Classical FP problem in python : Hamming problem

[Bengt Richter]

On 25 Jan 2005 08:30:03 GMT, Nick Craig-Wood <nick at craig-wood.com> wrote:

>Francis Girard <francis.girard at free.fr> wrote:
>>  def hamming():
>>    def _hamming():
>>      yield 1
>>      hamming2 = hammingGenerators[0]
>>      hamming3 = hammingGenerators[1]
>>      hamming5 = hammingGenerators[2]
>>      for n in imerge(imap(lambda h: 2*h, iter(hamming2)),
>>                      imerge(imap(lambda h: 3*h, iter(hamming3)),
>>                             imap(lambda h: 5*h, iter(hamming5)))):
>>        yield n
>>    hammingGenerators = tee(_hamming(), 4)
>>    return hammingGenerators[3]
>
>If you are after readability, you might prefer this...
>
>def hamming():
>  def _hamming():
>    yield 1
>    for n in imerge(imap(lambda h: 2*h, iter(hamming2)),
>                    imerge(imap(lambda h: 3*h, iter(hamming3)),
>                           imap(lambda h: 5*h, iter(hamming5)))):
>      yield n
>  hamming2, hamming3, hamming5, result = tee(_hamming(), 4)
>  return result
>
>PS interesting thread - never heard of Hamming sequences before!

Are the long words really that helpful?

    def hamming():
      def _hamming():
        yield 1
        for n in imerge(imap(lambda h: 2*h, iter(hg2)),
                        imerge(imap(lambda h: 3*h, iter(hg3)),
                               imap(lambda h: 5*h, iter(hg5)))):
          yield n
      hg2, hg3, hg5, result = tee(_hamming(), 4) # four hamming generators
      return result

Regards,
Bengt Richter