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我有一个 DAG 表示为节点及其后继边。使用简单的递归函数可以将其具体化为嵌套数据结构。
#tree1.pl
#!/usr/bin/env perl
use 5.028; use strictures; use Moops; use Kavorka qw(fun); use List::AllUtils qw(first);
class Node :ro {
has label => isa => Str;
has children => isa => ArrayRef[Str];
}
fun N($label, $children) {
return Node->new(label => $label, children => $children);
}
# list is really flat, but
# indentation outlines desired tree structure
our @dag = (
N(N0 => ['N1']),
N(N1 => ['N2']),
N(N2 => ['N3']),
N(N3 => ['N4', 'N5']),
N(N4 => []),
N(N5 => []),
);
fun tree(Node $n) {
return bless [
map {
my $c = $_;
tree(first {
$_->label eq $c
} @dag)
} $n->children->@*
] => $n->label;
}
tree($dag[0]);
# bless([ #N0
# bless([ #N1
# bless([ #N2
# bless([ #N3
# bless([] => 'N4'),
# bless([] => 'N5'),
# ] => 'N3')
# ] => 'N2')
# ] => 'N1')
# ] => 'N0')
our @dag = (
N(N0 => ['N1']),
N(N1 => ['N2']),
︙
N(N1 => ['N6', 'N5']),
︙
our @dag = (
N(N0 => ['N1']),
N([N1 => 'red'] => ['N2']),
︙
N([N1 => 'blue'] => ['N6', 'N5']),
︙
#tree2.pl
#!/usr/bin/env perl
use 5.028; use strictures; use Moops; use Kavorka qw(fun); use List::AllUtils qw(first);
class Node :ro {
has label => isa => Str;
has col => isa => Maybe[Str];
has children => isa => ArrayRef[Str];
has col_seen => is => 'rw', isa => Int;
}
fun N($c_l, $children) {
return ref $c_l
? Node->new(label => $c_l->[0], col => $c_l->[1], children => $children)
: Node->new(label => $c_l, children => $children);
}
# indentation outlines desired tree structure
our @dag = (
### start 1st tree
N(N0 => ['N1']),
N([N1 => 'red'] => ['N2']),
N(N2 => ['N3']),
N(N3 => ['N4', 'N5']),
N(N4 => []),
N(N5 => []),
### end 1st tree
### start 2nd tree
# N0
N([N1 => 'blue'] => ['N6', 'N5']),
N(N6 => ['N7']),
N(N7 => ['N4']),
# N4
# N5
### end 2nd tree
);
fun tree(Node $n) {
return bless [
map {
my $c = $_;
my @col = map { $_->col } grep { $_->label eq $c } @dag;
if (@col > 1) {
$n->col_seen($n->col_seen + 1);
die 'exhausted' if $n->col_seen > @col;
tree(first {
$_->label eq $c && $_->col eq $col[$n->col_seen - 1]
} @dag);
} else {
tree(first { $_->label eq $c } @dag);
}
} $n->children->@*
] => $n->label;
}
tree($dag[0]);
# bless([ #N0
# bless([ #N1
# bless([ #N2
# bless([ #N3
# bless([] => 'N4'),
# bless([] => 'N5')
# ] => 'N3')
# ] => 'N2')
# ] => 'N1')
# ] => 'N0')
tree($dag[0]);
# bless([ #N0
# bless([ #N1
# bless([ #N6
# bless([ #N7
# bless([] => 'N4')
# ] => 'N7')
# ] => 'N6'),
# bless([] => 'N5')
# ] => 'N1')
# ] => 'N0')
tree($dag[0]);
# exhausted
#tree3.pl
︙
our @dag = (
N(N0 => ['N1']),
N([N1 => 'red'] => ['N2']),
N(N2 => ['N3']),
N(N3 => ['N4', 'N5']),
N(N4 => []),
N(N5 => []),
# N0
N([N1 => 'blue'] => ['N6', 'N5']),
N(N6 => ['N7']),
N(N7 => ['N8', 'N4']),
N([N8 => 'purple'] => ['N5']),
# N5
N([N8 => 'orange'] => []),
N([N8 => 'cyan'] => ['N5', 'N5']),
# N5
# N5
# N4
# N5
);
︙
tree($dag[0]);
# bless([ #N0
# bless([ #N1
# bless([ #N2
# bless([ #N3
# bless([] => 'N4'),
# bless([] => 'N5')
# ] => 'N3')
# ] => 'N2')
# ] => 'N1')
# ] => 'N0')
tree($dag[0]);
# bless([ #N0
# bless([ #N1
# bless([ #N6
# bless([ #N7
# bless([ #N8
# bless([] => 'N5')
# ] => 'N8'),
# bless([] => 'N4')
# ] => 'N7')
# ] => 'N6'),
# bless([] => 'N5')
# ] => 'N1')
# ] => 'N0')
tree($dag[0]);
# exhausted
最佳答案
以下是您想要完成的目标吗? ( python 3)
from collections import defaultdict
from itertools import product
class bless:
def __init__(self, label, children):
self.label = label
self.children = children
def __repr__(self):
return self.__str__()
# Just pretty-print stuff
def __str__(self):
formatter = "\n{}\n" if self.children else "{}"
formatted_children = formatter.format(",\n".join(map(str, self.children)))
