I want to find the key corresponding to the min or max value of a dictionary in julia. In Python I would to the following:
my_dict = {1:20, 2:10}
min(my_dict, my
You could use reduce like this, which will return the key of the first smallest value in d
:
reduce((x, y) -> d[x] ≤ d[y] ? x : y, keys(d))
This only works for non-empty Dict
s, though. (But the notion of the “key of the minimal value of no values” does not really make sense, so that case should usually be handled seperately anyway.)
Edit regarding efficiency.
Consider these definitions (none of which handle empty collections)...
m1(d) = reduce((x, y) -> d[x] ≤ d[y] ? x : y, keys(d))
m2(d) = collect(keys(d))[indmin(collect(values(d)))]
function m3(d)
minindex(x, y) = d[x] ≤ d[y] ? x : y
reduce(minindex, keys(d))
end
function m4(d)
minkey, minvalue = next(d, start(d))[1]
for (key, value) in d
if value < minvalue
minkey = key
minvalue = value
end
end
minkey
end
...along with this code:
function benchmark(n)
d = Dict{Int, Int}(1 => 1)
m1(d); m2(d); m3(d); m4(d); m5(d)
while length(d) < n
setindex!(d, rand(-n:n), rand(-n:n))
end
@time m1(d)
@time m2(d)
@time m3(d)
@time m4(d)
end
Calling benchmark(10000000)
will print something like this:
1.455388 seconds (30.00 M allocations: 457.748 MB, 4.30% gc time)
0.380472 seconds (6 allocations: 152.588 MB, 0.21% gc time)
0.982006 seconds (10.00 M allocations: 152.581 MB, 0.49% gc time)
0.204604 seconds
From this we can see that m2
(from user3580870's answer) is indeed faster than my original solution m1
by a factor of around 3 to 4, and also uses less memory. This is appearently due to the function call overhead, but also the fact that the λ expression in m1
is not optimized very well. We can alleviate the second problem by defining a helper function like in m3
, which is better than m1
, but not as good as m2
.
However, m2
still allocates O(n) memory, which can be avoided: If you really need the efficiency, you should use an explicit loop like in m4
, which allocates almost no memory and is also faster.