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python - Pandas 按字典列表分组

问题描述

我有如下数据，我正在尝试将数据分组为日期名称和小时。

    [
        {
            "avg": 52,
            "hour": 9,
            "dayname": "Friday"
        },
        {
            "avg": 1,
            "hour": 10,
            "dayname": "Friday"
        },
        {
            "avg": 12,
            "hour": 11,
            "dayname": "Friday"
        },
        {
            "avg": 3,
            "hour": 12,
            "dayname": "Friday"
        },
        {
            "avg": 12,
            "hour": 09,
            "dayname": "Saturday"
        },
        {
            "avg": 30,
            "hour": 10,
            "dayname": "Saturday"
        },
        {
            "avg": 66,
            "hour": 11,
            "dayname": "Saturday"
        },
        {
            "avg": 45,
            "hour": 12,
            "dayname": "Saturday"
        }
]

我想要最终的OP：

hour Friday  Saturday
9     52       12
10     1       30
11    12       16
12     3       45

这是我尝试过的代码：

 cur = mysql.connection.cursor()
    sql = "select avg(value) avg, hour, dayname from table;"
    cur.execute(sql)
    row_headers = [x[0] for x in cur.description] #this will extract row headers
    rv = cur.fetchall()
    json_result = []
    for result in rv:
        json_result.append(dict(zip(row_headers, result)))
#    resultfromdb= json.dumps(json_result)
finalresult = #how to get the expected op

 return finalresult

从那里如何使用熊猫进行分组并获得最终结果？

标签： pythonpython-3.xpandaspandas-groupby

解决方案

您可以将字典列表直接输入pandas然后操作：

df = pd.DataFrame(lst)

res = df.pivot_table(index='hour', columns='dayname', values='avg', aggfunc=np.sum)\
        .reset_index()

res.columns.name = ''

print(res)

   hour  Friday  Saturday
0     9      52        12
1    10       1        30
2    11      12        66
3    12       3        45

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