How to find the item in a list whose f(item) is the smallest?

I have a list, l and a function f. f is not strictly increasing or decreasing. How can I find the item in the list whose f(item) is the smallest? For example, let's say the list is:

l = [1, 2, 3, 4]

and list(f(x)for x in l) is:

[2, 9, 0, 3]

f(3) is smaller than f of any of the other ones, so it should print "3". What's the shortest way to do this? I initially tried:

min(f(x) for x in l)

But this gives 0, not 3. If I was shooting for readability, not brevity, I would do:

index = 0
smallest = f(l)
for i in range(len(l)):
value = f(l[i])
if value < smallest:
smallest = value
index = i

This is fine, but horrendous for code-golf. Even if it was golfed

i,s=0,f(l)
for x in range(len(l)):
v=f(l[x])
if v<s:s,i=v,x

This is a bad solution. The shortest solution I can think of is:

g=[f(x)for x in l];print(l[g.index(min(g))])

(44 bytes) How can I golf this down further?

• Just min(l,key=f). – vaultah May 17 '16 at 19:42
• @vaultah Post that as an answer. – NoOneIsHere May 17 '16 at 19:46
• @KevinLau-notKenny This is a tips question. That is on-topic. – NoOneIsHere May 17 '16 at 19:50
• @trichoplax actually I think it's this one: meta.codegolf.stackexchange.com/a/1724/31625 – FryAmTheEggman May 17 '16 at 19:51
• This is an interesting topic. Perhaps make it general, not limited to Python? It may be interesting to see how to do that in different languages – Luis Mendo May 17 '16 at 20:29