[Tutor] totalViruses[i] /= float(numTrials),

[Prasad, Ramit]

Alan Gauld wrote:
> On 24/04/13 20:32, Oscar Benjamin wrote:
> > On 24 April 2013 20:07, Alan Gauld <alan.gauld at btinternet.com> wrote:
> >> On 24/04/13 16:52, Dave Angel wrote:
> >>
> >>>> Does it mean? ;  totalViruses[i] = totalViruses[i]/float(numTrials)
> >>>>
> >>> As the others have said, that's exactly right, at least if
> >>> totalViruses[i] is immutable, like an int or a float.
> >>
> >>
> >> What difference does immutability make here?
> 
> > The subtle distinction that Dave is referring to is about modifying an
> > object in place vs rebinding a name to a newly created object.
> 
> Sure, the bejhaviour with regard to the objects after the opreation is
> slightly different. But it makes noi dsifference to the duality of
> 
> x += n
> v
> x = x + n
> 
> The mutability issues are identical.
> 
> > Which behaviour occurs is entirely up to the author of the __iadd__
> > method of the class in question. This method can modify and return
> > self or it can return a different object.

Actually being able to return anything is true of even non iXXX 
methods like __div__. That's why without code, it is impossible
to say for sure, but probably it does what is expected.

>>> class Foo(object):
...     def __div__(self, num):
...         return 'shanana'*num # Bad!
...     
>>> z  = Foo()
>>> z /= 2
>>> z
'shananashanana'

> 
> This does make a difference because iadd (etc) can make a difference and
> I had forgotten that there are a separate set of operator methods for
> the ixxx operations, I was assuming that ixxx called the standard xxx.
> So in the special case of a user type implementing xxx differently to
> ixxx the behaviour might differ. But for all other cases I can think of
> the duality will still work? Apart from this exception
> 
> x += y
> 
> will yield the same result as
> 
> x = x + y
> 
> I think....

The only usecase (among built-ins) that I could find to contradict 
the statement, but I think it is a special case.

>>> a = [1,2,3] 
>>> b = (a,'c')
>>> b[0]+= [4]
TypeError: 'tuple' object does not support item assignment

>>> print b
([1, 2, 3, 4], 'c')
>>> b[0] = b[0] + [5]
TypeError: 'tuple' object does not support item assignment

>>> print b
([1, 2, 3, 4], 'c')


~Ramit


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