如何在熊猫数据框中执行转置功能

Question

我是 python 新手，所以请帮助我。

我有一个数据框

df

        id    status     emailaddress  ownername   Key   
        a     pending    a.gmail.com   as          1
        a     pending     a.gmail.com   as         2
        a     submitted   a.gmail.com   as         3
        a     submitted   a.gmail.com   as         4
        b     pending    b.gmail.com    bs         1
        b     pending     b.gmail.com   bs         2
        b     pending     b.gmail.com   bs         3
        b     pending     b.gmail.com   bs         4
        b     pending    b.gmail.com    bs         1
        c     submitted   c.gmail.com   bs         1
        c     submitted   c.gmail.com   bs         3
        c     submitted   c.gmail.com   bs         4

我想根据每个 id 的状态计算 Key 列的计数。

此外，如果状态任何一个状态 = 未决，则该用户的整体状态将在摘要表中处于未决状态。

Total Key = count (Key) where status(pending)

CompletedKey = count (Key) where status(submitted)

预期的 df_summary

        id    status     emailaddress  ownername   TotalKey   CompletedKey  RemainingKey   
        a      pending    a.gmail.com   as         4           2              2
        b      pending    b.gmail.com   bs         5           0              4
        c      submitted  c.gmail.com   cs         3           3              0
        

我的代码

    dfResponse = df.groupby(['id','ownername','status','emailaddress'])['Key'].count().reset_index(name="TotalKey")     

Answer 1

使用agg 计算计数器并与您的原始数据框合并。要保持正确的状态，请创建一个有序的分类 dtype，以确保在对值进行排序时首先出现 pending 状态。

>>> df.astype({'status': pd.CategoricalDtype(['pending', 'submitted'], ordered=True)}) \
      .merge(df.groupby('id')
               .agg(TotalKey=('Key', 'count'),
                    CompletedKey=('status', lambda x: sum(x == 'submitted')), 
                    RemainingKey=('status', lambda x: sum(x == 'pending'))),
             on='id') \
      .sort_values('status') \
      .drop_duplicates('id') \
      .drop(columns='Key')


  id     status emailaddress ownername  TotalKey  CompletedKey  RemainingKey
0  a    pending  a.gmail.com        as         4             2             2
4  b    pending  b.gmail.com        bs         5             0             5
9  c  submitted  c.gmail.com        bs         3             3             0

Answer 2

使用assign 为您要执行的不同Key 列创建布尔列，然后您可以在groupby 之后使用sum。对于状态列，保持where 挂起并在groupby 之后使用first 来获取值挂起，当至少有一个在组中时，然后fillna 通过提交缺少状态 - 因为如果没有，那么只有提交给这个组。

dfResponse = (
    df.assign(
        TotalKey=True, 
        CompletedKey=lambda x: x['status'].eq('submitted'),
        RemainingKey=lambda x: ~x['CompletedKey'], # assuming it is either submitted or pending
        status=lambda x: x['status'].where(x['RemainingKey'])) # keep only the pending values
      .groupby(['id','ownername','emailaddress']) # remove status from groupby
      .agg({'status':'first', # get pending if any pending in the group
            'TotalKey':sum, # sum because these columns are boolean
            'CompletedKey':sum,
            'RemainingKey':sum})
      .fillna({'status':'submitted'}) # fill missing status when no pending
      .reset_index()
)

print(dfResponse)
  id ownername emailaddress     status  TotalKey  CompletedKey  RemainingKey
0  a        as  a.gmail.com    pending         4             2             2
1  b        bs  b.gmail.com    pending         5             0             5
2  c        bs  c.gmail.com  submitted         3             3             0

Answer 3

您可以结合.groupby（确定每个待处理/提交组的计数）+.pivot_table（重塑/求和数据帧）：

x = df.groupby(["id", "status"], as_index=False).agg(
    {
        "status": "first",
        "emailaddress": "first",
        "ownername": "first",
        "Key": "count",
    }
)
x = x.pivot_table(
    index=["id", "emailaddress", "ownername", "status"],
    columns="status",
    aggfunc="sum",
)
x.columns = x.columns.map("_".join)
x = (
    x.rename(
        columns={"Key_pending": "RemainingKey", "Key_submitted": "CompletedKey"}
    )
    .fillna(0)
    .astype(int)
)
x["TotalKey"] = x.sum(1)
print(x.reset_index().sort_values(["id", "status"]).drop_duplicates("id"))

打印：

  id emailaddress ownername     status  RemainingKey  CompletedKey  TotalKey
0  a  a.gmail.com        as    pending             2             0         2
2  b  b.gmail.com        bs    pending             5             0         5
3  c  c.gmail.com        bs  submitted             0             3         3