python-2.7 - 如何在 python pandas 中处理这种复杂的逻辑？

Question

我有一些数据，比如遵循结构。它用于python pandas Data Frame，我将其命名为df。

Data1,Data2,Flag
2016-04-29,00:40:15,1
2016-04-29,00:40:24,2
2016-04-29,00:40:35,2
2015-04-29,00:40:36,2
2015-04-29,00:40:43,2
2015-04-29,00:40:45,2
2015-04-29,00:40:55,1
2015-04-29,00:41:05,1
2015-04-29,00:41:16,1
2015-04-29,00:41:17,2
.....................
.....................
2016-11-29,11:52:36,2
2016-11-29,11:52:43,2
2016-11-29,11:52:45,2
2016-11-29,11:52:55,1

我想得到满足以下要求的数据。

1. 如您所知，第一个数据的时间序列是 2016-04-29,00:40:15。我想在此数据框中获取比 Primer 的数据大 18 秒的下一个数据。我将获得第二个数据:2016-04-29,00:40:35,2第三条数据是:2015-04-29,00:40:55,1
2. 如果下一个数据的标志与引物的数据不同。无论是否超过 18 秒，我都会获取此数据。

针对以上两个需求，我将获取如下数据:

Data1,Data2,Flag
2016-04-29,00:40:15,1
2016-04-29,00:40:24,2
2015-04-29,00:40:43,2
2015-04-29,00:40:55,1
2015-04-29,00:41:16,1
2015-04-29,00:41:17,2
.....................

Answer 1

在这里，试试这个:

df['Data2'] = pd.to_timedelta(df['Data2'])

tdf = df.copy()
sel_idx = []
while len(tdf) > 0:
    sel_idx.extend([tdf.index[0]])
    cond1 = tdf['Data2'] > tdf.loc[sel_idx[-1], 'Data2'] + pd.to_timedelta(18, 's')
    cond2 = (tdf['Flag'] != tdf.loc[sel_idx[-1], 'Flag']) & (tdf['Data2'] > tdf.loc[sel_idx[-1], 'Data2'])
    tdf = tdf[cond1 | cond2]
df.loc[sel_idx, :]

---

测试

代码:

import pandas as pd
from io import StringIO

data = StringIO("""Data1,Data2,Flag
2016-04-29,00:40:15,1
2016-04-29,00:40:24,2
2016-04-29,00:40:35,2
2015-04-29,00:40:36,2
2015-04-29,00:40:43,2
2015-04-29,00:40:45,2
2015-04-29,00:40:55,1
2015-04-29,00:41:05,1
2015-04-29,00:41:16,1
2015-04-29,00:41:17,2
2016-11-29,11:52:36,2
2016-11-29,11:52:43,2
2016-11-29,11:52:45,2
2016-11-29,11:52:55,1""")

df = pd.read_csv(data)
df['Data2'] = pd.to_timedelta(df['Data2'])
print("Input\n", df)

tdf = df.copy()
sel_idx = []
while len(tdf) > 0:
    sel_idx.extend([tdf.index[0]])
    cond1 = tdf['Data2'] > tdf.loc[sel_idx[-1], 'Data2'] + pd.to_timedelta(18, 's')
    cond2 = (tdf['Flag'] != tdf.loc[sel_idx[-1], 'Flag']) & (tdf['Data2'] > tdf.loc[sel_idx[-1], 'Data2'])
    tdf = tdf[cond1 | cond2]
print("Ouput\n", df.loc[sel_idx, :])

输出:

Input
    Data1       Data2       Flag
0   2016-04-29  00:40:15    1
1   2016-04-29  00:40:24    2
2   2016-04-29  00:40:35    2
3   2015-04-29  00:40:36    2
4   2015-04-29  00:40:43    2
5   2015-04-29  00:40:45    2
6   2015-04-29  00:40:55    1
7   2015-04-29  00:41:05    1
8   2015-04-29  00:41:16    1
9   2015-04-29  00:41:17    2
10  2016-11-29  11:52:36    2
11  2016-11-29  11:52:43    2
12  2016-11-29  11:52:45    2
13  2016-11-29  11:52:55    1

Output
    Data1       Data2       Flag
0   2016-04-29  00:40:15    1
1   2016-04-29  00:40:24    2
4   2015-04-29  00:40:43    2
6   2015-04-29  00:40:55    1
8   2015-04-29  00:41:16    1
9   2015-04-29  00:41:17    2
10  2016-11-29  11:52:36    2
13  2016-11-29  11:52:55    1