I have a df as shown below.
df:
ID open_date limit
1 2020-06-03 100
1 2020-06-23 500
1 2019-06-29 300
1 2018-06-29 400
From the above I would like to calculate a column named age_in_days.
age_in_days is the number of days from open_date to 2020-06-30.
Expected output
ID open_date limit age_in_days
1 2020-06-03 100 27
1 2020-06-23 500 7
1 2019-06-29 300 367
1 2018-06-29 400 732
Make sure open_date in datetime dtype and subtract it from 2020-06-30
df['open_date'] = pd.to_datetime(df.open_date)
df['age_in_days'] = (pd.Timestamp('2020-06-30') - df.open_date).dt.days
Out[209]:
ID open_date limit age_in_days
0 1 2020-06-03 100 27
1 1 2020-06-23 500 7
2 1 2019-06-29 300 367
3 1 2018-06-29 400 732
Related
I have a table like below :
ID
Amount
Date
1
500
2022-01-03
1
200
2022-01-04
1
500
2022-01-05
1
340
2022-01-06
1
500
2022-01-25
1
500
2022-01-26
1
567
2022-01-27
1
500
2022-01-28
1
598
2022-01-31
1
500
2022-02-01
1
787
2022-02-02
1
500
2022-02-03
1
5340
2022-02-04
PROBLEM :-
So I have to calculate average of column where StartDate = 03/01/2022 (3rd Jan 2022) and for each month it would be like for January Average of Amount from StartDate to 25th Jan, then for Feb Startdate to 22nd Feb, so this date logic is also there
SET #Last = (SELECT DATEADD(DAY, CASE DATENAME(WEEKDAY, #Date)
WHEN 'Sunday' THEN -6
When 'Saturday' THEN -5
ELSE -7 END, DATEDIFF(DAY, 0, #Date)))
RETURN #Last
ID
Amount
Date
Last
1
500
2022-01-03
2022-01-25
1
500
2022-01-04
2022-01-25
1
340
2022-01-05
2022-01-25
1
500
2022-01-06
2022-01-25
1
567
2022-01-25
2022-01-25
1
500
2022-01-26
2022-01-25
1
500
2022-01-27
2022-01-25
1
40
2022-01-28
2022-01-25
1
500
2022-01-31
2022-01-25
1
589
2022-02-01
2022-02-22
1
540
2022-02-02
2022-02-22
1
500
2022-02-03
2022-02-22
1
5340
2022-02-04
2022-02-22
Like the above table..
Now if I calculate Avg(Amount), from 3rd jan to 25th Jan for Jan and 3rd Jan to 22nd Feb and so on.. It's not giving correct average, like it is calculating the rest of the days amount also. Also grouping by is grouping month wise not as where clause
Select Avg(Amount) from Table
where Date BETWEEN #StartDate AND Last
StartDate is fixed # 3rd Jan.
This is not giving the correct Avg. Any other way I could get the required data?
Given are two series, like this:
#period1
DATE
2020-06-22 310.62
2020-06-26 300.05
2020-09-23 322.64
2020-10-30 326.54
#period2
DATE
2020-06-23 312.05
2020-09-02 357.70
2020-10-12 352.43
2021-01-25 384.39
These two series are correlated to each other, i.e. they each mark either the beginning or the end of a date period. The first series marks the end of a period1 period, the second series marks the end of period2 period. The end of a period2 period is at the same time also the start of a period1 period, and vice versa.
I've been looking for a way to aggregate these periods as date ranges, but apparently this is not easily possible with Pandas dataframes. Suggestions extremely welcome.
In the easiest case, the output layout should reflect the end dates of periods, which period type it was, and the amount of change between start and stop of the period.
Explicit output:
DATE CHG PERIOD
2020-06-22 NaN 1
2020-06-23 1.43 2
2020-06-26 12.0 1
2020-09-02 57.65 2
2020-09-23 35.06 1
2020-10-12 29.79 2
2020-10-30 25.89 1
2021-01-25 57.85 2
However, if there is any possibility of actually grouping by a date range consisting of start AND stop date, that would be much more favorable
Thank you!
