do.call("rbind", list(data, frames)) 但也通过其原始数据帧对每一行进行索引

Question

df1 <- data.frame(a = 1:2, b = 3:4)
df2 <- data.frame(a = 5:6, b = 7:8)

# A common method loses the origin of each row.
do.call("rbind", list(df1, df2))
##   a b
## 1 1 3
## 2 2 4
## 3 5 7
## 4 6 8

# Whereas here, X1 records which data frame each row originated in.
library(plyr)
adply(list(df1, df2), 1)
##   X1 a b
## 1  1 1 3
## 2  1 2 4
## 3  2 5 7
## 4  2 6 8

有没有其他方法可以做到这一点，也许更有效？

Answer 1

这是一种方法。

library(dplyr)
library(tidyr)

foo <- list(df1, df2)

unnest(foo, names) %>%
mutate(names = gsub("^X", "", names))

#  names a b
#1     1 1 3
#2     1 2 4
#3     2 5 7
#4     2 6 8

Answer 2

带底座：

df1 <- data.frame(a = 1:2, b = 3:4)
df2 <- data.frame(a = 5:6, b = 7:8)

frames <- list(df1, df2)

do.call(rbind, lapply(seq_along(frames), function(x) {
  frames[[x]]$X1 <- x
  frames[[x]]
}))

##   a b X1
## 1 1 3  1
## 2 2 4  1
## 3 5 7  2
## 4 6 8  2

顺便说一句，如果您想了解 plyr 这对 (plyr::adply)、(plyr:::splitter_a) 和 (plyr::ldply) 有何影响。与那个答案相比，这些答案微不足道:-)

do.call("rbind", list(data, frames)) 但也通过其原始数据帧对每一行进行索引

do.call("rbind", list(data, frames)) but also index each row by its original data frame

r

dplyr