How to create a time threshold based column given a time gap?
up vote
1
down vote
favorite
I have a pandas dataframe with several columns, however for visual purposes consider the columns Id and timestamp. As you can see the pandas dataframe is sorted by Id column.
Id timestamp
11 2018-10-19 13:00:00
11 2018-10-19 13:05:00
11 2018-10-19 13:06:00
11 2018-10-19 13:07:00
11 2018-10-19 13:30:00
11 2018-10-19 13:31:00
11 2018-10-19 13:32:00
11 2018-10-19 13:55:00
11 2018-10-19 13:54:00
11 2018-10-21 20:47:09
11 2018-10-21 20:48:27
11 2018-10-21 20:48:45
11 2018-10-21 20:48:52
12 2018-10-09 20:30:46
12 2018-10-09 20:30:22
12 2018-10-09 20:30:05
12 2018-10-09 20:29:44
12 2018-10-09 20:29:31
13 2018-10-19 18:49:08
13 2018-10-19 18:49:13
13 2018-10-11 18:46:15
14 2018-10-11 10:46:40
14 2018-10-23 10:39:52
How can create create another ID column based on 10 minutes time gaps? That is for every timestamp 10 minutes threshold create a new different `ID_2:
Id timestamp ID_2
11 2018-10-19 13:00:00 01
11 2018-10-19 13:05:00 01
11 2018-10-19 13:06:00 01
11 2018-10-19 13:07:00 01
11 2018-10-19 13:30:00 02
11 2018-10-19 13:31:00 02
11 2018-10-19 13:32:00 02
11 2018-10-19 13:55:00 03
11 2018-10-19 13:54:00 03
11 2018-10-21 20:47:09 04
11 2018-10-21 20:48:27 04
11 2018-10-21 20:48:45 04
11 2018-10-21 20:48:52 04
12 2018-10-09 20:30:46 04
12 2018-10-09 20:30:22 04
12 2018-10-09 20:30:05 04
12 2018-10-09 20:29:44 05
12 2018-10-09 20:29:31 05
13 2018-10-19 18:49:08 06
13 2018-10-19 18:49:13 06
13 2018-10-11 18:46:15 07
14 2018-10-11 10:46:40 07
I tried to detect the time gaps as follows:
df['col_new'] = (df['timestamp'].diff()).dt.seconds > 600
However, I do not understand how to apply a backward fill in order to create the IDs. Therefore, how can I detect time gaps and assign them a new id?
python python-3.x pandas datetime
asked Nov 7 at 9:17
anon
1457
add a comment |
up vote
1
down vote
favorite
I have a pandas dataframe with several columns, however for visual purposes consider the columns Id and timestamp. As you can see the pandas dataframe is sorted by Id column.
Id timestamp
11 2018-10-19 13:00:00
11 2018-10-19 13:05:00
11 2018-10-19 13:06:00
11 2018-10-19 13:07:00
11 2018-10-19 13:30:00
11 2018-10-19 13:31:00
11 2018-10-19 13:32:00
11 2018-10-19 13:55:00
11 2018-10-19 13:54:00
11 2018-10-21 20:47:09
11 2018-10-21 20:48:27
11 2018-10-21 20:48:45
11 2018-10-21 20:48:52
12 2018-10-09 20:30:46
12 2018-10-09 20:30:22
12 2018-10-09 20:30:05
12 2018-10-09 20:29:44
12 2018-10-09 20:29:31
13 2018-10-19 18:49:08
13 2018-10-19 18:49:13
13 2018-10-11 18:46:15
14 2018-10-11 10:46:40
14 2018-10-23 10:39:52
How can create create another ID column based on 10 minutes time gaps? That is for every timestamp 10 minutes threshold create a new different `ID_2:
Id timestamp ID_2
11 2018-10-19 13:00:00 01
11 2018-10-19 13:05:00 01
11 2018-10-19 13:06:00 01
11 2018-10-19 13:07:00 01
11 2018-10-19 13:30:00 02
11 2018-10-19 13:31:00 02
11 2018-10-19 13:32:00 02
11 2018-10-19 13:55:00 03
11 2018-10-19 13:54:00 03
11 2018-10-21 20:47:09 04
11 2018-10-21 20:48:27 04
11 2018-10-21 20:48:45 04
11 2018-10-21 20:48:52 04
12 2018-10-09 20:30:46 04
12 2018-10-09 20:30:22 04
12 2018-10-09 20:30:05 04
12 2018-10-09 20:29:44 05
12 2018-10-09 20:29:31 05
13 2018-10-19 18:49:08 06
13 2018-10-19 18:49:13 06
13 2018-10-11 18:46:15 07
14 2018-10-11 10:46:40 07
I tried to detect the time gaps as follows:
df['col_new'] = (df['timestamp'].diff()).dt.seconds > 600
However, I do not understand how to apply a backward fill in order to create the IDs. Therefore, how can I detect time gaps and assign them a new id?
