Merge data sets by row differening columns [duplicate]

Answer 1

Let us say you have your data frames in a nice list:

l <- list(
    data.frame(x2=rnorm(3),x1=rnorm(3)),
    data.frame(x1=rnorm(3),x2=rnorm(3),x3=rnorm(3),x4=rnorm(3),x5=rnorm(3)),
    data.frame(x5=rnorm(3),x2=rnorm(3),x3=rnorm(3),x4=rnorm(3),x1=rnorm(3)),
    data.frame(x5=rnorm(3),x2=rnorm(3),x3=rnorm(3),x4=rnorm(3)),
    data.frame(x2=rnorm(3),x1=rnorm(3),x3=rnorm(3),x4=rnorm(3))
)

Grab the first one and (as @joran suggested) merge all the rest to that with eg. a lucid loop:

r <- l[[1]]
for (i in 2:length(l)) {
    r <- merge(r, l[[i]], all=TRUE)
}

Check out r:

> r
         x2        x3       x4       x1        x5
1  -1.72436 -0.774652  3.10001  0.23249 -1.278216
2  -1.25640        NA       NA  0.32997        NA
3  -1.00652 -0.946254  1.17313       NA  2.014517
4  -0.53770 -0.466626 -0.63369 -1.48375 -1.135515
5  -0.49787        NA       NA -0.34020        NA
6  -0.49704 -0.054175  0.85477       NA  0.831706
7   0.13027  0.421750 -0.18126 -0.65452  0.476576
8   0.18519 -1.006994  0.15141  0.66808        NA
9   0.33954 -0.224478  1.38596       NA  0.145807
10  0.57782  1.126430 -0.89582  0.80199        NA
11  0.59149 -0.447669  0.74855 -1.65790  0.059767
12  0.61374  0.751528 -1.93715  0.40125 -0.148243
13  0.89399  0.758481 -0.94801  0.05084        NA
14  0.94200        NA       NA  0.24945        NA
15  0.99509  0.586097 -0.91455 -0.49909  0.823696

---

I didn't like that loop, so wrote some recursive stuff:

> merger <- function(l) lapply(2:length(l), function(x) merge(l[[x-1]], l[[x]], all=TRUE))
> while (length(l) != 1) l<-merger(l)
> l
[[1]]
         x2       x1        x3       x4        x5
1  -1.72436  0.23249 -0.774652  3.10001 -1.278216
2  -1.25640  0.32997        NA       NA        NA
3  -1.00652       NA -0.946254  1.17313  2.014517
4  -0.53770 -1.48375 -0.466626 -0.63369 -1.135515
5  -0.49787 -0.34020        NA       NA        NA
6  -0.49704       NA -0.054175  0.85477  0.831706
7   0.13027 -0.65452  0.421750 -0.18126  0.476576
8   0.18519  0.66808 -1.006994  0.15141        NA
9   0.33954       NA -0.224478  1.38596  0.145807
10  0.57782  0.80199  1.126430 -0.89582        NA
11  0.59149 -1.65790 -0.447669  0.74855  0.059767
12  0.61374  0.40125  0.751528 -1.93715 -0.148243
13  0.89399  0.05084  0.758481 -0.94801        NA
14  0.94200  0.24945        NA       NA        NA
15  0.99509 -0.49909  0.586097 -0.91455  0.823696

Answer 2

I had to read your question quite a few times before I understood what you were looking for, but maybe you want rbind.fill from plyr:

d1 <- data.frame(x1,x2,x3,x4,x5)
d2 <- data.frame(x1,x3,x4,x5)
d3 <- data.frame(x2,x3,x4,x5)
d4 <- data.frame(x1,x2,x3,x4,x5)

> rbind.fill(d1,d4,d2,d3)
     x1   x2        x3         x4         x5
1     A    a 1.1216923  0.9236393  0.2749292
2     B    b 1.1913278  1.1145664 -0.5070576
3     C    c 0.2837657 -0.6631544 -1.0675885
4     A    a 1.1216923  0.9236393  0.2749292
5     B    b 1.1913278  1.1145664 -0.5070576
6     C    c 0.2837657 -0.6631544 -1.0675885
7     A <NA> 1.1216923  0.9236393  0.2749292
8     B <NA> 1.1913278  1.1145664 -0.5070576
9     C <NA> 0.2837657 -0.6631544 -1.0675885
10 <NA>    a 1.1216923  0.9236393  0.2749292
11 <NA>    b 1.1913278  1.1145664 -0.5070576
12 <NA>    c 0.2837657 -0.6631544 -1.0675885

Answer 3

Using data.table::rbindlist with fill = TRUE option:

data.table::rbindlist(
  list(data.frame(x1,x2,x3,x4,x5), 
       data.frame(x1,x3,x4,x5),
       data.frame(x2,x3,x4,x5), 
       data.frame(x1,x2,x3,x4,x5)),
  fill = TRUE)