code: def word_frequency(word, lang, wordlist='best', minimum=0.):
      
      args = (word, lang, wordlist, minimum)
      try:
          return _wf_cache[args]
      except KeyError:
          if len(_wf_cache) >= CACHE_SIZE:
              _wf_cache.clear()
          _wf_cache[args] = _word_frequency(*args)
          return _wf_cache[args]
docs: Get the frequency of `word` in the language with code `lang`, from the
      specified `wordlist`.
  
      These wordlists can be specified:
  
      - 'large': a wordlist built from at least 5 sources, containing word
        frequencies of 10^-8 and higher
      - 'small': a wordlist built from at least 3 sources, containing word
        frquencies of 10^-6 and higher
      - 'best': uses 'large' if available, and 'small' otherwise
  
      The value returned will always be at least as large as `minimum`.
      You could set this value to 10^-8, for example, to return 10^-8 for
      unknown words in the 'large' list instead of 0, avoiding a discontinuity.