python python-3.x pandas datetime
asked Nov 7 at 9:17
anon
1457
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a pandas dataframe with several columns, however for visual purposes consider the columns Id and timestamp. As you can see the pandas dataframe is sorted by Id column.
Id timestamp
11 2018-10-19 13:00:00
11 2018-10-19 13:05:00
11 2018-10-19 13:06:00
11 2018-10-19 13:07:00
11 2018-10-19 13:30:00
11 2018-10-19 13:31:00
11 2018-10-19 13:32:00
11 2018-10-19 13:55:00
11 2018-10-19 13:54:00
11 2018-10-21 20:47:09
11 2018-10-21 20:48:27
11 2018-10-21 20:48:45
11 2018-10-21 20:48:52
12 2018-10-09 20:30:46
12 2018-10-09 20:30:22
12 2018-10-09 20:30:05
12 2018-10-09 20:29:44
12 2018-10-09 20:29:31
13 2018-10-19 18:49:08
13 2018-10-19 18:49:13
13 2018-10-11 18:46:15
14 2018-10-11 10:46:40
14 2018-10-23 10:39:52
How can create create another ID column based on 10 minutes time gaps? That is for every timestamp 10 minutes threshold create a new different `ID_2:
Id timestamp ID_2
11 2018-10-19 13:00:00 01
11 2018-10-19 13:05:00 01
11 2018-10-19 13:06:00 01
11 2018-10-19 13:07:00 01
11 2018-10-19 13:30:00 02
11 2018-10-19 13:31:00 02
11 2018-10-19 13:32:00 02
11 2018-10-19 13:55:00 03
11 2018-10-19 13:54:00 03
11 2018-10-21 20:47:09 04
11 2018-10-21 20:48:27 04
11 2018-10-21 20:48:45 04
11 2018-10-21 20:48:52 04
12 2018-10-09 20:30:46 04
12 2018-10-09 20:30:22 04
12 2018-10-09 20:30:05 04
12 2018-10-09 20:29:44 05
12 2018-10-09 20:29:31 05
13 2018-10-19 18:49:08 06
13 2018-10-19 18:49:13 06
13 2018-10-11 18:46:15 07
14 2018-10-11 10:46:40 07
I tried to detect the time gaps as follows:
df['col_new'] = (df['timestamp'].diff()).dt.seconds > 600
However, I do not understand how to apply a backward fill in order to create the IDs. Therefore, how can I detect time gaps and assign them a new id?
python python-3.x pandas datetime
asked Nov 7 at 9:17
anon
1457
I have a pandas dataframe with several columns, however for visual purposes consider the columns Id and timestamp. As you can see the pandas dataframe is sorted by Id column.
