How to sort a dict by the values inside its keys?

Question

So I have something like this:

[
{
    "arg1":
          {"valueone" : 2, "valuetwo": 5"},
},
{
    "arg2":
          {"valueone" : 7, "valuetwo": 3"},
},
{
    "arg3":
          {"valueone" : 3, "valuetwo": 7"},
},
{
    "arg4":
          {"valueone" : 4, "valuetwo": 6"},
},
{
    "arg5":
          {"valueone" : 9, "valuetwo": 5"},
},

]

I want to sort them by their valueone s. I have a knowledge of OrderedDict, however I got some troubles to do this.

An expected result would be:

[
{
    "arg5":
          {"valueone" : 9, "valuetwo": 5"},
},
{
    "arg2":
          {"valueone" : 7, "valuetwo": 3"},
},
{
    "arg4":
          {"valueone" : 4, "valuetwo": 6"},
},
{
    "arg3":
          {"valueone" : 3, "valuetwo": 7"},
},

{
    "arg1":
          {"valueone" : 2, "valuetwo": 5"},
},


]

Any helps would be appreciated.

Answer 1

You could use the key parameter of sorted combined with reverse=True, for instance:

data = [
    {"arg1": {"valueone": 2, "valuetwo": "5"}},
    {"arg2": {"valueone": 7, "valuetwo": "3"}},
    {"arg3": {"valueone": 3, "valuetwo": "7"}},
    {"arg4": {"valueone": 4, "valuetwo": "6"}},
    {"arg5": {"valueone": 9, "valuetwo": "5"}}
]

result = sorted(data, key=lambda x: next(x["valueone"] for x in x.values()), reverse=True)
print(result)

Output

[{'arg5': {'valuetwo': '5', 'valueone': 9}}, {'arg2': {'valuetwo': '3', 'valueone': 7}}, {'arg4': {'valuetwo': '6', 'valueone': 4}}, {'arg3': {'valuetwo': '7', 'valueone': 3}}, {'arg1': {'valuetwo': '5', 'valueone': 2}}]