pandas 列中的连续值

Question

我有熊猫df['realize']

time                      realize
2016-01-18 08:25:00     -46.369083
2016-01-19 14:30:00     -819.010738
2016-01-20 11:10:00    -424.955847
2016-01-21 07:15:00     27.523859
2016-01-21 16:10:00     898.522762
2016-01-25 00:00:00    761.063545

哪里time：

df.index = df['time']
df.index = pd.to_datetime(df.index)

哪里df['realize']：

In: type(df['realize'])
Out: pandas.core.series.Series

我想计算连续值，规则很简单 ( df['realize'] > 0, df['realize'] < 0)

预计输出：

time                      realize    Consecutive
2016-01-18 08:25:00     -46.369083    1
2016-01-19 14:30:00     -819.010738   2
2016-01-20 11:10:00    -424.955847    3
2016-01-21 07:15:00     27.523859     1
2016-01-21 16:10:00     898.522762    2
2016-01-25 00:00:00    761.063545     3

我阅读了有关循环的主题，但没有找到我需要的内容。预先感谢您的帮助。

Answer 1

您可以执行以下操作：

g = df.realize.gt(0).astype(int).diff().fillna(0).abs().cumsum()
df['Consecutive'] = df.groupby(g).realize.cumcount().add(1)

               time     realize       Consecutive
0 2016-01-18 08:25:00  -46.369083            1
1 2016-01-19 14:30:00 -819.010738            2
2 2016-01-20 11:10:00 -424.955847            3
3 2016-01-21 07:15:00   27.523859            1
4 2016-01-21 16:10:00  898.522762            2
5 2016-01-25 00:00:00  761.063545            3

其中使用的石斑鱼是通过取DataFrame.diff布尔系列的一阶差分 ( ) 来获得的，指示是否realize大于0：

diff = df.realize.gt(0).astype(int).diff().fillna(0).abs()
df.assign(diff = diff, grouper = g)

         time            realize     Consecutive diff  grouper
0 2016-01-18 08:25:00  -46.369083            1   0.0      0.0
1 2016-01-19 14:30:00 -819.010738            2   0.0      0.0
2 2016-01-20 11:10:00 -424.955847            3   0.0      0.0
3 2016-01-21 07:15:00   27.523859            1   1.0      1.0
4 2016-01-21 16:10:00  898.522762            2   0.0      1.0
5 2016-01-25 00:00:00  761.063545            3   0.0      1.0