Alternative to expand.grid for data.frames

Question

I have a data.frame df and I want that every row in this df is duplicated lengthTime times and that a new column is added that counts

Answer 1

This works:

REP <- rep(1:nrow(df), 3)
df2 <- data.frame(df[REP, ], Time = rep(1:3, each = 9))
rownames(df2) <- NULL
df2

Answer 2

Why not just something like df[rep(1:nrow(df),times = 3),] to extend the data frame, and then add the extra column just as you have above, with df$Time <- rep(1:lengthTime, each=nrRow)?

Answer 3

Quick update

There is now also the crossing() function in package tidyr which can be used instead of merge, is somewhat faster, and returns a tbl_df / tibble.

data.frame(time=1:10) %>% merge(iris, by=NULL) 

data.frame(time=1:10) %>% tidyr::crossing(iris) 

Answer 4

A data.table solution:

> library(data.table)
>  ( df <- data.frame(ID   = rep(letters[1:3], each=3),
+                  CatA = rep(1:3, times = 3),
+                  CatB = letters[1:9]) )
  ID CatA CatB
1  a    1    a
2  a    2    b
3  a    3    c
4  b    1    d
5  b    2    e
6  b    3    f
7  c    1    g
8  c    2    h
9  c    3    i
> ( DT <- data.table(df)[, lapply(.SD, function(x) rep(x,3))][, Time:=rep(1:3, each=nrow(df0))] )
    ID CatA CatB Time
 1:  a    1    a    1
 2:  a    2    b    1
 3:  a    3    c    1
 4:  b    1    d    1
 5:  b    2    e    1
 6:  b    3    f    1
 7:  c    1    g    1
 8:  c    2    h    1
 9:  c    3    i    1
10:  a    1    a    2
11:  a    2    b    2
12:  a    3    c    2
13:  b    1    d    2
14:  b    2    e    2
15:  b    3    f    2
16:  c    1    g    2
17:  c    2    h    2
18:  c    3    i    2
19:  a    1    a    3
20:  a    2    b    3
21:  a    3    c    3
22:  b    1    d    3
23:  b    2    e    3
24:  b    3    f    3
25:  c    1    g    3
26:  c    2    h    3
27:  c    3    i    3

Another one :

> library(data.table)
>  ( df <- data.frame(ID   = rep(letters[1:3], each=3),
+                  CatA = rep(1:3, times = 3),
+                  CatB = letters[1:9]) )
> DT <- data.table(df)
> rbindlist(lapply(1:3, function(i) cbind(DT, Time=i)))
    ID CatA CatB Time
 1:  a    1    a    1
 2:  a    2    b    1
 3:  a    3    c    1
 4:  b    1    d    1
 5:  b    2    e    1
 6:  b    3    f    1
 7:  c    1    g    1
 8:  c    2    h    1
 9:  c    3    i    1
10:  a    1    a    2
11:  a    2    b    2
12:  a    3    c    2
13:  b    1    d    2
14:  b    2    e    2
15:  b    3    f    2
16:  c    1    g    2
17:  c    2    h    2
18:  c    3    i    2
19:  a    1    a    3
20:  a    2    b    3
21:  a    3    c    3
22:  b    1    d    3
23:  b    2    e    3
24:  b    3    f    3
25:  c    1    g    3
26:  c    2    h    3
27:  c    3    i    3

Answer 5

It's been a while since this question was posted, but I recently came across it looking for just the thing in the title, namely, an expand.grid that works for data frames. The posted answers address the OP's more specific question, so in case anyone is looking for a more general solution for data frames, here's a slightly more general approach:

expand.grid.df <- function(...) Reduce(function(...) merge(..., by=NULL), list(...))

# For the example in the OP
expand.grid.df(df, data.frame(1:lengthTime))

# More generally
df1 <- data.frame(A=1:3, B=11:13)
df2 <- data.frame(C=51:52, D=c("Y", "N"))
df3 <- data.frame(E=c("+", "-"))
expand.grid.df(df1, df2, df3)

Answer 6

You can also just do a simple merge by NULL (which will cause merge to do simple combinatorial data replication):

merge(data.frame(time=1:lengthTime), iris, by=NULL)