Re: "Collapsing" a list into a list of changes

[Steven Bethard]

John Lenton wrote:
For example, the fastest way
to get the factorial of a (small enough) number in pure python is
  factorial = lambda n: reduce(operator.mul, range(1, n+1))
Gah!  I'll never understand why people use lambda, which is intended to 
create _anonymous_ functions, to create named functions!  This code is 
basically equivalent to:

    def factorial(n):
        return reduce(operator.mul, range(1, n+1))
except that factorial.__name__ is now the meaningful 'factorial', 
instead of '<lambda>'.  See "Inappropriate use of Lambda" in
    http://www.python.org/moin/DubiousPython

My personal beef with the inappropriate use of lambda aside, and 
ignoring the fact that the reduce solution as written fails for 
factorial(0), some timings are in order.  Given fact.py:

----------------------------------------------------------------------
import operator
def f1(n):
    result = 1
    for x in xrange(1, n+1):
        result *= x
    return result
def f2(n):
    return reduce(operator.mul, range(1, n+1))
def f3(n):
    return reduce(operator.mul, xrange(1, n+1))
----------------------------------------------------------------------
I get the following timeit results (with Python 2.4):
$ python -m timeit -s "import fact" "[fact.f1(x) for x in xrange(1, 100)]"
100 loops, best of 3: 2.53 msec per loop
$ python -m timeit -s "import fact" "[fact.f2(x) for x in xrange(1, 100)]"
100 loops, best of 3: 2.46 msec per loop
$ python -m timeit -s "import fact" "[fact.f3(x) for x in xrange(1, 100)]"
100 loops, best of 3: 2.17 msec per loop
So it does look like reduce is actually faster in this case, especially 
if you use xrange instead of range.

Thanks for the example!
Steve
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