我有一个日期时间索引不规则的熊猫数据框。现在，我想基于连续的连续观察为数据帧建立索引。换句话说，我只想保留存在x或多个连续观察值的值。
请看以下示例：
idx = pd.DatetimeIndex(['2003-04-11', '2003-04-12', '2003-04-13','2003-04-17','2003-05-02', '2003-05-03', '2003-05-04','2003-07-23', '2003-07-24'])
df = pd.DataFrame(np.random.random((9,2)),index=idx)
df
0 1
2003-04-11 0.954287 0.331016
2003-04-12 0.553477 0.858590
2003-04-13 0.179510 0.103970
2003-04-17 0.608664 0.746860
2003-05-02 0.691829 0.081192
2003-05-03 0.790748 0.319989
2003-05-04 0.955903 0.668918
2003-07-23 0.630201 0.297902
2003-07-24 0.692403 0.847222
来自的3个连续观测值2003-04-11 ~ 13，然后是一个单独的观测值2003-04-17，比来自的3个连续观测值大，2003-05-02 ~ 04并且以的2个连续观测值结尾2003-07-23 ~ 24。
如何索引连续3天或更长时间的这些观察？在此示例中，应保留以下观察结果：
0 1
2003-04-11 0.954287 0.331016
2003-04-12 0.553477 0.858590
2003-04-13 0.179510 0.103970
2003-05-02 0.691829 0.081192
2003-05-03 0.790748 0.319989
2003-05-04 0.955903 0.668918
尽管答案被接受，但是您可以尝试其他方法：
df1 = df.loc[df.groupby((~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int).cumsum() ).transform(len).iloc[:, 0] == 3]
print df1
0 1
2003-04-11 0.350339 0.904514
2003-04-12 0.903141 0.423335
2003-04-13 0.394534 0.803299
2003-05-02 0.158032 0.565684
2003-05-03 0.715311 0.772509
2003-05-04 0.136462 0.533705
一步步：
print ~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))
#2003-04-11 True
#2003-04-12 False
#2003-04-13 False
#2003-04-17 True
#2003-05-02 True
#2003-05-03 False
#2003-05-04 False
#2003-07-23 True
#2003-07-24 False
#dtype: bool
print (~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int)
#2003-04-11 1
#2003-04-12 0
#2003-04-13 0
#2003-04-17 1
#2003-05-02 1
#2003-05-03 0
#2003-05-04 0
#2003-07-23 1
#2003-07-24 0
#dtype: int32
print (~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int).cumsum()
#2003-04-11 1
#2003-04-12 1
#2003-04-13 1
#2003-04-17 2
#2003-05-02 3
#2003-05-03 3
#2003-05-04 3
#2003-07-23 4
#2003-07-24 4
#dtype: int32
print df.groupby((~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int).cumsum()).transform(len)
# 0 1
#2003-04-11 3 3
#2003-04-12 3 3
#2003-04-13 3 3
#2003-04-17 1 1
#2003-05-02 3 3
#2003-05-03 3 3
#2003-05-04 3 3
#2003-07-23 2 2
#2003-07-24 2 2
print df.groupby((~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int).cumsum()).transform(len).iloc[:, 0]
#2003-04-11 3
#2003-04-12 3
#2003-04-13 3
#2003-04-17 1
#2003-05-02 3
#2003-05-03 3
#2003-05-04 3
#2003-07-23 2
#2003-07-24 2
#Name: 0, dtype: float64
print df.groupby((~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int).cumsum()).transform(len).iloc[:, 0] == 3
#2003-04-11 True
#2003-04-12 True
#2003-04-13 True
#2003-04-17 False
#2003-05-02 True
#2003-05-03 True
#2003-05-04 True
#2003-07-23 False
#2003-07-24 False
#Name: 0, dtype: bool
print df.loc[df.groupby((~(df.index.to_series().diff() == pd.Timedelta(1, unit='d'))).astype(int).cumsum()).transform(len).iloc[:, 0] == 3]
# 0 1
#2003-04-11 0.120301 0.635707
#2003-04-12 0.747283 0.681601
#2003-04-13 0.118192 0.777899
#2003-05-02 0.481396 0.294547
#2003-05-03 0.619790 0.058048
#2003-05-04 0.179386 0.348843
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我来说两句