Id timestamp
11 2018-10-19 13:00:00
11 2018-10-19 13:05:00
11 2018-10-19 13:06:00
11 2018-10-19 13:07:00
11 2018-10-19 13:30:00
11 2018-10-19 13:31:00
11 2018-10-19 13:32:00
11 2018-10-19 13:55:00
11 2018-10-19 13:54:00
11 2018-10-21 20:47:09
11 2018-10-21 20:48:27
11 2018-10-21 20:48:45
11 2018-10-21 20:48:52
12 2018-10-09 20:30:46
12 2018-10-09 20:30:22
12 2018-10-09 20:30:05
12 2018-10-09 20:29:44
12 2018-10-09 20:29:31
13 2018-10-19 18:49:08
13 2018-10-19 18:49:13
13 2018-10-11 18:46:15
14 2018-10-11 10:46:40
14 2018-10-23 10:39:52
How can create create another ID column based on 10 minutes time gaps? That is for every timestamp 10 minutes threshold create a new different `ID_2:
Id timestamp ID_2
11 2018-10-19 13:00:00 01
11 2018-10-19 13:05:00 01
11 2018-10-19 13:06:00 01
11 2018-10-19 13:07:00 01
11 2018-10-19 13:30:00 02
11 2018-10-19 13:31:00 02
11 2018-10-19 13:32:00 02
11 2018-10-19 13:55:00 03
11 2018-10-19 13:54:00 03
11 2018-10-21 20:47:09 04
11 2018-10-21 20:48:27 04
11 2018-10-21 20:48:45 04
11 2018-10-21 20:48:52 04
12 2018-10-09 20:30:46 04
12 2018-10-09 20:30:22 04
12 2018-10-09 20:30:05 04
12 2018-10-09 20:29:44 05
12 2018-10-09 20:29:31 05
13 2018-10-19 18:49:08 06
13 2018-10-19 18:49:13 06
13 2018-10-11 18:46:15 07
14 2018-10-11 10:46:40 07
I tried to detect the time gaps as follows:
df['col_new'] = (df['timestamp'].diff()).dt.seconds > 600
However, I do not understand how to apply a backward fill in order to create the IDs. Therefore, how can I detect time gaps and assign them a new id?
python python-3.x pandas datetime
python python-3.x pandas datetime
asked Nov 7 at 9:17
anon
1457
asked Nov 7 at 9:17
anon
1457
asked Nov 7 at 9:17
anon
1457
asked Nov 7 at 9:17
anon
1457
asked Nov 7 at 9:17
anon
1457
1457
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
I believe you need floor with factorize, last add zfill:
df['timestamp'] = pd.to_datetime(df['timestamp'])
a = pd.factorize(df['timestamp'].dt.floor('10Min'))[0] + 1
df['col_new'] = pd.Series(a, index=df.index).astype(str).str.zfill(2)
print (df)
Id timestamp ID_2 col_new
0 11 2018-10-19 13:00:00 01 01
1 11 2018-10-19 13:05:00 01 01
2 11 2018-10-19 13:06:00 01 01
3 11 2018-10-19 13:07:00 01 01
4 11 2018-10-19 13:30:00 02 02
5 11 2018-10-19 13:31:00 02 02
6 11 2018-10-19 13:32:00 02 02
7 11 2018-10-19 13:55:00 03 03
8 11 2018-10-19 13:54:00 03 03
9 11 2018-10-21 20:47:09 04 04
10 11 2018-10-21 20:48:27 04 04
11 11 2018-10-21 20:48:45 04 04
12 11 2018-10-21 20:48:52 04 04
13 12 2018-10-09 20:30:46 04 05
14 12 2018-10-09 20:30:22 04 05
15 12 2018-10-09 20:30:05 04 05
16 12 2018-10-09 20:29:44 05 06
17 12 2018-10-09 20:29:31 05 06
18 13 2018-10-19 18:49:08 06 07
19 13 2018-10-19 18:49:13 06 07
20 13 2018-10-11 18:46:15 07 08
