.x,y), .names = "{.col}_cor"))
# A tibble: 4 x 5
g v1 v2 v1_cor v2_cor
<dbl> <int> <int> <dbl> <dbl>
1 1 10 20 -0.591 -0.591
2 1 11 21 -0.591 -0.591
3 2 12 22 -0.591 -0.591
4 3 13 23 -0.591 -0.591
summarise()
If you only want the cor()
output in the results, you can use summarise
gdf %>% summarize(across(v1:v2, ~ cor(.x,y)))
# A tibble: 1 x 2
v1 v2
<dbl> <dbl>
1 -0.591 -0.591
CodePudding user response:
Using base R:
cor(gdf[,-1], y)
#> [,1]
#> v1 0.5080586
#> v2 0.5080586
Another possible solution, based on purrr::map_dfc
:
library(tidyverse)
gdf <-
tibble(g = c(1, 1, 2, 3), v1 = 10:13, v2 = 20:23)
set.seed(123)
y <- rnorm(4)
map_dfc(gdf[,-1], ~ cor(.x, y))
#> # A tibble: 1 × 2
#> v1 v2
#> <dbl> <dbl>
#> 1 0.508 0.508