print out Hash key and value pairs recursively - ruby

Question

I am trying to define a function that it can print out any hash values in a tree format. The function will do something like this:
From
{"parent1"=>
{"child1" => { "grandchild1" => 1,
"grandchild2" => 2},
"child2" => { "grandchild3" => 3,
"grandchild4" => 4}}
}
To
parent1:
child1:
grandchild1:1
grandchild2:2
child2:
grandchild3:3
grandchild4:4
And this is my code so far:
def readprop(foo)
level = ''
if foo.is_a?(Hash)
foo.each_key {|key| if foo[key].nil? == false
puts level + key + ":"
level += " "
readprop(foo[key])
end
}
else
puts level + foo
level = level[0,level.length - 2]
end
end
and it will give me a bad format like this:
parent1:
child1:
grandchild1:
1
grandchild2:
2
child2:
grandchild3:
3
grandchild4:
4

Answer 1

You are almost there. One way to solve it is to make level a part of the recursive function parameters. x is the hash in the question.
Simple recursive version:
def print_hash(h,spaces=4,level=0)
h.each do |key,val|
format = "#{' '*spaces*level}#{key}: "
if val.is_a? Hash
puts format
print_hash(val,spaces,level+1)
else
puts format + val.to_s
end
end
end
print_hash(x)
#parent1:
# child1:
# grandchild1: 1
# grandchild2: 2
# child2:
# grandchild3: 3
# grandchild4: 4
In this case you could also convert it to YAML (as mentioned in a comment above)
require 'YAML'
puts x.to_yaml
#---
#parent1:
# child1:
# grandchild1: 1
# grandchild2: 2
# child2:
# grandchild3: 3
# grandchild4: 4

Answer 2

I would use recursion, but there is another way that might be of interest to some. Below I've used a "pretty printer", awesome-print, to do part of the formatting (the indentation in particular), saving the result to a string, and then applied a couple of gsub's to the string to massage the results into the desired format.
Suppose your hash were as follows:
h = { "parent1"=>
{ "child1" => { "grandchild11" => 1,
"grandchild12" => { "great grandchild121" => 3 } },
"child2" => { "grandchild21" => { "great grandchild211" =>
{ "great great grandchild2111" => 4 } },
"grandchild22" => 2 }
}
}
We could then do the following.
require 'awesome_print'
puts str = h.awesome_inspect(indent: -5, index: false, plain: true).
gsub(/^\s*(?:{|},?)\s*\n|[\"{}]/, '').
gsub(/\s*=>\s/, ':')
prints
parent1:
child1:
grandchild11:1,
grandchild12:
great grandchild121:3
child2:
grandchild21:
great grandchild211:
great great grandchild2111:4
grandchild22:2
The steps:
str = h.awesome_inspect(indent: -5, index: false, plain: true)
puts str prints
{
"parent1" => {
"child1" => {
"grandchild11" => 1,
"grandchild12" => {
"great grandchild121" => 3
}
},
"child2" => {
"grandchild21" => {
"great grandchild211" => {
"great great grandchild2111" => 4
}
},
"grandchild22" => 2
}
}
}
s1 = str.gsub(/^\s*(?:{|},?)\s*\n|[\"{}]/, '')
puts s1 prints
parent1 =>
child1 =>
grandchild11 => 1,
grandchild12 =>
great grandchild121 => 3
child2 =>
grandchild21 =>
great grandchild211 =>
great great grandchild2111 => 4
grandchild22 => 2
s2 = s1.gsub(/\s*=>\s/, ':')
puts s2 prints the result above.

Answer 3

Not exactly what you require but I will submit this answer as I think you may find it useful:
require 'yaml'
hash = {"parent1"=> {"child1" => { "grandchild1" => 1,"grandchild2" => 2},
"child2" => { "grandchild3" => 3,"grandchild4" => 4}}}
puts hash.to_yaml
prints:
---
parent1:
child1:
grandchild1: 1
grandchild2: 2
child2:
grandchild3: 3
grandchild4: 4

Answer 4

See Ruby Recursive Tree
Suppose we have
#$ mkdir -p foo/bar
#$ mkdir -p baz/boo/bee
#$ mkdir -p baz/goo
We can get
{
"baz"=>{
"boo"=>{
"bee"=>{}},
"goo"=>{}},
"foo"=>{
"bar"=>{}}}
We can traverse the tree as the following. So, here's a way to make a Hash based on directory tree on disk:
Dir.glob('**/*'). # get all files below current dir
select{|f|
File.directory?(f) # only directories we need
}.map{|path|
path.split '/' # split to parts
}.inject({}){|acc, path| # start with empty hash
path.inject(acc) do |acc2,dir| # for each path part, create a child of current node
acc2[dir] ||= {} # and pass it as new current node
end
acc
}
Thanks to Mladen Jablanović in the other answer for this concept.