Use list comprehension to build a tuple
How can I use list comprehension to build a tuple of 2-tuple from a list. It would be equivalent to
tup = ()
for element in alist:
tup = tup + ((element.foo, element.bar),)
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
You may use the following expression
tup = *[(element.foo, element.bar) for element in alist]
This will first of all generate a list of tuples and then it will convert that list of tuples into a tuple of tuples.