original_docstring: r"""
          Return ``True`` if ``self`` is a subword of ``other``, and ``False`` otherwise.
  
          A finite word `u` is a *subword* of a finite word `v` if `u` is a
          subsequence of `v`. See Chapter 6 on Subwords in [Lot1997]_.
  
          Some references define subword as a consecutive subsequence. Use
          :meth:`is_factor` if this is what you need.
  
          INPUT:
  
          ``other`` -- a finite word
  
          EXAMPLES::
  
              sage: Word('bb').is_subword_of(Word('ababa'))
              True
              sage: Word('bbb').is_subword_of(Word('ababa'))
              False
  
          ::
  
              sage: Word().is_subword_of(Word('123'))
              True
              sage: Word('123').is_subword_of(Word('3211333213233321'))
              True
              sage: Word('321').is_subword_of(Word('11122212112122133111222332'))
              False
  
          .. SEEALSO::
  
              :meth:`longest_common_subword`
              :meth:`number_of_subword_occurrences`
              :meth:`is_factor`
          
docstring: 
  
  
  
  Some references define subword as a consecutive subsequence. Use
docstring_tokens: Some references define subword as a consecutive subsequence . Use
code: def is_subword_of(self, other):
          r"""
          Return ``True`` if ``self`` is a subword of ``other``, and ``False`` otherwise.
  
          A finite word `u` is a *subword* of a finite word `v` if `u` is a
          subsequence of `v`. See Chapter 6 on Subwords in [Lot1997]_.
  
          Some references define subword as a consecutive subsequence. Use
          :meth:`is_factor` if this is what you need.
  
          INPUT:
  
          ``other`` -- a finite word
  
          EXAMPLES::
  
              sage: Word('bb').is_subword_of(Word('ababa'))
              True
              sage: Word('bbb').is_subword_of(Word('ababa'))
              False
  
          ::
  
              sage: Word().is_subword_of(Word('123'))
              True
              sage: Word('123').is_subword_of(Word('3211333213233321'))
              True
              sage: Word('321').is_subword_of(Word('11122212112122133111222332'))
              False
  
          .. SEEALSO::
  
              :meth:`longest_common_subword`
              :meth:`number_of_subword_occurrences`
              :meth:`is_factor`
          """
          its = iter(self)
          try:
              s = next(its)
              for e in other:
                  if s == e:
                      s = next(its)
              else:
                  return False
          except StopIteration:
              return True
code_tokens: def is_subword_of ( self , other ) : its = iter ( self ) try : s = next ( its ) for e in other : if s == e : s = next ( its ) else : return False except StopIteration : return True
short_docstring: r"""
  Return ``True`` if ``self`` is a subword of ``other``, and ``False`` otherwise.
short_docstring_tokens: r " " " Return ` ` True ` ` if ` ` self ` ` is a subword of ` ` other ` ` and ` ` False ` ` otherwise .
comment: r"""
          Return ``True`` if ``self`` is a subword of ``other``, and ``False`` otherwise.
  
          A finite word `u` is a *subword* of a finite word `v` if `u` is a
          subsequence of `v`. See Chapter 6 on Subwords in [Lot1997]_.
  
          Some references define subword as a consecutive subsequence. Use
          :meth:`is_factor` if this is what you need.
  
          INPUT:
  
          ``other`` -- a finite word
  
          EXAMPLES::
  
              sage: Word('bb').is_subword_of(Word('ababa'))
              True
              sage: Word('bbb').is_subword_of(Word('ababa'))
              False
  
          ::
  
              sage: Word().is_subword_of(Word('123'))
              True
              sage: Word('123').is_subword_of(Word('3211333213233321'))
              True
              sage: Word('321').is_subword_of(Word('11122212112122133111222332'))
              False
  
          .. SEEALSO::
  
              :meth:`longest_common_subword`
              :meth:`number_of_subword_occurrences`
              :meth:`is_factor`
          """