"Steve R. Hastings" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> On Sun, 26 Mar 2006 10:34:16 -0700, Steven Bethard wrote:
> > What's the original?
>
>
>
> def minimum(cmp, lst):
> """minimum(cmp, lst)
>
> Returns the minimal element in non-empty list LST with elements
> compared via CMP() which should return values with the same semantics
> as Python's cmp(). If there are several minimal elements, the last
> one is returned.
> """
>
> if not lst:
> raise ValueError, 'empty list'
>
> minval = lst[0]
>
> for i in xrange(1, len(lst)):
> v = lst[i]
> if cmp(minval, v) > 0:
> minval = v
>
> return minval
>
>
>
> This is from Google's "goopy" package.
>
> http://goog-goopy.sourceforge.net/
>
> --
> Steve R. Hastings "Vita est"
> [EMAIL PROTECTED] http://www.blarg.net/~steveha
>
The doc string is not correct. If there are several minimal elements, the
*first* one is returned. See this example:
import math
class Vector(object):
def __init__(self,*x):
# x is list of n-dimensional coordinates
self.x = x
def magnitude(self):
return math.sqrt(sum(x**2 for x in self.x))
def __cmp__(self,other):
return cmp(self.magnitude(), other.magnitude())
def __str__(self):
return "Vector(%s)" % ",".join(str(x) for x in self.x)
a = Vector(0,1)
b = Vector(0,0)
c = Vector(1,0)
# these print 1, -1, and 0 to show that cmp() and Vector are doing the right
thing
print cmp(a,b)
print cmp(b,a)
print cmp(a,c)
print min(a,b) # gives Vector(0,0), or b
print max(a,b,c) # gives Vector(0,1) or a
print min(a,c) # gives Vector(0,1) or a
print minimum(Vector.__cmp__, [a,b]) # gives Vector(0,0), or b
print minimum(Vector.__cmp__, [a,c]) # gives Vector(0,1) or a
-- Paul
--
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