R中按組划分的spearman相關性

Question

你如何在 R 中按組計算 Spearman 相關性。我發現以下鏈接按組討論 Pearson 相關性。 但是當我嘗試用 spearman 替換類型時，它不起作用。

https://stats.stackexchange.com/questions/4040/r-compute-correlation-by-group

Answer 1

對於基本的 R 解決方案，這個怎么樣：

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

然后，如果您想要 data.frame 形式的結果：

data.frame(group = dimnames(r)[[1]], corr = as.vector(r))
#   group      corr
# 1    G1 0.4060606
# 2    G2 0.1272727

編輯：如果您更喜歡基於plyr的解決方案，這里是一個：

library(plyr)
ddply(df, .(group), summarise, "corr" = cor(var1, var2, method = "spearman"))

Answer 2

很老的問題，但這個tidy和broom解決方案非常簡單。 因此，我必須分享方法：

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 

由於某些時間/ dplyr版本，您需要編寫此代碼以獲得上述結果並且沒有錯誤：

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)

Answer 3

這是另一種方法：

# 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)

比較我的方法、Josh 的方法和使用 rbenchmark 的 plyr 解決方案：

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())

這給出了輸出

> 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

所以看起來我的方法稍微快一點但不是很多。 我認為 Josh 的方法更直觀一些。 plyr 解決方案是最容易編碼的，但它不是最快的（但它確實更方便）！

Answer 4

如果您想為大量組提供有效的解決方案，那么data.table是您要走的路。

library(data.table)
DT <- as.data.table(df)
setkey(DT, group)
DT[,list(corr = cor(var1,var2,method = 'spearman')), by = group]