Pandas: create a period based on date column - pandas

I have a dataframe
ID datetime
11 01-09-2021 10:00:00
11 01-09-2021 10:15:15
11 01-09-2021 15:00:00
12 01-09-2021 15:10:00
11 01-09-2021 18:00:00
I need to add period based just on datetime if it increases to 2 hours
ID datetime period
11 01-09-2021 10:00:00 1
11 01-09-2021 10:15:15 1
11 01-09-2021 15:00:00 2
12 01-09-2021 15:10:00 2
11 01-09-2021 18:00:00 3
And the same thing but based on ID and datetime
ID datetime period
11 01-09-2021 10:00:00 1
11 01-09-2021 10:15:15 1
11 01-09-2021 15:00:00 2
12 01-09-2021 15:10:00 1
11 01-09-2021 18:00:00 3
How can I do that?

You can get difference by Series.diff, convert to hours Series.dt.total_seconds, comapre for 2 and add cumulative sum:
df['period'] = df['datetime'].diff().dt.total_seconds().div(3600).gt(2).cumsum().add(1)
print (df)
ID datetime period
0 11 2021-01-09 10:00:00 1
1 11 2021-01-09 10:15:15 1
2 11 2021-01-09 15:00:00 2
3 12 2021-01-09 15:10:00 2
4 11 2021-01-09 18:00:00 3
Similar idea per groups:
f = lambda x: x.diff().dt.total_seconds().div(3600).gt(2).cumsum().add(1)
df['period'] = df.groupby('ID')['datetime'].transform(f)
print (df)
ID datetime period
0 11 2021-01-09 10:00:00 1
1 11 2021-01-09 10:15:15 1
2 11 2021-01-09 15:00:00 2
3 12 2021-01-09 15:10:00 1
4 11 2021-01-09 18:00:00 3

Related

Daily calculations in intraday data

Let's say I have a DataFrame with date_time index:
date_time a b
2020-11-23 04:00:00 10 5
2020-11-23 05:00:00 11 5
2020-11-23 06:00:00 12 5
2020-11-24 04:30:00 13 6
2020-11-24 05:30:00 14 6
2020-11-24 06:30:00 15 6
2020-11-25 06:00:00 16 7
2020-11-25 07:00:00 17 7
2020-11-25 08:00:00 18 7
"a" column is intraday data (every row - different value). "b" column - DAILY data - same data during the current day.
I need to make some calculations with "b" (daily) column and create "c" column with the result. For example, sum for two last days.
Result:
date_time a b c
2020-11-23 04:00:00 10 5 NaN
2020-11-23 05:00:00 11 5 NaN
2020-11-23 06:00:00 12 5 NaN
2020-11-24 04:30:00 13 6 11
2020-11-24 05:30:00 14 6 11
2020-11-24 06:30:00 15 6 11
2020-11-25 06:00:00 16 7 13
2020-11-25 07:00:00 17 7 13
2020-11-25 08:00:00 18 7 13
I guesss I should use something like
df['c'] = df.resample('D').b.rolling(3).sum ...
but I got "NaN" values in "c".
Could you help me? Thanks!
One thing you can do is to drop duplicates on the date and work on that:
# get the dates
df['date'] = df['date_time'].dt.normalize()
df['c'] = (df.drop_duplicates('date')['b'] # drop duplicates on dates
.rolling(2).sum() # rolling sum
)
df['c'] = df['c'].ffill() # fill the missing data
Output:
date_time a b date c
0 2020-11-23 04:00:00 10 5 2020-11-23 NaN
1 2020-11-23 05:00:00 11 5 2020-11-23 NaN
2 2020-11-23 06:00:00 12 5 2020-11-23 NaN
3 2020-11-24 04:30:00 13 6 2020-11-24 11.0
4 2020-11-24 05:30:00 14 6 2020-11-24 11.0
5 2020-11-24 06:30:00 15 6 2020-11-24 11.0
6 2020-11-25 06:00:00 16 7 2020-11-25 13.0
7 2020-11-25 07:00:00 17 7 2020-11-25 13.0
8 2020-11-25 08:00:00 18 7 2020-11-25 13.0

