如何计算时差（日期和时间在同一列中）并将此差异放在单独的列中？

Question

遇到了问题。有一个数据框，我需要在其中计算每个用户的操作之间经过了多少时间，并在此表的单独列中指出这种差异。原来是用DateTime单独计算时间，但是在表中怎么做呢？

table = pd.DataFrame({
    'user': ['Steve', 'Steve', 'Steve', 'Jack', 'Jack', 'Jack'],
    'country':['UK', 'UK', 'UK', 'CH', 'CH', 'CH'],
    'date': ['2018-01-15 00:05:07', '2018-01-15 00:06:14', '2018-01-15 00:08:36',
             '2018-01-15 00:14:51', '2018-01-15 00:15:18', '2018-01-15 00:17:24']
})
table.set_index('country', inplace=True)

for i in table.groupby(['country', 'user']):
    print(i)

在单独的列中，您应该得到：

给杰克

1. 00:14:51
2. 00:00:27
3. 00:02:06

给史蒂夫

1. 00:05:07
2. 00:01:07
3. 00:02:22

Answer 1

使用DataFrameGroupBy.diff 表示每组的差异 - 输出是时间增量：

table['date'] = pd.to_datetime(table['date'])

s = pd.to_timedelta(table['date'].dt.strftime('%H:%M:%S'))
table['new'] = table.groupby(['country', 'user'])['date'].diff().fillna(s)

print (table)
          user                date             new
country                                           
UK       Steve 2018-01-15 00:05:07 0 days 00:05:07
UK       Steve 2018-01-15 00:06:14 0 days 00:01:07
UK       Steve 2018-01-15 00:08:36 0 days 00:02:22
CH        Jack 2018-01-15 00:14:51 0 days 00:14:51
CH        Jack 2018-01-15 00:15:18 0 days 00:00:27
CH        Jack 2018-01-15 00:17:24 0 days 00:02:06

如果需要将时间增量转换为字符串：

def format_timedelta(x):
    ts = x.total_seconds()
    hours, remainder = divmod(ts, 3600)
    minutes, seconds = divmod(remainder, 60)
    return ('{:02d}:{:02d}:{:02d}').format(int(hours), int(minutes), int(seconds)) 

table['new'] = table['new'].apply(format_timedelta)

print (table)
          user                date       new
country                                     
UK       Steve 2018-01-15 00:05:07  00:05:07
UK       Steve 2018-01-15 00:06:14  00:01:07
UK       Steve 2018-01-15 00:08:36  00:02:22
CH        Jack 2018-01-15 00:14:51  00:14:51
CH        Jack 2018-01-15 00:15:18  00:00:27
CH        Jack 2018-01-15 00:17:24  00:02:06

---

我认为0 每组的第一个值更好：

table['date'] = pd.to_datetime(table['date'])

table['new'] = table.groupby(['country', 'user'])['date'].diff().fillna(pd.Timedelta(0))

print (table)
          user                date             new
country                                           
UK       Steve 2018-01-15 00:05:07 0 days 00:00:00
UK       Steve 2018-01-15 00:06:14 0 days 00:01:07
UK       Steve 2018-01-15 00:08:36 0 days 00:02:22
CH        Jack 2018-01-15 00:14:51 0 days 00:00:00
CH        Jack 2018-01-15 00:15:18 0 days 00:00:27
CH        Jack 2018-01-15 00:17:24 0 days 00:02:06