21 14 2018-10-11 18:46:40 07 08
Detail:
print (df['timestamp'].dt.floor('10Min'))
0 2018-10-19 13:00:00
1 2018-10-19 13:00:00
2 2018-10-19 13:00:00
3 2018-10-19 13:00:00
4 2018-10-19 13:30:00
5 2018-10-19 13:30:00
6 2018-10-19 13:30:00
7 2018-10-19 13:50:00
8 2018-10-19 13:50:00
9 2018-10-21 20:40:00
10 2018-10-21 20:40:00
11 2018-10-21 20:40:00
12 2018-10-21 20:40:00
13 2018-10-09 20:30:00
14 2018-10-09 20:30:00
15 2018-10-09 20:30:00
16 2018-10-09 20:20:00
17 2018-10-09 20:20:00
18 2018-10-19 18:40:00
19 2018-10-19 18:40:00
20 2018-10-11 18:40:00
21 2018-10-11 18:40:00
Name: timestamp, dtype: datetime64[ns]
answered Nov 7 at 9:36
jezrael
306k20240316
2
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
1
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I believe you need floor with factorize, last add zfill:
df['timestamp'] = pd.to_datetime(df['timestamp'])
a = pd.factorize(df['timestamp'].dt.floor('10Min'))[0] + 1
df['col_new'] = pd.Series(a, index=df.index).astype(str).str.zfill(2)
print (df)
Id timestamp ID_2 col_new
0 11 2018-10-19 13:00:00 01 01
1 11 2018-10-19 13:05:00 01 01
2 11 2018-10-19 13:06:00 01 01
3 11 2018-10-19 13:07:00 01 01
4 11 2018-10-19 13:30:00 02 02
5 11 2018-10-19 13:31:00 02 02
6 11 2018-10-19 13:32:00 02 02
7 11 2018-10-19 13:55:00 03 03
8 11 2018-10-19 13:54:00 03 03
9 11 2018-10-21 20:47:09 04 04
10 11 2018-10-21 20:48:27 04 04
11 11 2018-10-21 20:48:45 04 04
12 11 2018-10-21 20:48:52 04 04
13 12 2018-10-09 20:30:46 04 05
14 12 2018-10-09 20:30:22 04 05
15 12 2018-10-09 20:30:05 04 05
16 12 2018-10-09 20:29:44 05 06
17 12 2018-10-09 20:29:31 05 06
18 13 2018-10-19 18:49:08 06 07
19 13 2018-10-19 18:49:13 06 07
20 13 2018-10-11 18:46:15 07 08
21 14 2018-10-11 18:46:40 07 08
Detail:
print (df['timestamp'].dt.floor('10Min'))
0 2018-10-19 13:00:00
1 2018-10-19 13:00:00
2 2018-10-19 13:00:00
3 2018-10-19 13:00:00
4 2018-10-19 13:30:00
5 2018-10-19 13:30:00
6 2018-10-19 13:30:00
7 2018-10-19 13:50:00
8 2018-10-19 13:50:00
9 2018-10-21 20:40:00
10 2018-10-21 20:40:00
11 2018-10-21 20:40:00
12 2018-10-21 20:40:00
13 2018-10-09 20:30:00
14 2018-10-09 20:30:00
15 2018-10-09 20:30:00
16 2018-10-09 20:20:00
17 2018-10-09 20:20:00
18 2018-10-19 18:40:00
19 2018-10-19 18:40:00
20 2018-10-11 18:40:00
21 2018-10-11 18:40:00
Name: timestamp, dtype: datetime64[ns]
answered Nov 7 at 9:36
jezrael
306k20240316
2
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
1
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
add a comment |
up vote
3
down vote
accepted
I believe you need floor with factorize, last add zfill:
df['timestamp'] = pd.to_datetime(df['timestamp'])
a = pd.factorize(df['timestamp'].dt.floor('10Min'))[0] + 1
