Use list comprehension to build a tuple

tup = tuple((element.foo, element.bar) for element in alist)

Technically, it's a generator expression. It's like a list comprehension, but it's evaluated lazily and won't need to allocate memory for an intermediate list.

For completeness, the list comprehension would look like this:

tup = tuple([(element.foo, element.bar) for element in alist])

 

PS: attrgetter is not faster (alist has a million items here):

In [37]: %timeit tuple([(element.foo, element.bar) for element in alist])
1 loops, best of 3: 165 ms per loop

In [38]: %timeit tuple((element.foo, element.bar) for element in alist)
10 loops, best of 3: 155 ms per loop

In [39]: %timeit tuple(map(operator.attrgetter('foo','bar'), alist))
1 loops, best of 3: 283 ms per loop

In [40]: getter = operator.attrgetter('foo','bar')

In [41]: %timeit tuple(map(getter, alist))
1 loops, best of 3: 284 ms per loop

In [46]: %timeit tuple(imap(getter, alist))
1 loops, best of 3: 264 ms per loop