根据Python中单独列表中键的顺序，按键对字典列表进行排序

Question

I have a list of dictionaries which looks something like this:

[
    {
        "format": "format1",
        "code": "tr"
    },
    {
        "format": "format2",
        "code": "hc"
    },
    {
        "format": "format3",
        "code": "bx"
    },
    {
        "format": "format4",
        "code": "mm"
    },
    {
        "format": "format5",
        "code": "el"
    }
]

I need to order this list based on the value of the code key, but the order of the codes is determined by a separate list:

code_order = ["mm", "hc", "el", "tr", "bx"]

So the final list should look like this:

[
    {
        "format": "format4",
        "code": "mm"
    },
    {
        "format": "format2",
        "code": "hc"
    },
    {
        "format": "format5",
        "code": "el"
    },
    {
        "format": "format1",
        "code": "tr"
    },
    {
        "format": "format3",
        "code": "bx"
    }
]

Does anyone have any suggestions on how to achieve this? I'm having a difficult time figuring out how to do this kind of sort.

Answer 1

Python 2.7+:

lookup = {s: i for i, s in enumerate(code_order)}
print(sorted(l, key=lambda o: lookup[o['code']]))

Older:

lookup = dict((s, i) for i, s in enumerate(code_order))
print sorted(l, key=lambda o: lookup[o['code']])

Answer 2

If l is your list of dicts, then

sorted(l, key=lambda d: code_order.index(d['code']))

should do the trick. Read that as:

key is the function that looks up code in the given dict d , then checks the index of that code in code_order , so the final sort is by those indices.

(If code_order gets really large, then keep in mind that list.index takes linear time so you'd better replace it by a dict. But for this short code_order , it shouldn't matter.)