Tricky hash initialization

ruby-talk
1

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

···

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2

Ray Pereda wrote:

How do I this for arbitrarily deep hashes?

Hash.new {|h,k| h[k] = Hash.new(&h.default_proc)}

HTH,
Sebastian

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3

Sebastian Hungerecker 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

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4

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=>{}}}}

···

On Aug 15, 2008, at 1:32, Ray Pereda wrote:

--
Mikael Høilund
http://hoilund.org/

5

Hi --

···

On Fri, 15 Aug 2008, Ray Pereda wrote:

Sebastian Hungerecker 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"=>{}}}}

I think it's a variable clobbering issue. Try this:

hash = Hash.new {|h,k| h[k] = Hash.new(&h.default_proc) }

David

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Rails training from David A. Black and Ruby Power and Light:
  * Advancing With Rails August 18-21 Edison, NJ
  * Co-taught by D.A. Black and Erik Kastner
See http://www.rubypal.com for details and updates!

6

I should add: the same-variable-name version works in Ruby 1.9.

David

···

On Fri, 15 Aug 2008, David A. Black wrote:

Hi --

On Fri, 15 Aug 2008, Ray Pereda wrote:

Sebastian Hungerecker 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"=>{}}}}

I think it's a variable clobbering issue. Try this:

hash = Hash.new {|h,k| h[k] = Hash.new(&h.default_proc) }

David

--
Rails training from David A. Black and Ruby Power and Light:
* Advancing With Rails August 18-21 Edison, NJ
* Co-taught by D.A. Black and Erik Kastner
See http://www.rubypal.com for details and updates!

--
Rails training from David A. Black and Ruby Power and Light:
  * Advancing With Rails August 18-21 Edison, NJ
  * Co-taught by D.A. Black and Erik Kastner
See http://www.rubypal.com for details and updates!