如何对嵌套字典列表中的值进行排序、分组和聚合？

Question

given the list of dictionaries below, I want to do the following things:

1: Sort the following data by key (top level)'name'
2: Sort the by the nested key "name" under key "items"
3: Group values under items by aggregation interval for example "1d"
4: Get again the min max and avg result from step number 3\

Atm, I resolve this by iter down to the values and group them with pandas, aggregate again min max and avg from result. This way feels really tricky, and the performance is not given.

Can someone help me out?

[
    {
        '_id': 2,
        'name': 'b',
        'device': 'b',
        'items': [
            {
                'item_id': 'item_id_2', 'name': 'item_2', 'unit': 'b/s',
                'values': [
                    {'time': datetime.datetime(2022, 9, 5, 15, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 16, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 17, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 18, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 19, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 20, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                ]
            }
        ]
    },
    {
        '_id': 1,
        'name': 'a',
        'device': 'a',
        'items': [
            {
                'item_id': 'item_id_1', 'name': 'item_1', 'unit': 'b/s',
                'values': [
                    {'time': datetime.datetime(2022, 9, 5, 15, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 16, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 17, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 18, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 19, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                    {'time': datetime.datetime(2022, 9, 5, 20, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                ]
            }
        ]
    }
]

As for the result, I would expect something like this:

[
    {
        '_id': 1,
        'name': 'a',
        'device': 'a',
        'items': [
            {
                'item_id': 'item_id_1', 'name': 'item_1', 'unit': 'b/s',
                'values': [
                    {'time': datetime.datetime(2022, 9, 5, 0, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                ]
            }
        ]
    },
    {
        '_id': 1,
        'name': 'b',
        'device': 'b',
        'items': [
            {
                'item_id': 'item_id_2', 'name': 'item_2', 'unit': 'b/s',
                'values': [
                    {'time': datetime.datetime(2022, 9, 5, 0, 0), 'min': 0.0, 'max': 1.0, 'avg': 0.5},
                ]
            }
        ]
    }
]

Answer 1

With the initial list of dicts that you provided and that I choose to call data , here is one way to do it:

df = pd.DataFrame(data)

# First, sort values
df = df.assign(temp=df["items"].apply(lambda x: x[0]["name"])).pipe(
    lambda df_: df_.sort_values(by="temp").drop(columns="temp").reset_index(drop=True)
)

# Get aggregated as new column 'temp'
dfs = df["items"].apply(lambda x: pd.DataFrame(x[0].pop("values", None)))
df["temp"] = pd.Series(
    [
        {
            k: v[0]
            for k, v in df.set_index("time")
            .resample("D")
            .mean()
            .reset_index()
            .to_dict(orient="list")
            .items()
        }
        for df in dfs
    ]
)
df["items"] = df["items"].apply(lambda x: x[0])

# Merge intermediate dictionaries
df["items"] = df.apply(lambda x: x["items"] | {"values": [x["temp"]]}, axis=1)
df = df.drop(columns="temp")

And so:

print(df.to_json(orient="records"))
# Output
[
    {
        "_id": 1,
        "name": "a",
        "device": "a",
        "items": {
            "item_id": "item_id_1",
            "name": "item_1",
            "unit": "b\\/s",
            "values": [{"time": 1662336000000, "min": 0.0, "max": 1.0, "avg": 0.5}],
        },
    },
    {
        "_id": 2,
        "name": "b",
        "device": "b",
        "items": {
            "item_id": "item_id_2",
            "name": "item_2",
            "unit": "b\\/s",
            "values": [{"time": 1662336000000, "min": 0.0, "max": 1.0, "avg": 0.5}],
        },
    },
]