How to sort default dict with list values in python

Question

I have a defaultdict with list as

defaultdict(<class 'list'>, {'SOL200122': ['125', '135', '145', '170', '120', '130', '140', '150', '160']}

I want it to be sorted like

defaultdict(<class 'list'>, {'SOL200122': ['120', '125', '130', '135', '140', '145', '150', '160', '170']}

Have tried

sorted(myDict.items(), key=lambda k_v: (k_v[1][2]), reverse=True)

But not working

Answer 1

Use list.sort with key=int (thanks @Metapod, @deceze et al):

for v in d.values():
    v.sort(key=int)

If you want a one-liner (which makes no sense in this case because for-loop is very nice), a monstrosity like this one could be one I guess:

dict(zip(d.keys(), map(lambda x: sorted(x, key=int), d.values())))

Output:

{'SOL200122': ['120', '125', '130', '135', '140', '145', '150', '160', '170']}

Answer 2

try this:

for v in d.values():
    v.sort(key=int)

output:

{'SOL200122': ['120', '125', '130', '135', '140', '145', '150', '160', '170']}