How to calculate the tree the results by combining individual leaf paths? - r

Question

Let's say I have an input file where each line contains the path from the root (A) to a leaf
echo "A\tB\tC\nA\tB\tD\nA\tE" > lines.txt
A B C
A B D
A E
How can I easily generate the resulting tree?: (A(B(C,D),E))
I'd like to use GNU tools (awk, sed, etc.) because they tend to work better with large files, but an R script would also work. The R input would be:
# lines <- lapply(readLines("lines.txt"), strsplit, " +")
lines <- list(list(c("A", "B", "C")), list(c("A", "B", "D")), list(c("A","E")))

Answer 1

In Perl:
#!/usr/bin/env perl
use strict;
my $t = {};
while (<>) {
my #a = split;
my $t1 = $t;
while (my $a = shift #a) {
$t1->{$a} = {} if not exists $t1->{$a};
$t1 = $t1->{$a};
}
}
print &p($t)."\n";
sub p {
my ($t) = #_;
return
unless keys %$t;
return '('
. join(',', map { $_ . p($t->{$_}) } sort keys %$t)
. ')';
}
This script returns:
% cat <<EOF | perl l.pl
A B C
A B D
A E
EOF
(A(B(C,D),E))
Note that this script, due to recursion in p is not at all suited for large datasets. But that can be easily resolved by turning that into a double for loop, like in the first while above.

Answer 2

Why do it the easy way, if you can use Bourne Shell script instead? Note, this is not even Bash, this is plain old Bourne shell, without arrays...
#!/bin/sh
#
# A B C
# A B D
# A E
#
# "" vs "A B C" -> 0->3, ident 0 -> -0+3 -> "(A(B(C"
# "A B C" vs "A B D" -> 3->3, ident 2 -> -1+1 -> ",D"
# "A B D" vs "A E" -> 3->2, ident 1 -> -2+1 -> "),E"
# "A E" vs. endc -> 2->0, ident 0 -> -2+0 -> "))"
#
# Result: (A(B(C,D),E))
#
# Input stream is a path per line, path segments separated with spaces.
process_line () {
local line2="$#"
n2=$#
set -- $line1
n1=$#
s=
if [ $n2 = 0 ]; then # last line (empty)
for s1 in $line1; do
s="$s)"
done
else
sep=
remainder=false
for s2 in $line2; do
if ! $remainder; then
if [ "$1" != $s2 ]; then
remainder=true
if [ $# = 0 ]; then # only children
sep='('
else # sibling to an existing element
sep=,
shift
for s1 in $#; do
s="$s)"
done
fi
fi
fi
if $remainder; then # Process remainder as mismatch
s="$s$sep$s2"
sep='('
fi
shift # remove the first element of line1
done
fi
result="$result$s"
}
result=
line1=
(
cat - \
| sed -e 's/[[:space:]]\+/ /' \
| sed -e '/^$/d' \
| sort -u
echo '' # last line marker
) | while read line2; do
process_line $line2
line1="$line2"
test -n "$line2" \
|| echo $result
done
This produces the correct answer for two different files (l.sh is the shell version, l.pl the version in Perl):
% for i in l l1; do cat $i; ./l.sh < $i; ./l.pl < $i; echo; done
A
A B
A B C D
A B E F
A G H
A G H I
(A(B(C(D),E(F)),G(H(I))))
(A(B(C(D),E(F)),G(H(I))))
A B C
A B D
A E
(A(B(C,D),E))
(A(B(C,D),E))
Hoohah!

Answer 3

Okay, so I think I got it:
# input
lines <- c(list(c("A", "B", "C")), list(c("A", "B", "D")), list(c("A","E")))
# generate children
generate_children <- function(lines){
children <- list()
for (line in lines) {
for (index in 1:(length(line)-1)){
parent <- line[index]
next_child <- line[index + 1]
if (is.null(children[[parent]])){
children[[parent]] <- next_child
} else {
if (next_child %notin% children[[parent]]){
children[[parent]] <- c(children[[parent]], next_child)
}
}
}
}
children
}
expand_children <- function(current_parent, children){
if (current_parent %in% names(children)){
expanded_children <- sapply(children[[current_parent]], function(current_child){
expand_children(current_child, children)
}, USE.NAMES = FALSE)
output <- setNames(list(expanded_children), current_parent)
} else {
output <- current_parent
}
output
}
children <- generate_children(lines)
root <- names(children)[1]
tree <- expand_children(root, children)
dput(tree)
# structure(list(A = structure(list(B = c("C", "D"), "E"), .Names = c("B",""))), .Names = "A")
Is there a simpler answer?