Python: Assigning “through” an iterator

问题

I have an iterator over a mutable sequence, e.g.

foo = [1,2,3,4,5]
for bar in foo:

Is there a way to write to the elements in foo by using the reference which is contained in the iterator? The naive assignment:

   bar = 42

does not work of course. Is it possible to use the "behind the curtain" reference to the sequence element which is in the iterator ?

PS: The simple solution with using an index

for i in range(len(a)):
   a[i] = 42

will not work for my case, as I can't expose the container name.

回答1:

From my understanding, your use case is something like this:

class Z:
    def __init__(self):
        self.a, self.b, self.c = 1,2,3

    def it(self):
        for x in self.a, self.b, self.c:
            yield x

z = Z()
for x in z.it():
    if x == 1:
       x = 42 # z.a should be 42? - doesn't work!

This isn't possible in python - there's no "pointer" or "reference" data type. You can work around this by yielding a setter function instead of (or along with) the value:

class Z:
    def __init__(self):
        self.a, self.b, self.c = 1,2,3

    def it(self):
        for x in 'abc':
            yield getattr(self, x), lambda y: setattr(self, x, y)

z = Z()
for x, setter in z.it():
    if x == 1:
       setter(42) # works!

回答2:

Use enumerate() to generate indices for you in the loop:

for i, bar in enumerate(foo):
    foo[i] = bar + 42

回答3:

You could use a list comprehension:

foo[:] = [expression for bar in foo]

or, if the assignment is too complicated for an expression, you could use a function, func:

foo[:] = [func(bar) for bar in foo]

---

If you only have an iterator, then there is no way to reassign values to the underlying container for there is no guarantee that there even is an underlying container:

def generator():
    for i in range(10):
        yield i

for bar in generator():
    # Nothing you put here can change what's going on in generator

来源：https://stackoverflow.com/questions/19562356/python-assigning-through-an-iterator