7 days hourly mean with pandas

I need some help calculating a 7 days mean for every hour.
The timeseries has a hourly resolution and I need the 7 days mean for each hour e.g. for 13 o'clock
date, x
2020-07-01 13:00 , 4
2020-07-01 14:00 , 3
.
.
.
2020-07-02 13:00 , 3
2020-07-02 14:00 , 7
.
.
.
I tried it with pandas and a rolling mean, but rolling includes last 7 days.
Thanks for any hints!
Add a new hour column, grouping by hour column, and then add
The average was calculated over 7 days. This is consistent with the intent of the question.
df['hour'] = df.index.hour
df = df.groupby(df.hour)['x'].rolling(7).mean().reset_index()
df.head(35)
hour level_1 x
0 0 2020-07-01 00:00:00 NaN
1 0 2020-07-02 00:00:00 NaN
2 0 2020-07-03 00:00:00 NaN
3 0 2020-07-04 00:00:00 NaN
4 0 2020-07-05 00:00:00 NaN
5 0 2020-07-06 00:00:00 NaN
6 0 2020-07-07 00:00:00 48.142857
7 0 2020-07-08 00:00:00 50.285714
8 0 2020-07-09 00:00:00 60.000000
9 0 2020-07-10 00:00:00 63.142857
10 1 2020-07-01 01:00:00 NaN
11 1 2020-07-02 01:00:00 NaN
12 1 2020-07-03 01:00:00 NaN
13 1 2020-07-04 01:00:00 NaN
14 1 2020-07-05 01:00:00 NaN
15 1 2020-07-06 01:00:00 NaN
16 1 2020-07-07 01:00:00 52.571429
17 1 2020-07-08 01:00:00 48.428571
18 1 2020-07-09 01:00:00 38.000000
19 2 2020-07-01 02:00:00 NaN
20 2 2020-07-02 02:00:00 NaN
21 2 2020-07-03 02:00:00 NaN
22 2 2020-07-04 02:00:00 NaN
23 2 2020-07-05 02:00:00 NaN
24 2 2020-07-06 02:00:00 NaN
25 2 2020-07-07 02:00:00 46.571429
26 2 2020-07-08 02:00:00 47.714286
27 2 2020-07-09 02:00:00 42.714286
28 3 2020-07-01 03:00:00 NaN
29 3 2020-07-02 03:00:00 NaN
30 3 2020-07-03 03:00:00 NaN
31 3 2020-07-04 03:00:00 NaN
32 3 2020-07-05 03:00:00 NaN
33 3 2020-07-06 03:00:00 NaN
34 3 2020-07-07 03:00:00 72.571429