df['col_new'] = pd.Series(a, index=df.index).astype(str).str.zfill(2)
print (df)
Id timestamp ID_2 col_new
0 11 2018-10-19 13:00:00 01 01
1 11 2018-10-19 13:05:00 01 01
2 11 2018-10-19 13:06:00 01 01
3 11 2018-10-19 13:07:00 01 01
4 11 2018-10-19 13:30:00 02 02
5 11 2018-10-19 13:31:00 02 02
6 11 2018-10-19 13:32:00 02 02
7 11 2018-10-19 13:55:00 03 03
8 11 2018-10-19 13:54:00 03 03
9 11 2018-10-21 20:47:09 04 04
10 11 2018-10-21 20:48:27 04 04
11 11 2018-10-21 20:48:45 04 04
12 11 2018-10-21 20:48:52 04 04
13 12 2018-10-09 20:30:46 04 05
14 12 2018-10-09 20:30:22 04 05
15 12 2018-10-09 20:30:05 04 05
16 12 2018-10-09 20:29:44 05 06
17 12 2018-10-09 20:29:31 05 06
18 13 2018-10-19 18:49:08 06 07
19 13 2018-10-19 18:49:13 06 07
20 13 2018-10-11 18:46:15 07 08
21 14 2018-10-11 18:46:40 07 08
Detail:
print (df['timestamp'].dt.floor('10Min'))
0 2018-10-19 13:00:00
1 2018-10-19 13:00:00
2 2018-10-19 13:00:00
3 2018-10-19 13:00:00
4 2018-10-19 13:30:00
5 2018-10-19 13:30:00
6 2018-10-19 13:30:00
7 2018-10-19 13:50:00
8 2018-10-19 13:50:00
9 2018-10-21 20:40:00
10 2018-10-21 20:40:00
11 2018-10-21 20:40:00
12 2018-10-21 20:40:00
13 2018-10-09 20:30:00
14 2018-10-09 20:30:00
15 2018-10-09 20:30:00
16 2018-10-09 20:20:00
17 2018-10-09 20:20:00
18 2018-10-19 18:40:00
19 2018-10-19 18:40:00
20 2018-10-11 18:40:00
21 2018-10-11 18:40:00
Name: timestamp, dtype: datetime64[ns]
answered Nov 7 at 9:36
jezrael
306k20240316
2
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
1
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I believe you need floor with factorize, last add zfill:
df['timestamp'] = pd.to_datetime(df['timestamp'])
a = pd.factorize(df['timestamp'].dt.floor('10Min'))[0] + 1
df['col_new'] = pd.Series(a, index=df.index).astype(str).str.zfill(2)
print (df)
Id timestamp ID_2 col_new
0 11 2018-10-19 13:00:00 01 01
1 11 2018-10-19 13:05:00 01 01
2 11 2018-10-19 13:06:00 01 01
3 11 2018-10-19 13:07:00 01 01
4 11 2018-10-19 13:30:00 02 02
5 11 2018-10-19 13:31:00 02 02
6 11 2018-10-19 13:32:00 02 02
7 11 2018-10-19 13:55:00 03 03
8 11 2018-10-19 13:54:00 03 03
9 11 2018-10-21 20:47:09 04 04
10 11 2018-10-21 20:48:27 04 04
11 11 2018-10-21 20:48:45 04 04
12 11 2018-10-21 20:48:52 04 04
13 12 2018-10-09 20:30:46 04 05
14 12 2018-10-09 20:30:22 04 05
15 12 2018-10-09 20:30:05 04 05
16 12 2018-10-09 20:29:44 05 06
17 12 2018-10-09 20:29:31 05 06
18 13 2018-10-19 18:49:08 06 07
19 13 2018-10-19 18:49:13 06 07
20 13 2018-10-11 18:46:15 07 08
21 14 2018-10-11 18:46:40 07 08
Detail:
print (df['timestamp'].dt.floor('10Min'))
0 2018-10-19 13:00:00
1 2018-10-19 13:00:00
2 2018-10-19 13:00:00
3 2018-10-19 13:00:00
4 2018-10-19 13:30:00
5 2018-10-19 13:30:00
6 2018-10-19 13:30:00
7 2018-10-19 13:50:00
8 2018-10-19 13:50:00
