Python 数据帧查找间隔并计算发生次数

我得到了一份不同事件的列表，其中有混合事件。例如，事件1可能会发生三次，然后另一个事件以及稍后的事件1将再次发生

我需要的是每个事件的间隔以及这些间隔内该事件的发生次数

values = {
        '2017-11-28 11:00': 'event1',
        '2017-11-28 11:01': 'event1',
        '2017-11-28 11:02': 'event1',
        '2017-11-28 11:03': 'event2',
        '2017-11-28 11:04': 'event2',
        '2017-11-28 11:05': 'event1',
        '2017-11-28 11:06': 'event1',
        '2017-11-28 11:07': 'event1',
        '2017-11-28 11:08': 'event3',
        '2017-11-28 11:09': 'event3',
        '2017-11-28 11:10': 'event2',
        }

import pandas as pd
df = pd.DataFrame.from_dict(values, orient='index').reset_index()
df.columns = ['time', 'event']
df['time'] = df['time'].apply(pd.to_datetime)
df.set_index('time', inplace=True)
df.sort_index(inplace=True)
df.head()

#%% first solution

# create intervals and count occurrences per interval
df['interval'] = (df['event'] != df['event'].shift(1)).astype(int).cumsum()
df['count'] = df.groupby(['event', 'interval']).cumcount() + 1

# now group by intervals
df.groupby('interval').last()

#%% second solution

df = df.reset_index()
# create intervals
df = df.groupby(df['event'].ne(df['event'].shift()).cumsum())
# calc start/end times and count occurances at the same time
df.apply(lambda x: pd.DataFrame({
                    'start':[x['time'].min()], 
                    'end':[x['time'].max()],
                    'event':[x['event'].iloc[0]],
                    'count':[len(x)]})).reset_index(drop=True)

预期结果是：

occurrences = [
        {'start':'2017-11-28 11:00',
         'end':'2017-11-28 11:02',
         'event':'event1',
         'count':3},
        {'start':'2017-11-28 11:03',
         'end':'2017-11-28 11:04',
         'event':'event2',
         'count':2},
        {'start':'2017-11-28 11:05',
         'end':'2017-11-28 11:07',
         'event':'event1',
         'count':3},
        {'start':'2017-11-28 11:08',
         'end':'2017-11-28 11:09',
         'event':'event3',
         'count':2},
        {'start':'2017-11-28 11:10',
         'end':'2017-11-28 11:10',
         'event':'event2',
         'count':1},
        ]

---

我正在考虑使用pd.merge\u asof查找间隔的开始/结束时间，并使用pd.cut（）进行分组和计数。但不知怎的，我被卡住了。感谢您的帮助。

尝试以下方法：

In [68]: x = df.reset_index()

In [69]: (x.groupby(x.event.ne(x.event.shift()).cumsum())
    ...:   .apply(lambda x:
    ...:             pd.DataFrame({
    ...:                 'start':[x['time'].min()],
    ...:                 'end':[x['time'].min()],
    ...:                 'event':[x['event'].iloc[0]],
    ...:                 'count':[len(x)]})
    ...:         )
    ...:   .reset_index(drop=True)
    ...:   .to_dict('r')
    ...: )
Out[69]:
[{'count': 3,
  'end': Timestamp('2017-11-28 11:00:00'),
  'event': 'event1',
  'start': Timestamp('2017-11-28 11:00:00')},
 {'count': 2,
  'end': Timestamp('2017-11-28 11:03:00'),
  'event': 'event2',
  'start': Timestamp('2017-11-28 11:03:00')},
 {'count': 3,
  'end': Timestamp('2017-11-28 11:05:00'),
  'event': 'event1',
  'start': Timestamp('2017-11-28 11:05:00')},
 {'count': 2,
  'end': Timestamp('2017-11-28 11:08:00'),
  'event': 'event3',
  'start': Timestamp('2017-11-28 11:08:00')},
 {'count': 1,
  'end': Timestamp('2017-11-28 11:10:00'),
  'event': 'event2',
  'start': Timestamp('2017-11-28 11:10:00')}]

