Pandas 使用嵌套数据框列的总和创建一列

Question

如何使用来自嵌套数据帧的值的总和将新列添加到数据帧，而不会丢失任何其他列和使用 Pandas 的嵌套数据？

具体来说，我想创建一个新列total_cost ，其中包含一行的所有嵌套数据帧的总和。

我设法使用一系列groupby和apply创建了以下数据groupby ：

  user_id   description                                       unit_summary
0     111  xxx  [{'total_period_cost': 100, 'unit_id': 'xxx', ...
1     222  xxx  [{'total_period_cost': 100, 'unit_id': 'yyy', ...

我正在尝试添加列total_cost ，它是每个嵌套数据帧的total_period_cost的总和（按user_id分组）。 如何实现以下数据框？

  user_id   description   total_cost                          unit_summary
0     111  xxx  300  [{'total_period_cost': 100, 'unit_id': 'xxx', ...
1     222  xxx  100  [{'total_period_cost': 100, 'unit_id': 'yyy', ...

我的代码：

import pandas as pd

series = [{
    "user_id":"111", 
    "description": "xxx",
    "unit_summary":[
        {
        "total_period_cost":100,
        "unit_id":"xxx",
        "cost_per_unit":50,
        "total_period_usage":2
        },
        {
        "total_period_cost":200,
        "unit_id":"yyy",
        "cost_per_unit":25,
        "total_period_usage": 8
        }
    ]
},
{
    "user_id":"222",
    "description": "xxx",
    "unit_summary":[
        {
            "total_period_cost":100,
            "unit_id":"yyy",
            "cost_per_unit":25,
            "total_period_usage": 4
        }
    ]
}]

df = pd.DataFrame(series)

print(df)
print(df.to_dict(orient='records'))

这是我用来实现series JSON 对象的 groupby..apply 代码示例：

import pandas as pd

series = [
    {"user_id":"111", "unit_id":"xxx","cost_per_unit":50, "total_period_usage": 1},
    {"user_id":"111", "unit_id":"xxx","cost_per_unit":50, "total_period_usage": 1},
    {"user_id":"111", "unit_id":"yyy","cost_per_unit":25, "total_period_usage": 8},
    {"user_id":"222", "unit_id":"yyy","cost_per_unit":25, "total_period_usage": 3},
    {"user_id":"222", "unit_id":"yyy","cost_per_unit":25, "total_period_usage": 1}
]

df = pd.DataFrame(series)

sumc = (
    df.groupby(['user_id', 'unit_id', 'cost_per_unit'], as_index=False)
        .agg({'total_period_usage': 'sum'})
)

sumc['total_period_cost'] = sumc.total_period_usage * sumc.cost_per_unit

sumc = (
    sumc.groupby(['user_id'])
        .apply(lambda x: x[['total_period_cost', 'unit_id', 'cost_per_unit', 'total_period_usage']].to_dict('r'))
        .reset_index()
)

sumc = sumc.rename(columns={0:'unit_summary'})

sumc['description'] = 'xxx'

print(sumc)
print(sumc.to_dict(orient='records'))

通过在 anky_91 的答案中添加以下内容来解决它：

def myf(x):
    return pd.DataFrame(x).loc[:,'total_period_cost'].sum()
# Sum all server sumbscriptions total_period_cost
sumc['total_period_cost'] = sumc['unit_summary'].apply(myf)

Answer 1

您可以将unit_summary列中的每一行读取为数据unit_summary并对所需的列求和：

方法一： apply

def myf(x):
    return pd.DataFrame(x).loc[:,'total_period_cost'].sum()
df['total_cost'] = df['unit_summary'].apply(myf)

print(df)

方法2：同样通过列表理解：

df['total_cost'] = [pd.DataFrame(i)['total_period_cost'].sum() 
                            for i in df['unit_summary'].tolist()]

方法3：使用explode ：

m = df['unit_summary'].explode()
df['total_cost'] = pd.DataFrame(m.tolist(),index=m.index)['total_period_cost'].sum(level=0)

---

  user_id description                                       unit_summary  \
0     111         xxx  [{'total_period_cost': 100, 'unit_id': 'xxx', ...   
1     222         xxx  [{'total_period_cost': 100, 'unit_id': 'yyy', ...   

   total_cost  
0         300  
1         100 

除了上述之外，从您的原始数据帧开始，我们还可以执行以下操作来实现所需的输出，但这不会为您提供带有 dicts ('unit_summary`) 的系列：

(df.assign(total_cost=df['cost_per_unit']*df['total_period_usage'])
  .groupby(['user_id'],as_index=False)['total_cost'].sum().assign(description='xxxx'))

---

  user_id  total_cost description
0     111         300        xxxx
1     222         100        xxxx