Add a column value with the other date time column at minutes level in pandas

I have a data frame as shown below
ID ideal_appt_time service_time
1 2020-01-06 09:00:00 22
2 2020-01-06 09:30:00 15
1 2020-01-08 14:00:00 42
2 2020-01-12 01:30:00 5
I would like to add service time in terms of minutes with ideal_appt_time and create new column called finish.
Expected Output:
ID ideal_appt_time service_time finish
1 2020-01-06 09:00:00 22 2020-01-06 09:22:00
2 2020-01-06 09:30:00 15 2020-01-06 09:45:00
1 2020-01-08 14:00:00 42 2020-01-08 14:42:00
2 2020-01-12 01:30:00 35 2020-01-12 02:05:00
Use to_timedelta for convert column to timedeltas by minutes and add to datetimes:
df['ideal_appt_time'] = pd.to_datetime(df['ideal_appt_time'])
df['finish'] = df['ideal_appt_time'] + pd.to_timedelta(df['service_time'], unit='Min')
print (df)
ID ideal_appt_time service_time finish
0 1 2020-01-06 09:00:00 22 2020-01-06 09:22:00
1 2 2020-01-06 09:30:00 15 2020-01-06 09:45:00
2 1 2020-01-08 14:00:00 42 2020-01-08 14:42:00
3 2 2020-01-12 01:30:00 5 2020-01-12 01:35:00
Data
df=pd.DataFrame({'ideal_appt_time':['2020-01-06 09:00:00','2020-01-06 09:30:00','2020-01-08 14:00:00','2020-01-12 01:30:00'],'service_time':[22,15,42,35]})
Another way out
df['finish'] = pd.to_datetime(df['ideal_appt_time']).add( df['service_time'].astype('timedelta64[m]'))
df
ideal_appt_time service_time finish
0 2020-01-06 09:00:00 22 2020-01-06 09:22:00
1 2020-01-06 09:30:00 15 2020-01-06 09:45:00
2 2020-01-08 14:00:00 42 2020-01-08 14:42:00
3 2020-01-12 01:30:00 35 2020-01-12 02:05:00

Pandas group by datetime within column level

I have a dataframe created by:
df = pd.DataFrame({})
df['Date'] = pd.to_datetime(np.arange(0,12), unit='h', origin='2018-08-01 06:00:00')
df['ship'] = [1,1,2,2,2,3,3,3,3,3,3,3] # ship ID number
dt_trip = 4 # maximum duration of each trip to be classified as the same trip
Date ship
0 2018-08-01 06:00:00 1
1 2018-08-01 07:00:00 1
2 2018-08-01 08:00:00 2
3 2018-08-01 09:00:00 2
4 2018-08-01 10:00:00 2
5 2018-08-01 11:00:00 3
6 2018-08-01 12:00:00 3
7 2018-08-01 13:00:00 3
8 2018-08-01 14:00:00 3
9 2018-08-01 15:00:00 3
10 2018-08-01 16:00:00 3
11 2018-08-01 17:00:00 3
I try to get a a new column which shows the trips of each ship. Each trip is defined by an interval of 4 hours with respect to the start of the trip. When a new ship number is on the next row, automatically a new trip should start (irrespective of the previous datetime). From a previous post I got a solution for the trips.
origin = df["Date"][0].hour
df["Trip"] = df.apply(lambda x: ((x["Date"].hour - origin) // dt_trip) + 1, axis=1)
df["Trip"] = df.groupby(['Trip','ship']).ngroup() +1 # trip starts at: 1
This solution takes a new trip when the ship-column changes its row. The only change I want to have is to change the origin to the datetime when a new trip starts. So index 4 should have Trip = 2, because the ship is the same and the time difference between the start of the trip (index=2). Now it looks at the first given datetime.
Desired solution looks like:
Date ship Trip Trip_desired
0 2018-08-01 06:00:00 1 1 1
1 2018-08-01 07:00:00 1 1 1
2 2018-08-01 08:00:00 2 2 2
3 2018-08-01 09:00:00 2 2 2
4 2018-08-01 10:00:00 2 3 2
5 2018-08-01 11:00:00 3 4 3
6 2018-08-01 12:00:00 3 4 3
7 2018-08-01 13:00:00 3 4 3
8 2018-08-01 14:00:00 3 5 3
9 2018-08-01 15:00:00 3 5 4
10 2018-08-01 16:00:00 3 5 4
11 2018-08-01 17:00:00 3 5 4
I would do:
total_time = df['Date'] - df.groupby('ship')['Date'].transform('min')
trips = total_time.dt.total_seconds().fillna(0)//(dt_trip*3600)
df['trip'] = df.groupby(['ship', trips]).ngroup()+1
Output:
Date ship trip
0 2018-08-01 06:00:00 1 1
1 2018-08-01 07:00:00 1 1
2 2018-08-01 08:00:00 2 2
3 2018-08-01 09:00:00 2 2
4 2018-08-01 10:00:00 2 2
5 2018-08-01 11:00:00 3 3
6 2018-08-01 12:00:00 3 3
7 2018-08-01 13:00:00 3 3
8 2018-08-01 14:00:00 3 3
9 2018-08-01 15:00:00 3 4
10 2018-08-01 16:00:00 3 4
11 2018-08-01 17:00:00 3 4