9 2018-10-21 20:40:00
10 2018-10-21 20:40:00
11 2018-10-21 20:40:00
12 2018-10-21 20:40:00
13 2018-10-09 20:30:00
14 2018-10-09 20:30:00
15 2018-10-09 20:30:00
16 2018-10-09 20:20:00
17 2018-10-09 20:20:00
18 2018-10-19 18:40:00
19 2018-10-19 18:40:00
20 2018-10-11 18:40:00
21 2018-10-11 18:40:00
Name: timestamp, dtype: datetime64[ns]
answered Nov 7 at 9:36
jezrael
306k20240316
I believe you need floor with factorize, last add zfill:
df['timestamp'] = pd.to_datetime(df['timestamp'])
a = pd.factorize(df['timestamp'].dt.floor('10Min'))[0] + 1
df['col_new'] = pd.Series(a, index=df.index).astype(str).str.zfill(2)
print (df)
Id timestamp ID_2 col_new
0 11 2018-10-19 13:00:00 01 01
1 11 2018-10-19 13:05:00 01 01
2 11 2018-10-19 13:06:00 01 01
3 11 2018-10-19 13:07:00 01 01
4 11 2018-10-19 13:30:00 02 02
5 11 2018-10-19 13:31:00 02 02
6 11 2018-10-19 13:32:00 02 02
7 11 2018-10-19 13:55:00 03 03
8 11 2018-10-19 13:54:00 03 03
9 11 2018-10-21 20:47:09 04 04
10 11 2018-10-21 20:48:27 04 04
11 11 2018-10-21 20:48:45 04 04
12 11 2018-10-21 20:48:52 04 04
13 12 2018-10-09 20:30:46 04 05
14 12 2018-10-09 20:30:22 04 05
15 12 2018-10-09 20:30:05 04 05
16 12 2018-10-09 20:29:44 05 06
17 12 2018-10-09 20:29:31 05 06
18 13 2018-10-19 18:49:08 06 07
19 13 2018-10-19 18:49:13 06 07
20 13 2018-10-11 18:46:15 07 08
21 14 2018-10-11 18:46:40 07 08
Detail:
print (df['timestamp'].dt.floor('10Min'))
0 2018-10-19 13:00:00
1 2018-10-19 13:00:00
2 2018-10-19 13:00:00
3 2018-10-19 13:00:00
4 2018-10-19 13:30:00
5 2018-10-19 13:30:00
6 2018-10-19 13:30:00
7 2018-10-19 13:50:00
8 2018-10-19 13:50:00
9 2018-10-21 20:40:00
10 2018-10-21 20:40:00
11 2018-10-21 20:40:00
12 2018-10-21 20:40:00
13 2018-10-09 20:30:00
14 2018-10-09 20:30:00
15 2018-10-09 20:30:00
16 2018-10-09 20:20:00
17 2018-10-09 20:20:00
18 2018-10-19 18:40:00
19 2018-10-19 18:40:00
20 2018-10-11 18:40:00
21 2018-10-11 18:40:00
Name: timestamp, dtype: datetime64[ns]
answered Nov 7 at 9:36
jezrael
306k20240316
edited Nov 7 at 9:49
answered Nov 7 at 9:36
jezrael
306k20240316
answered Nov 7 at 9:36
jezrael
306k20240316
answered Nov 7 at 9:36
jezrael
306k20240316
306k20240316
2
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
1
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
add a comment |
2
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
1
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
2
2
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
wow, this is superbly elegant
– Pankaj Joshi
Nov 7 at 9:48
1
1
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
@PankajJoshi - Thank you.
– jezrael
Nov 7 at 9:48
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Post as a guest
Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