或者，如果希望将

时间

列作为字符串，请执行以下操作：

In [75]: (x.groupby(x.event.ne(x.event.shift()).cumsum())
    ...:   .apply(lambda x:
    ...:             pd.DataFrame({
    ...:                 'start':[x['time'].min().strftime('%Y-%m-%d %H:%M:%S')],
    ...:                 'end':[x['time'].min().strftime('%Y-%m-%d %H:%M:%S')],
    ...:                 'event':[x['event'].iloc[0]],
    ...:                 'count':[len(x)]})
    ...:         )
    ...:   .reset_index(drop=True)
    ...:   .to_dict('r')
    ...: )
Out[75]:
[{'count': 3,
  'end': '2017-11-28 11:00:00',
  'event': 'event1',
  'start': '2017-11-28 11:00:00'},
 {'count': 2,
  'end': '2017-11-28 11:03:00',
  'event': 'event2',
  'start': '2017-11-28 11:03:00'},
 {'count': 3,
  'end': '2017-11-28 11:05:00',
  'event': 'event1',
  'start': '2017-11-28 11:05:00'},
 {'count': 2,
  'end': '2017-11-28 11:08:00',
  'event': 'event3',
  'start': '2017-11-28 11:08:00'},
 {'count': 1,
  'end': '2017-11-28 11:10:00',
  'event': 'event2',
  'start': '2017-11-28 11:10:00'}]

---

这里有两个解决方案。第一个是基于和提供的链接。它为间隔创建连续的数字，并累计统计这些间隔内的事件

values = {
        '2017-11-28 11:00': 'event1',
        '2017-11-28 11:01': 'event1',
        '2017-11-28 11:02': 'event1',
        '2017-11-28 11:03': 'event2',
        '2017-11-28 11:04': 'event2',
        '2017-11-28 11:05': 'event1',
        '2017-11-28 11:06': 'event1',
        '2017-11-28 11:07': 'event1',
        '2017-11-28 11:08': 'event3',
        '2017-11-28 11:09': 'event3',
        '2017-11-28 11:10': 'event2',
        }

import pandas as pd
df = pd.DataFrame.from_dict(values, orient='index').reset_index()
df.columns = ['time', 'event']
df['time'] = df['time'].apply(pd.to_datetime)
df.set_index('time', inplace=True)
df.sort_index(inplace=True)
df.head()

#%% first solution

# create intervals and count occurrences per interval
df['interval'] = (df['event'] != df['event'].shift(1)).astype(int).cumsum()
df['count'] = df.groupby(['event', 'interval']).cumcount() + 1

# now group by intervals
df.groupby('interval').last()

#%% second solution

df = df.reset_index()
# create intervals
df = df.groupby(df['event'].ne(df['event'].shift()).cumsum())
# calc start/end times and count occurances at the same time
df.apply(lambda x: pd.DataFrame({
                    'start':[x['time'].min()], 
                    'end':[x['time'].max()],
                    'event':[x['event'].iloc[0]],
                    'count':[len(x)]})).reset_index(drop=True)

第二种解决方案是基于上面给出的答案。与第一个想法类似，它还创建了间隔编号，但也找到了此类间隔的开始/结束时间戳

values = {
        '2017-11-28 11:00': 'event1',
        '2017-11-28 11:01': 'event1',
        '2017-11-28 11:02': 'event1',
        '2017-11-28 11:03': 'event2',
        '2017-11-28 11:04': 'event2',
        '2017-11-28 11:05': 'event1',
        '2017-11-28 11:06': 'event1',
        '2017-11-28 11:07': 'event1',
        '2017-11-28 11:08': 'event3',
        '2017-11-28 11:09': 'event3',
        '2017-11-28 11:10': 'event2',
        }

import pandas as pd
df = pd.DataFrame.from_dict(values, orient='index').reset_index()
df.columns = ['time', 'event']
df['time'] = df['time'].apply(pd.to_datetime)
df.set_index('time', inplace=True)
df.sort_index(inplace=True)
df.head()

#%% first solution

# create intervals and count occurrences per interval
df['interval'] = (df['event'] != df['event'].shift(1)).astype(int).cumsum()
df['count'] = df.groupby(['event', 'interval']).cumcount() + 1

# now group by intervals
df.groupby('interval').last()

#%% second solution

df = df.reset_index()
# create intervals
df = df.groupby(df['event'].ne(df['event'].shift()).cumsum())
# calc start/end times and count occurances at the same time
df.apply(lambda x: pd.DataFrame({
                    'start':[x['time'].min()], 
                    'end':[x['time'].max()],
                    'event':[x['event'].iloc[0]],
                    'count':[len(x)]})).reset_index(drop=True)

---

您是否只寻找基于熊猫的解决方案？对于非基于pandas的解决方案，您可以对其进行迭代。我打赌有一个很好的pandas解决方案。我发现这是基于迭代的。但是熊猫会更好。你甚至可以看看这里，@VivekHarikrishnan，这是个好主意。良好的起点。问题在于如何按时间间隔对它们进行分组——也就是说，您如何知道何时达到了时间间隔的最高累计计数？附言：我找到了解决办法。如果你发你的，我会接受的。根据@chrisb在@Vivek Harkrishnan提到的问题中的回答，你可以在

block

上执行

groupby

，然后选择

min（'time'）

作为

start

和

max（'time'）

as

end