How to get count incremental by date

I am trying to get a count of rows with incremental dates.
My table looks like this:
ID name status create_date
1 John AC 2016-01-01 00:00:26.513
2 Jane AC 2016-01-02 00:00:26.513
3 Kane AC 2016-01-02 00:00:26.513
4 Carl AC 2016-01-03 00:00:26.513
5 Dave AC 2016-01-04 00:00:26.513
6 Gina AC 2016-01-04 00:00:26.513
Now what I want to return from the SQL is something like this:
Date Count
2016-01-01 1
2016-01-02 3
2016-01-03 4
2016-01-04 6
You can make use of COUNT() OVER () without PARTITION BY,by using ORDER BY. It will give you the cumulative sum.Use DISTINCT to filter out the duplicate values.
SELECT DISTINCT CAST(create_date AS DATE) [Date],
COUNT(create_date) OVER (ORDER BY CAST(create_date AS DATE)) as [COUNT]
FROM [YourTable]
SELECT create_date, COUNT(create_date) as [COUNT]
FROM (
SELECT CAST(create_date AS DATE) create_date
FROM [YourTable]
) T
GROUP BY create_date
Per your description, you need a continuous dates list, Does it make sense?
This sample only generating one-month data.
CREATE TABLE #tt(ID INT, name VARCHAR(10), status VARCHAR(10), create_date DATETIME)
INSERT INTO #tt
SELECT 1,'John','AC','2016-01-01 00:00:26.513' UNION
SELECT 2,'Jane','AC','2016-01-02 00:00:26.513' UNION
SELECT 3,'Kane','AC','2016-01-02 00:00:26.513' UNION
SELECT 4,'Carl','AC','2016-01-03 00:00:26.513' UNION
SELECT 5,'Dave','AC','2016-01-04 00:00:26.513' UNION
SELECT 6,'Gina','AC','2016-01-04 00:00:26.513' UNION
SELECT 7,'Tina','AC','2016-01-08 00:00:26.513'
SELECT * FROM #tt
SELECT CONVERT(DATE,DATEADD(d,sv.number,n.FirstDate)) AS [Date],COUNT(n.num) AS [Count]
FROM master.dbo.spt_values AS sv
LEFT JOIN (
SELECT MIN(t.create_date)OVER() AS FirstDate,DATEDIFF(d,MIN(t.create_date)OVER(),t.create_date) AS num FROM #tt AS t
) AS n ON n.num<=sv.number
WHERE sv.type='P' AND sv.number>=0 AND MONTH(DATEADD(d,sv.number,n.FirstDate))=MONTH(n.FirstDate)
GROUP BY CONVERT(DATE,DATEADD(d,sv.number,n.FirstDate))
Date Count
---------- -----------
2016-01-01 1
2016-01-02 3
2016-01-03 4
2016-01-04 6
2016-01-05 6
2016-01-06 6
2016-01-07 6
2016-01-08 7
2016-01-09 7
2016-01-10 7
2016-01-11 7
2016-01-12 7
2016-01-13 7
2016-01-14 7
2016-01-15 7
2016-01-16 7
2016-01-17 7
2016-01-18 7
2016-01-19 7
2016-01-20 7
2016-01-21 7
2016-01-22 7
2016-01-23 7
2016-01-24 7
2016-01-25 7
2016-01-26 7
2016-01-27 7
2016-01-28 7
2016-01-29 7
2016-01-30 7
2016-01-31 7
2017-01-01 7
2017-01-02 7
