ruby,recursion , How to flatten a structure of embedded Set and Hash


How to flatten a structure of embedded Set and Hash

Question:

Tag: ruby,recursion

I would like to convert an embedding structure into a flat one.

Here some examples:

t(
  Set[]
) # => Set[]

t(
  Set[ 'foo' ]
) # => Set[ ['foo'] ]

t(
  Set[ 'foo0',
       'foo1',
       'foo2' ]
) # => Set[ ['foo0'], ['foo1'], ['foo2'] ]

t(
  Set[ 'foo' => Set[ 'bar' ] ]
) # => Set[ ['foo', 'bar'] ]

t(
  Set[ 'foo' => Set[ 'bar0', 'bar1', 'bar2' ] ]
) # => Set[ ['foo', 'bar0'],
  #         ['foo', 'bar1'],
  #         ['foo', 'bar2'] ]

t(
  Set[ {'foo' => Set[ 'bar0', 'bar1', 'bar2' ]},
       {'FOO' => Set[ 'BAR0', 'BAR1', 'BAR2' ]} ]
) # => Set[ ['foo', 'bar0'],
  #         ['foo', 'bar1'],
  #         ['foo', 'bar2'],
  #         ['FOO', 'BAR0'],
  #         ['FOO', 'BAR1'],
  #         ['FOO', 'BAR2'] ]

t(
  Set[ {'foo' => Set[ {'FOO' => Set[ 'BAR' ]} ]} ]
) # => Set[ ['foo', 'FOO', 'BAR'] ]

t(
  Set[ {'foo' => Set[ {'FOO' => Set[ 'BAR' ]} ]},
       'baz' ]
) # => Set[ ['foo', 'FOO', 'BAR'],
  #         'baz' ]

t(
  Set[ {'foo' => Set[ {'FOO' => Set[ 'BAR0', 'BAR1' ]} ]},
       'baz' ]
) # => Set[ ['foo', 'FOO', 'BAR0'],
  #         ['foo', 'FOO', 'BAR1'],
  #         ['baz'] ]

t(
  Set[ {'foo' => Set[ {'FOO' => Set[ 'BAR0', {'BAR1' => Set[ {'abc' => Set[ 'def' ]} ]} ]} ]},
       'baz' ]
) # => Set[ ['foo', 'FOO', 'BAR0'],
  #         ['foo', 'FOO', 'BAR1', 'abc', 'def'],
  #         ['baz'] ]

I think we should use resistivity to convert the given structure. But I have no idea for an implementation. Please feel free to any example of t function.

Edit:

In other words, I would like to develop the embedding structure into a flatten structure.

We can see the embedding structure such as an expression with factors. And if we multiply them, the flatten structure is returned.


Answer:

Recursion is your friend:

require 'set'
def t_h(inp, prefix = [])
    if (inp.is_a?(Hash))
        result = []
        inp.each do |k,v|
            pprefix = prefix.dup
            result << t_h(v, pprefix << k)
        end
        return result.flatten(1)
    elsif (inp.is_a?(Set))
        result = []
        inp.each do |el|
            result << t_h(el, prefix)
        end
        return result.flatten(1)
    else
        pprefix = prefix.dup
        return [ pprefix << inp ]
    end
end

def t(inp)
    Set.new(t_h(inp))
end

# examples
t(Set[ 'foo' => Set[ 'bar' ] ])
t(Set[ 'foo' => Set[ 'bar0', 'bar1', 'bar2' ] ])
t(
  Set[ {'foo' => Set[ 'bar0', 'bar1', 'bar2' ]},
       {'FOO' => Set[ 'BAR0', 'BAR1', 'BAR2' ]} ]
)

Code is far from optimal, but you get the idea.


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