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熊猫带重置取累积和

cumsum pandas python

Justin · 技术社区 · 9 年前

问题

密码

cb_arbitrage['shift'] = cb_arbitrage.index.shift(1, freq='T')

                        cccccccc     bbbbbbbb  cb_spread         shift
timestamp                                                                   
2017-07-07 18:23:00  2535.002000  2524.678462  10.323538 2017-07-07 18:24:00
2017-07-07 18:24:00  2535.007826  2523.297619  11.710207 2017-07-07 18:25:00
2017-07-07 18:25:00  2535.004167  2524.391000  10.613167 2017-07-07 18:26:00
2017-07-07 18:26:00  2534.300000  2521.838667  12.461333 2017-07-07 18:27:00
2017-07-07 18:27:00  2530.231429  2520.195625  10.035804 2017-07-07 18:28:00
2017-07-07 18:28:00  2529.444667  2518.782143  10.662524 2017-07-07 18:29:00
2017-07-07 18:29:00  2528.988000  2518.802963  10.185037 2017-07-07 18:30:00
2017-07-07 18:59:00  2514.403367  2526.473333  12.069966 2017-07-07 19:00:00
2017-07-07 19:01:00  2516.410000  2528.980000  12.570000 2017-07-07 19:02:00

cb_arbitrage['shift'] = cb_arbitrage['shift'].shift(1)
cb_arbitrage['shift'][0] = cb_arbitrage.index[0]
cb_arbitrage['count'] = 0

                        cccccccc     bbbbbbbb  cb_spread               shift  count
timestamp                                                                          
2017-07-07 18:23:00  2535.002000  2524.678462  10.323538 2017-07-07 18:23:00      0
2017-07-07 18:24:00  2535.007826  2523.297619  11.710207 2017-07-07 18:24:00      0
2017-07-07 18:25:00  2535.004167  2524.391000  10.613167 2017-07-07 18:25:00      0
2017-07-07 18:26:00  2534.300000  2521.838667  12.461333 2017-07-07 18:26:00      0
2017-07-07 18:27:00  2530.231429  2520.195625  10.035804 2017-07-07 18:27:00      0
2017-07-07 18:28:00  2529.444667  2518.782143  10.662524 2017-07-07 18:28:00      0
2017-07-07 18:29:00  2528.988000  2518.802963  10.185037 2017-07-07 18:29:00      0
2017-07-07 18:59:00  2514.403367  2526.473333  12.069966 2017-07-07 18:30:00      0
2017-07-07 19:01:00  2516.410000  2528.980000  12.570000 2017-07-07 19:00:00      0

然后,循环计算累积和,并重置:

count = 0
for i, row in cb_arbitrage.iterrows():

    if i == cb_arbitrage.loc[i]['shift']:
        count += 1
        cb_arbitrage.set_value(i, 'count', count)
    else:
        count = 1
        cb_arbitrage.set_value(i, 'count', count)

                        cccccccc     bbbbbbbb  cb_spread               shift  count
timestamp                                                                          
2017-07-07 18:23:00  2535.002000  2524.678462  10.323538 2017-07-07 18:23:00      1
2017-07-07 18:24:00  2535.007826  2523.297619  11.710207 2017-07-07 18:24:00      2
2017-07-07 18:25:00  2535.004167  2524.391000  10.613167 2017-07-07 18:25:00      3
2017-07-07 18:26:00  2534.300000  2521.838667  12.461333 2017-07-07 18:26:00      4
2017-07-07 18:27:00  2530.231429  2520.195625  10.035804 2017-07-07 18:27:00      5
2017-07-07 18:28:00  2529.444667  2518.782143  10.662524 2017-07-07 18:28:00      6
2017-07-07 18:29:00  2528.988000  2518.802963  10.185037 2017-07-07 18:29:00      7
2017-07-07 18:59:00  2514.403367  2526.473333  12.069966 2017-07-07 18:30:00      1
2017-07-07 19:01:00  2516.410000  2528.980000  12.570000 2017-07-07 19:00:00      1
2017-07-07 21:55:00  2499.904560  2510.814000  10.909440 2017-07-07 19:02:00      1
2017-07-07 21:56:00  2500.134615  2510.812857  10.678242 2017-07-07 21:56:00      2

1 回复 | 直到 9 年前

Ted Petrou 9 年前

您可以使用 diff 方法,该方法查找当前行和前一行之间的差异。然后,您可以检查并查看此差值是否等于一分钟。从这里开始,在数据中重置连胜有很多技巧。

我们首先取布尔级数的累积和,这使我们接近我们想要的。为了重置序列,我们将该累积和序列乘以原始布尔值,因为False的计算结果为0。

s = cb_arbitrage.timestamp.diff() == pd.Timedelta('1 minute')
s1 = s.cumsum()
s.mul(s1).diff().where(lambda x: x < 0).ffill().add(s1, fill_value=0) + 1

0     1.0
1     2.0
2     3.0
3     4.0
4     5.0
5     6.0
6     7.0
7     1.0
8     1.0
9     1.0
10    2.0