my_list = [1, 2] def f(): print my_list yield 11 print my_list yield 22 print my_list my_list[:] = f() print "finally ", print my_list output:
[1, 2] [1, 2] [1, 2] finally [11, 22] what I expected was:
[1, 2] [11, 2] [11, 22] finally [11, 22] Someone once told me slice assignment was in place. Obviously not. Is there an elegant way to achieve it?
Slice assignment is in-place, but the assignment doesn't happen until the entire generator is consumed. It doesn't assign one element at a time to the list. It reads them all and then sticks them all into the list at once.
(Note that it has to do it this way because you can assign a sequence to a slice even if the sequence is a different length than the original slice. You can do, e.g., x[2:3] = [1, 2, 3, 4, 5, 6]. There's no way to do this by replacing one element at a time, because there's no one-to-one mapping between the old slice and the new sequence that's replacing it.)
There isn't a way to achieve what you want by slice assignment, because slice assignment always works in this all-at-once way. You would have to iterate over the list and the generator in parallel and replace individual elements one at a time.
It is possible to do this, but not as directly.
Consider:
my_list = [1,2,3,4] def f(): print 'first: ', my_list yield 11 print 'second:', my_list yield 22 print 'third: ', my_list def inplace(l, func): gen=func() for i,x in enumerate(l): try: l[i]=gen.next() except StopIteration: if i<len(l): del l[i:] break inplace(my_list,f) print "final: ", my_list Prints:
first: [1, 2, 3, 4] second: [11, 2, 3, 4] third: [11, 22, 3, 4] final: [11, 22] Of course, the way it is written, the list will be the shorter of the original list or the elements in the generator.
