Walrus operator example in PEP572

问题

One of the examples given in PEP572 is

# Reuse a value that's expensive to compute
[y := f(x), y**2, y**3]

currently in python, you'd have to do one of the following:

# option 1
y = f(x)
[y, y**2, y**3]

or

# option 2 
[f(x), f(x)**2, f(x)**3]

the example implies that option 2 here could be improved, but I have never seen that recommended over the first option. Is there ever a reason why option 2 (and therefore the walrus operator) would be better than option 1?

回答1:

Just to make things clear:

[y := f(x), y**2, y**3]

is equivalent to:

y = f(x)
[y, y**2, y**3]

(f(x) is called only once)

but, in general, not this:

[f(x), f(x)**2, f(x)**3]

(f(x) is called three times)

because of potential f() side-effects (or potential unnecessary computational burden, if f() is a pure function).

So, in general, replacing [f(x), f(x)**2, f(x)**3] with [y := f(x), y**2, y**3] should be inspected carefully.

---

For example:

def f(x):
    print('Brooks was here.')
    return 2 * x


x = 1
y = f(x)
l1 = [y, y**2, y**3]

prints Brooks was here. once, while:

l2 = [f(x), f(x)**2, f(x)**3]

will print Brooks was here. three times. Of course, l1 == l2.

---

So, to answer your question more directly, you may want to use:

[f(x), f(x)**2, f(x)**3]

and not this

y = f(x)
[y, y**2, y**3]

when you are specifically interested in the side-effects, whatever that might be.

回答2:

A walrus with dynamic programming, it could be faster depend on f(x).

e = 1
[e := e * f(x) for i in range(1, 4)]

来源：https://stackoverflow.com/questions/57309129/walrus-operator-example-in-pep572