重复data.frame N次

Question

我有以下数据框架

data.frame(a = c(1,2,3),b = c(1,2,3))
  a b
1 1 1
2 2 2
3 3 3

我想把它变成

  a b
1 1 1
2 2 2
3 3 3
4 1 1
5 2 2
6 3 3
7 1 1
8 2 2
9 3 3

或重复N次.在R中有一个简单的功能吗？谢谢!

Answer 1

编辑:更新到更好的现代R答案.

你可以使用replicate(),然后rbind将结果重新组合在一起.rownames会自动更改为从1:nrows运行.

d <- data.frame(a = c(1,2,3),b = c(1,2,3))
n <- 3
do.call("rbind", replicate(n, d, simplify = FALSE))

更传统的方法是使用索引,但这里的rowname更改不是那么整洁(但更有用):

 d[rep(seq_len(nrow(d)), n), ]

以下是对上面的改进,前两个使用purrr函数式编程,惯用的purrr:

purrr::map_dfr(seq_len(3), ~d)

并且较少惯用的purrr(相同的结果,虽然更尴尬):

purrr::map_dfr(seq_len(3), function(x) d)

最后通过索引而不是列表应用使用dplyr:

d %>% slice(rep(row_number(), 3))

Answer 2

对于data.frame对象,这个解决方案比@mdsummer和@ wojciech-sobala快几倍.

d[rep(seq_len(nrow(d)), n), ]

对于data.table对象,@ mdsummer比转换后应用上述内容要快一些data.frame.对于大n,这可能会翻转. .

完整代码:

packages <- c("data.table", "ggplot2", "RUnit", "microbenchmark")
lapply(packages, require, character.only=T)

Repeat1 <- function(d, n) {
  return(do.call("rbind", replicate(n, d, simplify = FALSE)))
}

Repeat2 <- function(d, n) {
  return(Reduce(rbind, list(d)[rep(1L, times=n)]))
}

Repeat3 <- function(d, n) {
  if ("data.table" %in% class(d)) return(d[rep(seq_len(nrow(d)), n)])
  return(d[rep(seq_len(nrow(d)), n), ])
}

Repeat3.dt.convert <- function(d, n) {
  if ("data.table" %in% class(d)) d <- as.data.frame(d)
  return(d[rep(seq_len(nrow(d)), n), ])
}

# Try with data.frames
mtcars1 <- Repeat1(mtcars, 3)
mtcars2 <- Repeat2(mtcars, 3)
mtcars3 <- Repeat3(mtcars, 3)

checkEquals(mtcars1, mtcars2)
#  Only difference is row.names having ".k" suffix instead of "k" from 1 & 2
checkEquals(mtcars1, mtcars3)

# Works with data.tables too
mtcars.dt <- data.table(mtcars)
mtcars.dt1 <- Repeat1(mtcars.dt, 3)
mtcars.dt2 <- Repeat2(mtcars.dt, 3)
mtcars.dt3 <- Repeat3(mtcars.dt, 3)

# No row.names mismatch since data.tables don't have row.names
checkEquals(mtcars.dt1, mtcars.dt2)
checkEquals(mtcars.dt1, mtcars.dt3)

# Time test
res <- microbenchmark(Repeat1(mtcars, 10),
                      Repeat2(mtcars, 10),
                      Repeat3(mtcars, 10),
                      Repeat1(mtcars.dt, 10),
                      Repeat2(mtcars.dt, 10),
                      Repeat3(mtcars.dt, 10),
                      Repeat3.dt.convert(mtcars.dt, 10))
print(res)
ggsave("repeat_microbenchmark.png", autoplot(res))

Answer 3

该软件包dplyr包含bind_rows()直接组合列表中所有数据框的功能,因此无需do.call()与rbind()以下内容一起使用:

df <- data.frame(a = c(1, 2, 3), b = c(1, 2, 3))
library(dplyr)
bind_rows(replicate(3, df, simplify = FALSE))

对于大量重复bind_rows()也比以下快得多rbind():

library(microbenchmark)
microbenchmark(rbind = do.call("rbind", replicate(1000, df, simplify = FALSE)),
               bind_rows = bind_rows(replicate(1000, df, simplify = FALSE)),
               times = 20)
## Unit: milliseconds
##       expr       min        lq      mean   median        uq       max neval cld
##      rbind 31.796100 33.017077 35.436753 34.32861 36.773017 43.556112    20   b
##  bind_rows  1.765956  1.818087  1.881697  1.86207  1.898839  2.321621    20  a 

Answer 4

使用data.table 包，您可以将特殊符号.I与rep：

df <- data.frame(a = c(1,2,3), b = c(1,2,3))
dt <- as.data.table(df)

n <- 3

dt[rep(dt[, .I], n)]

这使：

   a b
1: 1 1
2: 2 2
3: 3 3
4: 1 1
5: 2 2
6: 3 3
7: 1 1
8: 2 2
9: 3 3

Answer 5

d <- data.frame(a = c(1,2,3),b = c(1,2,3))
r <- Reduce(rbind, list(d)[rep(1L, times=3L)])