in reply to Re^9: constructing a tree from csv file
in thread constructing a tree from csv file

I tried using "next" 
    while ($parent = $tree{$parent}{parent}[0]) {
        if ( $parent eq 'Q' ) {next;}

[download]
But its still getting hanged !!!!

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Re^11: constructing a tree from csv file
by choroba (Cardinal) on Nov 20, 2013 at 07:59 UTC
    The cycle might not contain Q.
    لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
[reply]
      Choroba the cycle has Q. Heres a snippet from the data im running the code on:- 
      __DATA__
child,    Parent,    relation
Q    M6    E
M6    Q    E
M10    Q    E
M28    M6    P

      [download]
      As you can see in line 1 & 2, its a cycle and hence leading to crashing of code. Because if i remove first/second line , the code is working fine. But I do want these relations also, so thats why I want the tree to stop tracing back once it has reached 'Q' for a particular child and not go beyond that.
[reply]
[d/l]
      Heres the code which is getting hanged (the M6<-Q relation is cyclical as pointed by you, and thus getting hanged). Please suggest :- 
      #!/usr/bin/perl
use warnings;
use strict;

my %group = (               # Hash table/dictionary for all the  group
+s
             'P'=> 'I_1',
             'Pl'=>'I_2',
             'P.P'=>'I_3',
             'P.Pl'=>'I_4',
             'Pl.P'=>'I_5',
             'Pl.Pl'=>'I_6',
             'P.P.P'=>'I_7',
             'P.P.Pl'=>'I_8',
             'P.Pl.P'=>'I_9',
             'P.Pl.Pl'=>'I_10',
             'Pl.P.P'=>'I_11',
             'Pl.P.Pl'=>'I_12',
             'Pl.Pl.P'=>'I_13',
             'Pl.Pl.Pl'=>'I_14',
             'E'=> 'II_15',
             'P.E'  => 'II_16',
             'Pl.E' => 'II_17',
             'P.P.E'=>'II_18',
             'P.Pl.E'=>'II_19',
             'Pl.P.E'=>'II_20',
             'Pl.Pl.E'=>'II_21',
             'E.P'=> 'III_22',
             'E.Pl'=>'III_23',
             'P.E.P'=>'III_24',
             'P.E.Pl'=>'III_25',
             'Pl.E.P'=>'III_26',
             'Pl.E.Pl'=>'III_27',
             'E.P.P'=>'III_28',
             'E.P.Pl'=>'III_29',
             'E.Pl.P'=>'III_30',
             'E.Pl.Pl'=>'III_31',
             'E.E'=>'IV_32',
             'P.E.E'=>'IV_33',
             'Pl.E.E'=>'IV_34',
             'E.P.E'=>'IV_35',
             'E.Pl.E'=>'IV_36',
             'E.E.P'=>'IV_37',
             'E.E.Pl'=>'IV_38',
             'E.E.E'=>'IV_39',
            );  


<DATA>; # Skip the headers (first row).

my %tree;
while (<DATA>) {   # parse through the input data and fill in our tree
+ data structure
    chomp;
    my ($child, $parent, $likelihood) = split /\t/;
    #$relation =~ s/,$//; # Would help if we have commas after relatio
+ns also
    $tree{$child}{parent} = [$parent, $likelihood];
}

for my $node (keys %tree) {
        my @path      = $node;
    my @likelihood = ($tree{$node}{parent}[1]);
    my $parent    = $node;
    
    while ($parent = $tree{$parent}{parent}[0]) {
        
        push @path, $parent;
        
        push @likelihood, $tree{$parent}{parent}[1];
        # if ( $parent eq 'Q' ) {next;}
    }
    my $path_string = join '.', reverse grep defined, @likelihood;

    print "$node:\t";
    print join '<-', @path;  # Join the likelihood path. Then if group
+ is found for a likelihood from the group hash table then print it, e
+lse quit
    print " = ", $path_string, ', ', $group{$path_string} // q();  
    print "\n";
}

__DATA__
child,    Parent,    likelihood
M7    Q    P
M22    Q    E
M6    Q    E
M28    M6    P
M7    M28    E
M54    M7    Pl
M213    M54    E
M206    M54    E
M194    M54    E
M53    M7    Pl
M102    M17    E
M92    M17    E
M93    M17    E
M94    M17    E
M95    M17    E
M34    M6    E
M35    M6    E
Q    M6    E
M10    Q    E

      [download]
[reply]
[d/l]
        Use last instead of next.
        لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
[reply]
      choroba , Another tricky issue with the program is, that suppose consider this :- 
      child    parent    likelihood
M7    Q    P
M7    M28    E
M28    M6    E
M6    Q    Pl

      [download]
      So in this case the output is coming to be M7:M7<-M28<-M6<-Q = E.E.Pl, whereas the correct output should be M7:M7<-Q which is the shortest path from M7 to Q.
[reply]
[d/l]
        In a tree, each node has just one parent. See Graph for handling general graphs.
        لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
[reply]

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