我想為每個獨特的組合做相關性，并按另一個變數分組。我的解決方案適用于非常小的資料集購買想象更多的列變得非常乏味。
set.seed((13))
df <- data.frame(group = rep(LETTERS[1:3], 3),
var1 = rnorm(9, 1),
var2 = rnorm(9, 2),
var3 = rnorm(9, 1))
df %>%
group_by(group) %>%
summarise(var1_var2 = cor(var1, var2),
var1_var3 = cor(var1, var3),
var2_var3 = cor(var2, var3))
我也試過這個，但它不起作用。
df %>%
group_by(group) %>%
summarise(cor = cor(df[,2:ncol(df)]))
uj5u.com熱心網友回復：
這是一個選項。繪制出所有組合，然后為每個組和每個 var 運行一個 cor 測驗，然后在最后旋轉更寬：
library(tidyverse)
map_dfr(unique(df$group), \(x){
data.frame(t(combn(c("var1", "var2", "var3"), 2))) |>
mutate(group = x)
}) |>
mutate(cor = pmap_dbl(list(X1, X2, group),
~cor(df[df$group == ..3, ..1],
df[df$group == ..3, ..2]))) |>
unite(test, X1, X2) |>
pivot_wider(names_from = test, values_from = cor)
#> # A tibble: 3 x 4
#> group var1_var2 var1_var3 var2_var3
#> <chr> <dbl> <dbl> <dbl>
#> 1 A 0.318 -0.476 -0.985
#> 2 B -0.373 -0.487 -0.628
#> 3 C 0.535 0.991 0.645
uj5u.com熱心網友回復：
另一種解決方案。這適用于任意數量的變數。
library(dplyr)
library(tidyr)
library(purrr)
library(tibble)
set.seed((13))
df <- data.frame(group = rep(LETTERS[1:3], 3),
var1 = rnorm(9, 1),
var2 = rnorm(9, 2),
var3 = rnorm(9, 2))
df %>%
select(-group) %>%
split(df$group) %>%
imap_dfr(
~ {
expand.grid(
first = names(.x),
second = names(.x),
stringsAsFactors = FALSE
) %>%
filter(first < second) %>%
rowwise() %>%
transmute(
group = .y,
pair = paste(first, second, sep = "_"),
cor = cor(.x[[first]], .x[[second]])
)
}
) %>%
pivot_wider(
names_from = "pair",
values_from = "cor"
)
# # A tibble: 3 × 4
# group var1_var2 var1_var3 var2_var3
# <chr> <dbl> <dbl> <dbl>
# 1 A 0.318 -0.476 -0.985
# 2 B -0.373 -0.487 -0.628
# 3 C 0.535 0.991 0.645
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標籤：r