frange() question

Michael J. Fromberger Michael.J.Fromberger at Clothing.Dartmouth.EDU
Fri Sep 21 14:54:20 CEST 2007


In article <fctrc1$doh$1 at news.nems.noaa.gov>,
 George Trojan <george.trojan at noaa.gov> wrote:

> A while ago I found somewhere the following implementation of frange():
> 
> def frange(limit1, limit2 = None, increment = 1.):
>      """
>      Range function that accepts floats (and integers).
>      Usage:
>      frange(-2, 2, 0.1)
>      frange(10)
>      frange(10, increment = 0.5)
>      The returned value is an iterator.  Use list(frange) for a list.
>      """
>      if limit2 is None:
>          limit2, limit1 = limit1, 0.
>      else:
>          limit1 = float(limit1)
>      count = int(math.ceil(limit2 - limit1)/increment)
>      return (limit1 + n*increment for n in range(count))
> 
> I am puzzled by the parentheses in the last line. Somehow they make 
> frange to be a generator:
>  >> print type(frange(1.0, increment=0.5))
> <type 'generator'>
> But I always thought that generators need a keyword "yield". What is 
> going on here?

Hi, George,

The expression returned is a "generator expression", 

  return (limit1 + n*increment for n in range(count))

Thus, although frange itself is not written as a generator, it does 
return a generator as its result.  The syntax is like that of list 
comprehensions; see:

  <http://docs.python.org/ref/genexpr.html>

Cheers,
-M

-- 
Michael J. Fromberger             | Lecturer, Dept. of Computer Science
http://www.dartmouth.edu/~sting/  | Dartmouth College, Hanover, NH, USA



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