下面是另一种更复杂的方法，并排排列记录并减去时间戳列。

df['Timestamp'] = pd.to_datetime(df['Timestamp'])

df = df.set_index('Id')
df['g'] = (df['Event_Type'] == 'Opened').cumsum()

df_open = df.query('Event_Type == "Opened"').groupby('g').head(1)
df_close = df.query('Event_Type == "Closed"').groupby('g').head(1)

df_result = df_open.merge(df_close, on='g', suffixes=('_Opened', '_Closed'))
df_result['Timedelta'] = df_result['Timestamp_Closed'] - df_result['Timestamp_Opened']

df_result

输出量：

Timestamp_Opened App_Name_Opened Event_Type_Opened  g    Timestamp_Closed App_Name_Closed Event_Type_Closed       Timedelta
0 2018-01-16 06:01:05       Instagram            Opened  1 2018-01-16 06:01:06       Instagram            Closed 0 days 00:00:01
1 2018-01-16 06:01:07       Instagram            Opened  2 2018-01-16 06:02:08       Instagram            Closed 0 days 00:01:01
2 2018-01-16 06:01:09       Instagram            Opened  5 2018-01-16 06:01:09       Instagram            Closed 0 days 00:00:00
3 2018-01-16 06:03:44       Instagram            Opened  6 2018-01-16 06:03:44       Instagram            Closed 0 days 00:00:00

试试看：

out, state = [], None
for i, e in zip(df["Id"], df["Event_Type"]):
    if e == "Opened":
        state = i
    elif e == "Closed" and state is not None:
        out.append([state, i])
        state = None

print(out)

图纸：

[[1, 2], [3, 6], [9, 10], [11, 12]]

要获取时差，请执行以下操作：

df["Timestamp"] = pd.to_datetime(df["Timestamp"])

out, state = [], None
for i, e in zip(df.index, df["Event_Type"]):
    if e == "Opened":
        state = i
    elif e == "Closed" and state is not None:
        out.append(df.loc[i, "Timestamp"] - df.loc[state, "Timestamp"])
        state = None

print(out)

图纸：

[Timedelta('0 days 00:00:01'), Timedelta('0 days 00:01:01'), Timedelta('0 days 00:00:00'), Timedelta('0 days 00:00:00')]

步骤1

将Timestamp列转换为日期时间d类型

df['Timestamp'] = pd.to_datetime(df['Timestamp'])

第二步

创建一个以信号“Closed”结尾的组。将划分的组命名为grp。

cond = df['Event_Type'].eq('Closed')
grp = cond.cumsum() - cond

步骤3

首先，删除df中带有Interaction的行。然后，计算grp中Timestamp列之间的diff(1)。最后，只保留Event_Type为Closed的列，并删除NaN。

(df[df['Event_Type'].ne('Interaction')]
  .groupby(grp)['Timestamp'].diff(1)[cond]
  .dropna())

产出：