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我有一组日期。我想从它们的前向邻居中减去它们以获得它们之间的增量。我的代码如下所示：

import pandas, numpy, StringIO


txt = '''ID,DATE
002691c9cec109e64558848f1358ac16,2003-08-13 00:00:00
002691c9cec109e64558848f1358ac16,2003-08-13 00:00:00
0088f218a1f00e0fe1b94919dc68ec33,2006-05-07 00:00:00
0088f218a1f00e0fe1b94919dc68ec33,2006-06-03 00:00:00
00d34668025906d55ae2e529615f530a,2006-03-09 00:00:00
00d34668025906d55ae2e529615f530a,2006-03-09 00:00:00
0101d3286dfbd58642a7527ecbddb92e,2007-10-13 00:00:00
0101d3286dfbd58642a7527ecbddb92e,2007-10-27 00:00:00
0103bd73af66e5a44f7867c0bb2203cc,2001-02-01 00:00:00
0103bd73af66e5a44f7867c0bb2203cc,2008-01-20 00:00:00
'''
df = pandas.read_csv(StringIO.StringIO(txt))
df = df.sort('DATE')
df.DATE = pandas.to_datetime(df.DATE)
grouped = df.groupby('ID')
df['X_SEQUENCE_GAP'] = pandas.concat([g['DATE'].sub(g['DATE'].shift(), fill_value=0) for title,g in grouped])

我得到了相当难以理解的结果。所以，我会去，我有一个逻辑错误。

我得到的结果如下：

                               ID                DATE       X_SEQUENCE_GAP
0  002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00 12277 days, 00:00:00
1  002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00             00:00:00
3  0088f218a1f00e0fe1b94919dc68ec33 2006-06-03 00:00:00    27 days, 00:00:00
2  0088f218a1f00e0fe1b94919dc68ec33 2006-05-07 00:00:00 13275 days, 00:00:00
5  00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00 13216 days, 00:00:00
4  00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00             00:00:00
6  0101d3286dfbd58642a7527ecbddb92e 2007-10-13 00:00:00 13799 days, 00:00:00
7  0101d3286dfbd58642a7527ecbddb92e 2007-10-27 00:00:00    14 days, 00:00:00
9  0103bd73af66e5a44f7867c0bb2203cc 2008-01-20 00:00:00  2544 days, 00:00:00
8  0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 00:00:00 11354 days, 00:00:00

我期待 exapme 0 和 1 都会有 0 结果。非常感谢任何帮助。

4

1 回答 1

4

这是在 0.11rc1 中（我认为不适用于以前的版本）当您移动日期时，第一个是 NaT（如 nan，但适用于 datetimes/timedeltas）

In [27]: df['X_SEQUENCE_GAP'] = grouped.apply(lambda g: g['DATE']-g['DATE'].shift())

In [30]: df.sort()
Out[30]: 
                                 ID                DATE      X_SEQUENCE_GAP
0  002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00                 NaT
1  002691c9cec109e64558848f1358ac16 2003-08-13 00:00:00            00:00:00
2  0088f218a1f00e0fe1b94919dc68ec33 2006-05-07 00:00:00                 NaT
3  0088f218a1f00e0fe1b94919dc68ec33 2006-06-03 00:00:00   27 days, 00:00:00
4  00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00                 NaT
5  00d34668025906d55ae2e529615f530a 2006-03-09 00:00:00            00:00:00
6  0101d3286dfbd58642a7527ecbddb92e 2007-10-13 00:00:00                 NaT
7  0101d3286dfbd58642a7527ecbddb92e 2007-10-27 00:00:00   14 days, 00:00:00
8  0103bd73af66e5a44f7867c0bb2203cc 2001-02-01 00:00:00                 NaT
9  0103bd73af66e5a44f7867c0bb2203cc 2008-01-20 00:00:00 2544 days, 00:00:00

然后你可以填充纳（但你必须做这个确认类型转换，因为一个 numpy 错误，将在 0.12 中得到修复）。

 In [57]: df['X_SEQUENCE_GAP'].sort_index().astype('timedelta64[ns]').fillna(0)
Out[57]: 
0              00:00:00
1              00:00:00
2              00:00:00
3     27 days, 00:00:00
4              00:00:00
5              00:00:00
6              00:00:00
7     14 days, 00:00:00
8              00:00:00
9   2544 days, 00:00:00
Name: X_SEQUENCE_GAP, dtype: timedelta64[ns]
于 2013-04-15T19:59:33.727 回答