return "bless([{}] => '{}')".format(formatted_children, self.label)
class Node:
def __init__(self, label, children):
self.label = label
self.children = children
class DAG:
def __init__(self, nodes):
self.nodes = nodes
# Add the root nodes to a singular, generated root node (for simplicity)
# This is not necessary to implement the color-separation logic,
# it simply lessens the number of edge cases I must handle to demonstate
# the logic. Your existing code will work fine without this "hack"
non_root = {child for node in self.nodes for child in node.children}
root_nodes = [node.label for node in self.nodes if node.label not in non_root]
self.root = Node("", root_nodes)
# Make a list of all the trees
self.tree_list = self.make_trees(self.root)
def tree(self):
if self.tree_list:
return self.tree_list.pop(0)
return list()
# This is the meat of the program, and is really the logic you are after
# Its a recursive function that parses the tree top-down from our "made-up"
# root, and makes <bless>s from the nodes. It returns a list of all separately
# colored trees, and if prior (recusive) calls already made multiple trees, it
# will take the cartesian product of each tree per label
def make_trees(self, parent):
# A defaultdict is just a hashtable that's empty values
# default to some data type (list here)
trees = defaultdict(list)
# This is some nasty, inefficient means of fetching the children
# your code already does this more efficiently in perl, and since it
# contributes nothing to the answer, I'm not wasting time refactoring it
for node in (node for node in self.nodes if node.label in parent.children):
# I append the tree(s) found in the child to the list of <label>s trees
trees[node.label] += self.make_trees(node)
# This line serves to re-order the trees since the dictionary doesn't preserve
# ordering, and also restores any duplicated that would be lost
values = [trees[label] for label in parent.children]
# I take the cartesian product of all the lists of trees each label
# is associated with in the dictionary. So if I have
# [N1-subtree] [red-N2-subtree, blue-N2-subtree] [N3-subtree]
# as children of N0, then I'll return:
# [bless(N0, [N1-st, red-N2-st, N3-st]), bless(N0, [N1-st, blue-N2-st, N3-st])]
return [bless(parent.label, prod) for prod in product(*values)]
if __name__ == "__main__":
N0 = Node('N0', ['N1'])
N1a = Node('N1', ['N2'])
N2 = Node('N2', ['N3'])
N3 = Node('N3', ['N4', 'N5'])
N4 = Node('N4', [])
N5 = Node('N5', [])
N1b = Node('N1', ['N6', 'N5'])
N6 = Node('N6', ['N7'])
N7 = Node('N7', ['N8', 'N4'])
N8a = Node('N8', ['N5'])
N8b = Node('N8', [])
N8c = Node('N8', ['N5', 'N5'])
dag = DAG([N0, N1a, N2, N3, N4, N5, N1b, N6, N7, N8a, N8b, N8c])
print(dag.tree())
print(dag.tree())
print(dag.tree())
print(dag.tree())
print(dag.tree())
print(dag.tree())
bless
对象(来自每个产品)。我返回这个列表,这个过程重复调用堆栈,直到我们最终得到完整的树列表。
bless
s,但是手动调整应该是微不足道的。唯一真正感兴趣的部分是
make_trees()
功能,其余的只是设置验证内容或使代码尽可能容易地与您的 perl 代码进行比较。
bless([
bless([
bless([
bless([
bless([
bless([] => 'N4'),
bless([] => 'N5')
] => 'N3')
] => 'N2')
] => 'N1')
] => 'N0')
] => '')
bless([
bless([
bless([
bless([
bless([
bless([
bless([] => 'N5')
] => 'N8'),
bless([] => 'N4')
] => 'N7')
] => 'N6'),
bless([] => 'N5')
] => 'N1')
] => 'N0')
] => '')
bless([
bless([
bless([
bless([
bless([
bless([] => 'N8'),
bless([] => 'N4')
] => 'N7')
] => 'N6'),
bless([] => 'N5')
] => 'N1')
] => 'N0')
] => '')
bless([
bless([
bless([
bless([
bless([
bless([
bless([] => 'N5'),
bless([] => 'N5')
] => 'N8'),
bless([] => 'N4')
] => 'N7')
] => 'N6'),
bless([] => 'N5')
] => 'N1')
] => 'N0')
] => '')
[]
[]
关于perl - 从 DAG 中提取树,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54010961/
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