Oliver Eichler wrote:
> what will be the fastest solution to following problem
>
> class a:
> def __init__(self):
> self.varA = 0
> self.varB = 0
> s.o. ...
>
>
> listA = [a]*10
>
>
> varA = []
> varB = []
>
> for l in listA:
> varA.append(l.varA)
> varB.append(l.varB)
> s.o. ...
>
>
> I could think of:
> [varA.append(l.varA) for l in listA]
> [varB.append(l.varB) for l in listA]
> s.o. ...
>
> But is this faster? After all I have to iterate over listA several times.
>
> Any better solutions?
A different approach, a "virtual list" could even be faster if client code
must access only a fraction of the items in the virtual list. If you are
lucky, you never have to iterate over the complete list.
>>> class A:
... def __init__(self, a):
... self.a = a
... def __repr__(self):
... return "A(%r)" % self.a
...
>>> items = map(A, range(10))
>>> items
[A(0), A(1), A(2), A(3), A(4), A(5), A(6), A(7), A(8), A(9)]
>>> class List:
... def __init__(self, items, getter):
... self.items = items
... self.getter = getter
... def __getitem__(self, index):
... return self.getter(self.items[index])
... def __iter__(self):
... return imap(self.getter, self.items)
...
>>> from operator import attrgetter
>>> from itertools import imap
>>> items_a = List(items, attrgetter("a"))
>>> items_a
<__main__.List instance at 0x402ae2cc>
>>> items_a[2]
2
>>> list(items_a)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> items[2] = A(42)
>>> items
[A(0), A(1), A(42), A(3), A(4), A(5), A(6), A(7), A(8), A(9)]
>>> items_a[2]
42
Peter
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