h = Hash.new {|h,k| h[k] = {} }
This is one level deep defaults to a fresh new hash.
h = Hash.new {|h,k| h[k] = Hash.new{ |h2,k2| h2[k2] = {} } }
This is two levels deep initialization to a fresh new hash.
How do I this for arbitrarily deep hashes?
h[“a”][“b”][“c”] = 3 #=> {“c”=>3}
I’m using this to build very clean implementations of state machine
building algorithms.
Your help is very much appreciated.
-Ray
Ray Pereda wrote:
How do I this for arbitrarily deep hashes?
Hash.new {|h,k| h[k] = Hash.new(&h.default_proc)}
HTH,
Sebastian
Sebastian H. wrote:
Hash.new {|h,k| h[k] = Hash.new(&h.default_proc)}
Thanks for the attempt. Seems on the right track but not quite.
h = Hash.new {|h,k| h[k] = Hash.new(&h.default_proc)}
=> {}
h[“a”][“b”][“c”]
=> {}
h
=> {“c”=>{}}
After h[“a”][“b”][“c”]
what needs to happen is h is equal to
{“a”=>{“b”=>{“c”=>{}}}}
Appreciating the Ruby Community’s Help,
Ray
On Aug 15, 2008, at 1:32, Ray Pereda wrote:
h = Hash.new {|h,k| h[k] = Hash.new(&h.default_proc)}
Don’t do that. The h inside the block is conflicting with the h
outside. Make sure the variables don’t collide, and it works:
the_hash = Hash.new {|h,k| h[k] = Hash.new(&h.default_proc)}
=> {}
the_hash[:a][:b][:c]
=> {}
the_hash
=> {:a=>{:b=>{:c=>{}}}}
I should add: the same-variable-name version works in Ruby 1.9.
David
Hi –
On Fri, 15 Aug 2008, Ray Pereda wrote:
=> {“c”=>{}}
After h[“a”][“b”][“c”]
what needs to happen is h is equal to
{“a”=>{“b”=>{“c”=>{}}}}
I think it’s a variable clobbering issue. Try this:
hash = Hash.new {|h,k| h[k] = Hash.new(&h.default_proc) }
David