p1 = pd.DataFrame(data={'Date': ['2020-06-22', '2020-06-26', '2020-09-23', '2020-10-30'], 'val':[310.62, 300.05, 322.64, 326.54]})
p2 = pd.DataFrame(data={'Date': ['2020-06-23', '2020-09-02', '2020-10-12', '2021-01-25'], 'val':[312.05, 357.7, 352.43, 384.39]})
p1['period'] = 1
p2['period'] = 2
df = p1.append(p2).sort_values('Date').reset_index(drop=True)
df['CHG'] = abs(df['val'].diff(periods=1))
df.drop('val', axis=1)
Output:
Date period CHG
0 2020-06-22 1 NaN
1 2020-06-23 2 1.43
2 2020-06-26 1 12.00
3 2020-09-02 2 57.65
4 2020-09-23 1 35.06
5 2020-10-12 2 29.79
6 2020-10-30 1 25.89
7 2021-01-25 2 57.85
EDIT: matching the format START - STOP - CHANGE - PERIOD
Starting from the above data frame:
df['Start'] = df.Date.shift(periods=1)
df.rename(columns={'Date': 'Stop'}, inplace=True)
df = df1[['Start', 'Stop', 'CHG', 'period']]
df
Output:
Start Stop CHG period
0 NaN 2020-06-22 NaN 1
1 2020-06-22 2020-06-23 1.43 2
2 2020-06-23 2020-06-26 12.00 1
3 2020-06-26 2020-09-02 57.65 2
4 2020-09-02 2020-09-23 35.06 1
5 2020-09-23 2020-10-12 29.79 2
6 2020-10-12 2020-10-30 25.89 1
7 2020-10-30 2021-01-25 57.85 2
# If needed:
df1.index = pd.to_datetime(df1.index)
df2.index = pd.to_datetime(df2.index)
df = pd.concat([df1, df2], axis=1)
df.columns = ['start','stop']
df['CNG'] = df.bfill(axis=1)['start'].diff().abs()
df['PERIOD'] = 1
df.loc[df.stop.notna(), 'PERIOD'] = 2
df = df[['CNG', 'PERIOD']]
print(df)
Output:
CNG PERIOD
Date
2020-06-22 NaN 1
2020-06-23 1.43 2
2020-06-26 12.00 1
2020-09-02 57.65 2
2020-09-23 35.06 1
2020-10-12 29.79 2
2020-10-30 25.89 1
2021-01-25 57.85 2
2021-01-29 14.32 1
2021-02-12 22.57 2
2021-03-04 15.94 1
2021-05-07 45.42 2
2021-05-12 16.71 1
2021-09-02 47.78 2
2021-10-04 24.55 1
2021-11-18 41.09 2
2021-12-01 19.23 1
2021-12-10 20.24 2
2021-12-20 15.76 1
2022-01-03 22.73 2
2022-01-27 46.47 1
2022-02-09 26.30 2
2022-02-23 35.59 1
2022-03-02 15.94 2
2022-03-08 21.64 1
2022-03-29 45.30 2
2022-04-29 49.55 1
2022-05-04 17.06 2
2022-05-12 36.72 1
2022-05-17 15.98 2
2022-05-19 18.86 1
2022-06-02 27.93 2
2022-06-17 51.53 1
I used Pandas’ resample function for calculating the sales of a list of proucts every 6 months.
I used the resample function for ‘6M’ and using apply({“column-name”:”sum”}).
Now I’d like to create a table with the sum of the sales for the first six months.
How can I extract the sum of the first 6 months, given that all products have records for more than 3 years, and none of them have the same start date?
Thanks in advance for any suggestions.
Here is an example of the data:
Product Date sales
Product 1 6/30/2017 20
12/31/2017 60
6/30/2018 50
12/31/2018 100
Product 2 1/31/2017 30
7/31/2017 150
1/31/2018 200
7/31/2018 300
1/31/2019 100
While waiting for your data, I worked on this. See if this is something that will be helpful for you.
