How to append group row into dataframe - r

Question

I have this df1:
A B C
1 2 3
5 7 9
where A B C are columns names.
I have another df2 with one column:
A
1
2
3
4
I would like to append df2 for each column of df1, creating this final dataframe:
A B C
1 2 3
5 7 9
1 1 1
2 2 2
3 3 3
4 4 4
is it possible to do it?

Answer 1

data.frame(sapply(df1, c, unlist(df2)), row.names = NULL)
# A B C
#1 1 2 3
#2 5 7 9
#3 1 1 1
#4 2 2 2
#5 3 3 3
#6 4 4 4
DATA
df1 = structure(list(A = c(1L, 5L), B = c(2L, 7L), C = c(3L, 9L)), .Names = c("A",
"B", "C"), class = "data.frame", row.names = c(NA, -2L))
df2 = structure(list(A = 1:4), .Names = "A", class = "data.frame", row.names = c(NA,
-4L))

Answer 2

We can replicate df2 for the number of columns of df1, unname it, then rbind it.
rbind(df1, unname(rep(df2, ncol(df1))))
# A B C
# 1 1 2 3
# 2 5 7 9
# 3 1 1 1
# 4 2 2 2
# 5 3 3 3
# 6 4 4 4
Data:
df1 <- structure(list(A = c(1L, 5L), B = c(2L, 7L), C = c(3L, 9L)), .Names = c("A",
"B", "C"), class = "data.frame", row.names = c(NA, -2L))
df2 <- structure(list(A = 1:4), .Names = "A", row.names = c(NA, -4L), class = "data.frame")

Answer 3

We can use base R methods
rbind(df1, setNames(as.data.frame(do.call(cbind, rep(list(df2$A), 3))), names(df1)))
# A B C
#1 1 2 3
#2 5 7 9
#3 1 1 1
#4 2 2 2
#5 3 3 3
#6 4 4 4
data
df1 <- structure(list(A = c(1L, 5L), B = c(2L, 7L), C = c(3L, 9L)), .Names = c("A",
"B", "C"), class = "data.frame", row.names = c(NA, -2L))
df2 <- structure(list(A = 1:4), .Names = "A", class = "data.frame",
row.names = c(NA, -4L))

Answer 4

Here is a base R method with rbind, rep, and setNames:
rbind(dat, setNames(data.frame(rep(dat1, ncol(dat))), names(dat)))
A B C
1 1 2 3
2 5 7 9
3 1 1 1
4 2 2 2
5 3 3 3
6 4 4 4
Edit: turns outdata.frame isn't necessary:
rbind(dat, setNames(rep(dat1, ncol(dat)), names(dat)))
will work.
data
dat <-
structure(list(A = c(1L, 5L), B = c(2L, 7L), C = c(3L, 9L)), .Names = c("A",
"B", "C"), class = "data.frame", row.names = c(NA, -2L))
dat1 <-
structure(list(A = 1:4), .Names = "A", row.names = c(NA, -4L),
class = "data.frame")

Answer 5

I just love R, here is yet another Base R solution but with mapply:
data.frame(mapply(c, df1, df2))
Result:
A B C
1 1 2 3
2 5 7 9
3 1 1 1
4 2 2 2
5 3 3 3
6 4 4 4
Note:
No need to deal with colnames like almost all the other solutions... The key to why this works is that "mapply calls FUN for the values of ... [each element]
(re-cycled to the length of the longest...[element]" (See ?mapply). In other words, df2$A is recycled to however many columns df1 has.
Data:
df1 = structure(list(A = c(1L, 5L), B = c(2L, 7L), C = c(3L, 9L)), .Names = c("A",
"B", "C"), class = "data.frame", row.names = c(NA, -2L))
df2 = structure(list(A = 1:4), .Names = "A", row.names = c(NA, -4L), class = "data.frame")

Answer 6

Data:
df1 <- data.frame(A=c(1,5),
B=c(2,7),
C=c(3,9))
df2 <- data.frame(A=c(1,2,3,4))
Solution:
df2 <- matrix(rep(df2$A, ncol(df1)), ncol=ncol(df1))
colnames(df2) <- colnames(df1)
rbind(df1,df2)
Result:
A B C
1 1 2 3
2 5 7 9
3 1 1 1
4 2 2 2
5 3 3 3
6 4 4 4

Answer 7

A solution from purrr, which uses map_dfc to loop through all columns in df1 to combine all the elements with df2$A.
library(purrr)
map_dfc(df1, ~c(., df2$A))
# A tibble: 6 x 3
A B C
<int> <int> <int>
1 1 2 3
2 5 7 9
3 1 1 1
4 2 2 2
5 3 3 3
6 4 4 4
Data
df1 <- structure(list(A = c(1L, 5L), B = c(2L, 7L), C = c(3L, 9L)), .Names = c("A",
"B", "C"), class = "data.frame", row.names = c(NA, -2L))
df2 <- structure(list(A = 1:4), .Names = "A", class = "data.frame",
row.names = c(NA, -4L))

Answer 8

By analogy with #useR's excellent Base R answer, here's a tidyverse solution:
library(purrr)
map2_df(df1, df2, c)
A B C
1 1 2 3
2 5 7 9
3 1 1 1
4 2 2 2
5 3 3 3
6 4 4 4
Here are a few other (less desirable) options from when I first answered this question.
library(dplyr)
bind_rows(df1, df2 %>% mutate(B=A, C=A))
Or, if we want to dynamically get the number of columns and their names from df1:
bind_rows(df1,
df2[,rep(1,ncol(df1))] %>% setNames(names(df1)))
And one more Base R method:
rbind(df1, setNames(df2[,rep(1,ncol(df1))], names(df1)))

Answer 9

For the sake of completeness, here is data.table approach which doesn't require to handle column names:
library(data.table)
setDT(df1)[, lapply(.SD, c, df2$A)]
A B C
1: 1 2 3
2: 5 7 9
3: 1 1 1
4: 2 2 2
5: 3 3 3
6: 4 4 4
Note that the OP has described df2 to consist only of one column.
There is also a base R version of this approach:
data.frame(lapply(df1, c, df2$A))
A B C
1 1 2 3
2 5 7 9
3 1 1 1
4 2 2 2
5 3 3 3
6 4 4 4
This is similar to d.b's approach but doesn't required to deal with column names.