Question

我有data.frame的列表。例如

set.seed(1)
my_list <- list()
ids = c("a","b","c","d","e")
for(i in 1:5){
  my_list[[i]] <- data.frame(id = ids, p = rnorm(length(ids)), m = rnorm(length(ids)), hp = runif(length(ids)), hm = runif(length(ids)), d = rnorm(length(ids)), a = rnorm(length(ids)))
}

我想要的是有效地为每个id（在“id”列中）计算列表中所有数据帧上的“p”，“m”，“d”和“a”列的方差。理想情况下，这将返回data.frame这样的（基于上面绘制的值）：

> result.df
  id     var_p     var_m      var_d     var_a
1  a 0.2371569 1.7810729 0.08264279 0.5074250
2  b 0.1091675 0.2107997 1.15051229 1.1578691
3  c 0.5385789 0.7650123 0.44215343 0.3137903
4  d 1.0174542 0.7818498 0.06414317 0.6079849
5  e 0.7343667 1.2870542 1.41615858 0.7362462

Answer 1

使用my_list

library(plyr)
df = do.call(rbind, my_list)
out = ddply(df, .(id), colwise(var, c('p','m','d','a')))

#> out
#  id         p         m          d         a
#1  a 0.2371569 1.7810729 0.08264279 0.5074250
#2  b 0.1091675 0.2107997 1.15051229 1.1578691
#3  c 0.5385789 0.7650123 0.44215343 0.3137903
#4  d 1.0174542 0.7818498 0.06414317 0.6079849
#5  e 0.7343667 1.2870542 1.41615858 0.7362462

或使用lapply和apply

的组合作为R替代方案

df = do.call(rbind, my_list)
df1 = do.call(rbind, 
      lapply(split(df, df$id), 
      function(x) apply(subset(x, select = c(p,m,d,a)), 2, var)))

out = transform(df1, id = row.names(df1))

#> out
#          p         m          d         a id
#a 0.2371569 1.7810729 0.08264279 0.5074250  a
#b 0.1091675 0.2107997 1.15051229 1.1578691  b
#c 0.5385789 0.7650123 0.44215343 0.3137903  c
#d 1.0174542 0.7818498 0.06414317 0.6079849  d
#e 0.7343667 1.2870542 1.41615858 0.7362462  e

或使用doBy

library(doBy)
df = do.call(rbind, my_list)
out = summaryBy( p + m + d + a ~ id , data = df, keep.names=TRUE, FUN = var)

#> out
#  id         p         m          d         a
#1  a 0.2371569 1.7810729 0.08264279 0.5074250
#2  b 0.1091675 0.2107997 1.15051229 1.1578691
#3  c 0.5385789 0.7650123 0.44215343 0.3137903
#4  d 1.0174542 0.7818498 0.06414317 0.6079849
#5  e 0.7343667 1.2870542 1.41615858 0.7362462

或使用sqldf

library(sqldf)
df = do.call(rbind, my_list)
out = sqldf("select id, variance(p), variance(m), 
             variance(d), variance(a) from df group by id")

#> out
#  id variance(p) variance(m) variance(d) variance(a)
#1  a   0.2371569   1.7810729  0.08264279   0.5074250
#2  b   0.1091675   0.2107997  1.15051229   1.1578691
#3  c   0.5385789   0.7650123  0.44215343   0.3137903
#4  d   1.0174542   0.7818498  0.06414317   0.6079849
#5  e   0.7343667   1.2870542  1.41615858   0.7362462

Answer 2

这是基础R方法

dat <- do.call(rbind,my_list)
aggregate( cbind(p,m,d,a) ~ id, var, data=dat)

给出了

  id         p         m          d         a
1  a 0.2371569 1.7810729 0.08264279 0.5074250
2  b 0.1091675 0.2107997 1.15051229 1.1578691
3  c 0.5385789 0.7650123 0.44215343 0.3137903
4  d 1.0174542 0.7818498 0.06414317 0.6079849
5  e 0.7343667 1.2870542 1.41615858 0.7362462

Answer 3

library(data.table)
rbindlist(my_list)[, lapply(.SD, var), by = id, .SDcols = c("p","m","d","a")]
#    id         p         m          d         a
# 1:  a 0.2371569 1.7810729 0.08264279 0.5074250
# 2:  b 0.1091675 0.2107997 1.15051229 1.1578691
# 3:  c 0.5385789 0.7650123 0.44215343 0.3137903
# 4:  d 1.0174542 0.7818498 0.06414317 0.6079849
# 5:  e 0.7343667 1.2870542 1.41615858 0.7362462

Answer 4

已更新以使用bind_rows()（根据@hadley建议，效率高于do.call(rbind,...)）

library(dplyr)
dat <- bind_rows(dat)[,c("id","p","m","d","a")]
dat %>% group_by(id) %>% summarise_each(funs(var))

#   id         p         m          d         a
# 1  a 0.2371569 1.7810729 0.08264279 0.5074250
# 2  b 0.1091675 0.2107997 1.15051229 1.1578691
# 3  c 0.5385789 0.7650123 0.44215343 0.3137903
# 4  d 1.0174542 0.7818498 0.06414317 0.6079849
# 5  e 0.7343667 1.2870542 1.41615858 0.7362462

在data.frames列表中的特定data.frame列上的高效函数

4 个答案: