平均值与tapply。唯一性(ave)==taply始终?如果是这样,为什么会有一个存在?

ave base R mean dplyr data.table

set.seed(24)
df1 <- data.frame(grp = sample(LETTERS, 1e6, replace = TRUE), val = rnorm(1e6))
system.time(with(df1, ave(val, grp)))
#   user  system elapsed 
#  0.070   0.004   0.073 

library(dplyr)
system.time(df1 %>%
              group_by(grp) %>%
              mutate(new = mean(val)))
#   user  system elapsed 
#  0.159   0.000   0.160 

library(data.table)
system.time(setDT(df1)[, new := mean(val), by = grp])
#  user  system elapsed 
#  0.056   0.000   0.057 

tapply tidyverse sort unique list

tapply(1:10, rep(LETTERS[1:3], c(3, 3, 4)), FUN = range)

ave(1:10, rep(LETTERS[1:3], c(3, 3, 4)), FUN = range)

by(x, y, FUN = sum)

ave tapply by

1. ave(x, y, FUN = sum) x y length(x)

2. tapply(x, y, sum)

3. by(x, y, sum)

dplyr

1. group_by mutate

   data.frame(x, y) %>% group_by(y) %>% mutate(x = sum(x)) %>% pull(x)
   

2. summarise

   data.frame(x, y) %>% group_by(y) %>% summarise(x = sum(x)) %>% pull(x)
   

vector data.frame matrix