Sort nested dictionary by value, and remainder by another value, in Python

Solution 1:

Use the key argument for sorted(). It lets you specify a function that, given the actual item being sorted, returns a value that should be sorted by. If this value is a tuple, then it sorts like tuples sort - by the first value, and then by the second value.

sorted(your_list, key=lambda x: (your_dict[x]['downloads'], your_dict[x]['date']))

Solution 2:

You can pass a key function to sorted which returns a tuple containing the two things you wish to sort on. Assuming that your big dictionary is called d:

def keyfunc(tup):
    key, d = tup
    return d["downloads"], d["date"]

items = sorted(d.items(), key = keyfunc)

You can do this with a lambda if you prefer, but this is probably more clear. Here's the equivalent lambda-based code:

items = sorted(d.items(), key = lambda tup: (tup[1]["downloads"], tup[1]["date"]))

Incidentally, since you mentioned that you wanted to sort by "downloads" first, the above two examples sort according to download counts in ascending order. However, from context it sounds like you might want to sort in decreasing order of downloads, in which case you'd say

return -d["downloads"], d["date"]

in your keyfunc. If you wanted something like sorting in ascending order for non-zero download numbers, then having all zero-download records after that, you could say something like

return (-d["downloads"] or sys.maxint), d["date"]