2017-01-03 7
2017-01-04 7
2017-01-05 7
2017-01-06 7
2017-01-07 7
2017-01-08 7
2017-01-09 7
2017-01-10 7
2017-01-11 7
2017-01-12 7
2017-01-13 7
2017-01-14 7
2017-01-15 7
2017-01-16 7
2017-01-17 7
2017-01-18 7
2017-01-19 7
2017-01-20 7
2017-01-21 7
2017-01-22 7
2017-01-23 7
2017-01-24 7
2017-01-25 7
2017-01-26 7
2017-01-27 7
2017-01-28 7
2017-01-29 7
2017-01-30 7
2017-01-31 7
2018-01-01 7
2018-01-02 7
2018-01-03 7
2018-01-04 7
2018-01-05 7
2018-01-06 7
2018-01-07 7
2018-01-08 7
2018-01-09 7
2018-01-10 7
2018-01-11 7
2018-01-12 7
2018-01-13 7
2018-01-14 7
2018-01-15 7
2018-01-16 7
2018-01-17 7
2018-01-18 7
2018-01-19 7
2018-01-20 7
2018-01-21 7
2018-01-22 7
2018-01-23 7
2018-01-24 7
2018-01-25 7
2018-01-26 7
2018-01-27 7
2018-01-28 7
2018-01-29 7
2018-01-30 7
2018-01-31 7
2019-01-01 7
2019-01-02 7
2019-01-03 7
2019-01-04 7
2019-01-05 7
2019-01-06 7
2019-01-07 7
2019-01-08 7
2019-01-09 7
2019-01-10 7
2019-01-11 7
2019-01-12 7
2019-01-13 7
2019-01-14 7
2019-01-15 7
2019-01-16 7
2019-01-17 7
2019-01-18 7
2019-01-19 7
2019-01-20 7
2019-01-21 7
2019-01-22 7
2019-01-23 7
2019-01-24 7
2019-01-25 7
2019-01-26 7
2019-01-27 7
2019-01-28 7
2019-01-29 7
2019-01-30 7
2019-01-31 7
2020-01-01 7
2020-01-02 7
2020-01-03 7
2020-01-04 7
2020-01-05 7
2020-01-06 7
2020-01-07 7
2020-01-08 7
2020-01-09 7
2020-01-10 7
2020-01-11 7
2020-01-12 7
2020-01-13 7
2020-01-14 7
2020-01-15 7
2020-01-16 7
2020-01-17 7
2020-01-18 7
2020-01-19 7
2020-01-20 7
2020-01-21 7
2020-01-22 7
2020-01-23 7
2020-01-24 7
2020-01-25 7
2020-01-26 7
2020-01-27 7
2020-01-28 7
2020-01-29 7
2020-01-30 7
2020-01-31 7
2021-01-01 7
2021-01-02 7
2021-01-03 7
2021-01-04 7
2021-01-05 7
2021-01-06 7
2021-01-07 7
2021-01-08 7
2021-01-09 7
2021-01-10 7
2021-01-11 7
2021-01-12 7
2021-01-13 7
2021-01-14 7
2021-01-15 7
2021-01-16 7
2021-01-17 7
2021-01-18 7
2021-01-19 7
2021-01-20 7
2021-01-21 7
2021-01-22 7
2021-01-23 7
2021-01-24 7
2021-01-25 7
2021-01-26 7
2021-01-27 7
2021-01-28 7
2021-01-29 7
2021-01-30 7
2021-01-31 7
select r.date,count(r.date) count
from
(
select id,name,substring(convert(nvarchar(50),create_date),1,10) date
from tblName
) r
group by r.date
In this code, in the subquery part,
I select the first 10 letter of date which is converted from dateTime to nvarchar so I make like '2016-01-01'. (which is not also necessary but for make code more readable I prefer to do it in this way).
Then with a simple group by I have date and date's count.