Question

我有一个熊猫数据框，它只有两列，像这样

          Timestamp       X
0   2017-01-01 00:00:00 18450
1   2017-01-01 00:10:00 13787
2   2017-01-01 00:20:00 3249
3   2017-01-01 00:30:00 44354
4   2017-01-01 00:40:00 50750

从月初到月底，时间戳列基本上间隔10分钟。要创建示例，可以使用以下代码。

l_data = pd.DataFrame()

l_data['Timestamp'] = pd.date_range(start=pd.Timestamp('2017-01-18 00:00:00'), end=pd.Timestamp('2017-01-20 00:00:00'), freq='10T')
l_data['X'] = random.sample(range(0, 100000), len(l_data))

我和我一样有字典

  {Timestamp('2017-01-18 01:37:19.160000'): Timestamp('2017-01-18 01:37:29.520000'),
 Timestamp('2017-01-18 01:41:04.880000'): Timestamp('2017-01-18 01:41:10.280000'),
 Timestamp('2017-01-18 21:33:52.800000'): Timestamp('2017-01-18 21:40:00.040000'),
 Timestamp('2017-01-18 21:40:02.120000'): Timestamp('2017-01-18 21:50:00.040000'),
 Timestamp('2017-01-18 21:50:02.120000'): Timestamp('2017-01-18 22:00:00.040000'),
 Timestamp('2017-01-18 22:00:02.120000'): Timestamp('2017-01-18 22:01:50.760000'),
 Timestamp('2017-01-18 22:20:22.760000'): Timestamp('2017-01-18 22:25:20.760000'),
 Timestamp('2017-01-18 22:35:52.800000'): Timestamp('2017-01-18 22:40:00.040000')}

字典中的键是开始时间，值是结束时间。我想基于L

中的dict创建一个名为l_data的列

如果键和值在dict中的时间大于5分钟，我必须将落在l_data中该范围内的时间戳记为1。

如何以一种直接的方式而不是使用多个循环的方式在熊猫中实现这一目标？

预期的输出将如下所示

126 1/18/2017 21:00 43401   0
127 1/18/2017 21:10 290     0
128 1/18/2017 21:20 92509   0
129 1/18/2017 21:30 64545   0
130 1/18/2017 21:40 47780   1
131 1/18/2017 21:50 53293   1
132 1/18/2017 22:00 45634   0
133 1/18/2017 22:10 51462   0
134 1/18/2017 22:20 44736   0
135 1/18/2017 22:30 11697   1
136 1/18/2017 22:40 82587   1
137 1/18/2017 22:50 76250   0
138 1/18/2017 23:00 33307   0
139 1/18/2017 23:10 25851   0
140 1/18/2017 23:20 71131   0
141 1/18/2017 23:30 88015   0
142 1/18/2017 23:40 45577   0
143 1/18/2017 23:50 76761   0
144 1/19/2017 0:00  45363   0

仅显示重要的行

Answer 1

我相信您需要：

d = { pd.Timestamp('2017-01-18 21:45:02.120000'): pd.Timestamp('2017-01-18 21:50:29.040000'),
pd.Timestamp('2017-01-18 21:51:02.120000'): pd.Timestamp('2017-01-18 22:52:00.040000'),
pd.Timestamp('2017-01-18 22:52:02.120000'): pd.Timestamp('2017-01-18 22:57:59.760000'),
pd.Timestamp('2017-01-18 23:41:52.800000'): pd.Timestamp('2017-01-18 23:43:00.040000'),
pd.Timestamp('2017-01-18 23:44:52.800000'): pd.Timestamp('2017-01-18 23:50:30.040000'),
pd.Timestamp('2017-01-19 01:10:32.800000'): pd.Timestamp('2017-01-19 01:11:30.040000'),
pd.Timestamp('2017-01-19 01:40:32.800000'): pd.Timestamp('2017-01-19 01:55:30.040000'),
pd.Timestamp('2017-01-19 01:57:32.800000'): pd.Timestamp('2017-01-19 02:04:30.040000')}

l_data = pd.DataFrame()
l_data['Timestamp'] = pd.date_range(start=pd.Timestamp('2017-01-18 20:00:00'), 
                                    end=pd.Timestamp('2017-01-19 04:00:00'), freq='10T')
l_data['expected'] = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

#print (l_data)

df = pd.DataFrame({'start': list(d.keys()),'end': list(d.values())})
#fikter by 5 minutes
df = df[(df['end'] - df['start']) > pd.Timedelta(5*60, 's')]

#correct 1 minutes end time
s = df['end'].dt.floor('10T')
df['end1'] = s.where((df['end'] - s) < pd.Timedelta(60, 's'), s + pd.Timedelta(10*60, 's'))
print (df)
                    start                     end                end1
0 2017-01-18 21:45:02.120 2017-01-18 21:50:29.040 2017-01-18 21:50:00
1 2017-01-18 21:51:02.120 2017-01-18 22:52:00.040 2017-01-18 23:00:00
2 2017-01-18 22:52:02.120 2017-01-18 22:57:59.760 2017-01-18 23:00:00
4 2017-01-18 23:44:52.800 2017-01-18 23:50:30.040 2017-01-18 23:50:00
6 2017-01-19 01:40:32.800 2017-01-19 01:55:30.040 2017-01-19 02:00:00
7 2017-01-19 01:57:32.800 2017-01-19 02:04:30.040 2017-01-19 02:10:00

#for each group resample by 10min and add missimg datetimes
v = (df.reset_index()[['start','end1','index']]
       .melt('index')
       .set_index('value')
       .groupby('index')
       .resample('10T')['index']
       .ffill()
       .dropna()
       .index
       .get_level_values(1)
       .unique()
       )
#print (v)

l_data['L'] = l_data['Timestamp'].isin(v).astype(int)
print (l_data.head(20))
             Timestamp  expected  L
0  2017-01-18 20:00:00         0  0
1  2017-01-18 20:10:00         0  0
2  2017-01-18 20:20:00         0  0
3  2017-01-18 20:30:00         0  0
4  2017-01-18 20:40:00         0  0
5  2017-01-18 20:50:00         0  0
6  2017-01-18 21:00:00         0  0
7  2017-01-18 21:10:00         0  0
8  2017-01-18 21:20:00         0  0
9  2017-01-18 21:30:00         0  0
10 2017-01-18 21:40:00         0  0
11 2017-01-18 21:50:00         1  1
12 2017-01-18 22:00:00         1  1
13 2017-01-18 22:10:00         1  1
14 2017-01-18 22:20:00         1  1
15 2017-01-18 22:30:00         1  1
16 2017-01-18 22:40:00         1  1
17 2017-01-18 22:50:00         1  1
18 2017-01-18 23:00:00         1  1
19 2017-01-18 23:10:00         0  0

根据字典选择熊猫数据框的行

1 个答案: