Question

Is it possible to have a list be evaluated lazily in Python?

For example

a = 1
list = [a]
print list
#[1]
a = 2
print list
#[1]

If the list was set to evaluate lazily then the final line would be [2]

Answer 1

The concept of “lazy” evaluation normally comes with functional languages — but in those you could not reassign two different values to the same identifier, so, not even there could your example be reproduced.

The point is not about laziness at all — it is that using an identifier is guaranteed to be identical to getting a reference to the same value that identifier is referencing, and re-assigning an identifier, a bare name, to a different value, is guaranteed to make the identifier refer to a different value from them on. The reference to the first value (object) is not lost.

Consider a similar example where re-assignment to a bare name is not in play, but rather any other kind of mutation (for a mutable object, of course — numbers and strings are immutable), including an assignment to something else than a bare name:

>>> a = [1]
>>> list = [a]
>>> print list
[[1]]
>>> a[:] = [2]
>>> print list
[[2]]

Since there is no a - ... that reassigns the bare name a, but rather an a[:] = ... that reassigns a‘s contents, it’s trivially easy to make Python as “lazy” as you wish (and indeed it would take some effort to make it “eager”!-)… if laziness vs eagerness had anything to do with either of these cases (which it doesn’t;-).

Just be aware of the perfectly simple semantics of “assigning to a bare name” (vs assigning to anything else, which can be variously tweaked and controlled by using your own types appropriately), and the optical illusion of “lazy vs eager” might hopefully vanish;-)