Set R data.table row order by chaining 2 columns

Here is an option using igraph with data.table:

#add id in front of cols to distinguishes them as vertices
cols <- paste0("col", 1L:2L)
dt[, (cols) := lapply(.SD, function(x) paste0(id, x)), .SDcols=cols]

#permutations of root nodes and leaf nodes
chains <- dt[, CJ(root=setdiff(col1, col2), leaf=setdiff(col2, col1)), id]

#find all paths from root nodes to leaf nodes
#note that igraph requires vertices to be of character type
library(igraph)
g <- graph_from_data_frame(dt[, .(col1, col2)])
l <- lapply(unlist(
  apply(chains, 1L, function(x) all_simple_paths(g, x[["root"]], x[["leaf"]])), 
  recursive=FALSE), names)
links <- data.table(g=rep(seq_along(l), lengths(l)), col1=unlist(l))

#look up edges
dt[links, on=.(col1), nomatch=0L]

output:

    id  col1  col2 source g
 1:  A    A0 A5703      b 1
 2:  A A5703 A5222      d 1
 3:  A A5222 A7521      e 1
 4:  A A7521 A7907      c 1
 5:  A A7915 A8004      a 2
 6:  B    B0 B1644      x 3
 7:  B B1644 B1625      y 3
 8:  B B1625 B1625      z 3
 9:  B B1625 B1625      w 3
10:  B B1625 B1505      v 3

data:

library(data.table)
dt <- data.table(id = c('A', 'A', 'A', 'A', 'A', 'B', 'B', 'B', 'B', 'B')
  , col1 = c(7521, 0, 7915, 5222, 5703, 1644, 1625, 0, 1625, 1625)
  , col2 = c(7907, 5703, 8004, 7521, 5222, 1625, 1625, 1644, 1625, 1505)
  , source = c('c', 'b', 'a', 'e', 'd', 'y', 'z', 'x', 'w', 'v'))

Here's another approach that:

1. Reorders the data which will place the 0 value first.
2. Loops through the rest of the values to return the index of where col2 matches col1.

setorder(dt, col1)

neworder = seq_len(nrow(dt))
init = 1L
col1 = dt[['col1']]; col2 = dt[['col2']]

for (i in seq_along(neworder)[-1L]) {
  ind = match(col2[init], col1)
  if (is.na(ind)) break
  neworder[i] = init = ind
}

dt[neworder]

##       id  col1  col2
##   <char> <num> <num>
##1:      A     0  5703
##2:      A  5703  5222
##3:      A  5222  7521
##4:      A  7521  7907
##5:      A  7915  8004

If you are doing it with grouping, you can wrap the loop within a dt[, .I[{...}, by = id]$V1 to return the indices. Or to make it look better, we can make a function.

recursive_order = function (x, y) {
  neworder = seq_len(length(x))
  init = 1L

  for (i in neworder[-1L]) {
    ind = match(y[init], x)
    if (is.na(ind)) break

    # Multiple matches which means all the maining matches are the same number
    if (ind == init) { 
      inds = which(x %in% y[init])
      l = length(inds)
      neworder[i:(i + l - 2L)] = inds[-1L]
      break
    }
    neworder[i] = init = ind
  }
  return(neworder)
}

dt <- data.table(id = c('A', 'A', 'A', 'A', 'A', 'B', 'B', 'B', 'B', 'B')
                 , col1 = c(7521, 0, 7915, 5222, 5703, 1644, 1625, 0, 1625, 1625)
                 , col2 = c(7907, 5703, 8004, 7521, 5222, 1625, 1625, 1644, 1625, 1505)
                 , source = c('c', 'b', 'a', 'e', 'd', 'y', 'z', 'x', 'w', 'v'))

setorder(dt, col1)
dt[dt[, .I[recursive_order(col1, col2)], by = id]$V1]

       id  col1  col2 source
    <char> <num> <num> <char>
 1:      A     0  5703      b
 2:      A  5703  5222      d
 3:      A  5222  7521      e
 4:      A  7521  7907      c
 5:      A  7915  8004      a
 6:      B     0  1644      x
 7:      B  1644  1625      y
 8:      B  1625  1625      z
 9:      B  1625  1625      w
10:      B  1625  1505      v