ruby - How to flatten a structure of embedded Set and Hash -


i convert embedding structure flat one.

  • an embedding structure set of 0 or more objects, such as: string or hash having string key , other embedding structure value.
  • a flat structure set of arrays or strings.

here 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 should use resistivity convert given structure. have no idea implementation. please feel free example of t function.

edit:

in other words, develop embedding structure flatten structure.

we can see embedding structure such expression factors. , if multiply them, flatten structure returned.

recursion friend:

require 'set' def t_h(inp, prefix = [])     if (inp.is_a?(hash))         result = []         inp.each |k,v|             pprefix = prefix.dup             result << t_h(v, pprefix << k)         end         return result.flatten(1)     elsif (inp.is_a?(set))         result = []         inp.each |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 far optimal, idea.


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