import pandas as pd
df = pd.DataFrame({'Date':['2018-01-10','2018-02-15','2018-03-18',
'2018-07-10','2018-09-12','2018-10-14',
'2018-11-16','2018-12-20','2019-01-10',
'2019-04-15','2019-06-12','2019-10-18',
'2019-12-02','2020-01-05','2020-02-25',
'2020-03-15','2020-04-11','2020-07-22'],
'Sales':[200,300,100,250,150,350,150,200,250,
200,300,100,250,150,350,150,200,250]})
#first breakdown the data by Yearly Quarters
df['YQtr'] = pd.PeriodIndex(pd.to_datetime(df.Date), freq='Q')
#next create a column to identify Half Yearly - H1 for Jan-Jun & H2 for Jul-Dec
df.loc[df['YQtr'].astype(str).str[-2:].isin(['Q1','Q2']),'HYear'] = df['YQtr'].astype(str).str[:-2]+'H1'
df.loc[df['YQtr'].astype(str).str[-2:].isin(['Q3','Q4']),'HYear'] = df['YQtr'].astype(str).str[:-2]+'H2'
#Do a cummulative sum on Half Year to get sales by H1 & H2 for each year
df['HYear_cumsum'] = df.groupby('HYear')['Sales'].cumsum()
#Now filter out only the rows with the max value. That's the H1 & H2 sales figure
df1 = df[df.groupby('HYear')['HYear_cumsum'].transform('max')== df['HYear_cumsum']]
print (df)
print (df1)
The output of this will be:
Source Data + Half Year cumulative sum:
Date Sales YQtr HYear HYear_cumsum
0 2018-01-10 200 2018Q1 2018H1 200
1 2018-02-15 300 2018Q1 2018H1 500
2 2018-03-18 100 2018Q1 2018H1 600
3 2018-07-10 250 2018Q3 2018H2 250
4 2018-09-12 150 2018Q3 2018H2 400
5 2018-10-14 350 2018Q4 2018H2 750
6 2018-11-16 150 2018Q4 2018H2 900
7 2018-12-20 200 2018Q4 2018H2 1100
8 2019-01-10 250 2019Q1 2019H1 250
9 2019-04-15 200 2019Q2 2019H1 450
10 2019-06-12 300 2019Q2 2019H1 750
11 2019-10-18 100 2019Q4 2019H2 100
12 2019-12-02 250 2019Q4 2019H2 350
13 2020-01-05 150 2020Q1 2020H1 150
14 2020-02-25 350 2020Q1 2020H1 500
15 2020-03-15 150 2020Q1 2020H1 650
16 2020-04-11 200 2020Q2 2020H1 850
17 2020-07-22 250 2020Q3 2020H2 250
The half year cumulative sum for each half year.
Date Sales YQtr HYear HYear_cumsum
2 2018-03-18 100 2018Q1 2018H1 600
7 2018-12-20 200 2018Q4 2018H2 1100
10 2019-06-12 300 2019Q2 2019H1 750
12 2019-12-02 250 2019Q4 2019H2 350
16 2020-04-11 200 2020Q2 2020H1 850
17 2020-07-22 250 2020Q3 2020H2 250
I will look at your sample data and work on it later tonight.
I have data in a DF (df1) that starts and ends like this below and I'm trying to shift the "0" and "1" columns below so that the date and time is moved back one hour so that the date and time start at hour == 0 not hour == 1.
data starts (df1) -
0 1 2 3 4 5 6 7
0 20160101 100 7.977169 109404.0 20160101 100 4.028678 814.0
1 20160101 200 8.420204 128546.0 20160101 200 4.673662 2152.0
2 20160101 300 9.515370 165931.0 20160101 300 8.019863 8100.0
data ends (df1) -
0 1 2 3 4 5 6 7
8780 20161231 2100 4.198906 11371.0 20161231 2100 0.995571 131.0
8781 20161231 2200 4.787433 19083.0 20161231 2200 1.029809 NaN
8782 20161231 2300 3.987506 9354.0 20161231 2300 0.900942 NaN
8783 20170101 0 3.284947 1815.0 20170101 0 0.899262 NaN
I need the date and time to start shifted back one hour so start time is hour begin not hour end -
0 1 2 3 4 5 6 7
0 20160101 000 7.977169 109404.0 20160101 100 4.028678 814.0
1 20160101 100 8.420204 128546.0 20160101 200 4.673662 2152.0
2 20160101 200 9.515370 165931.0 20160101 300 8.019863 8100.0
and ends like this with the date and time below -
0 1 2 3 4 5 6 7
8780 20161231 2000 4.198906 11371.0 20161231 2100 0.995571 131.0
8781 20161231 2100 4.787433 19083.0 20161231 2200 1.029809 NaN
8782 20161231 2200 3.987506 9354.0 20161231 2300 0.900942 NaN
8783 20161231 2300 3.284947 1815.0 20170101 0 0.899262 NaN
And, i have no real idea of how to accomplish this or how to research it. Thank you,
It would be better to create a proper datetime object then simply remove the hours as a sum which will handle any redaction in days. We can then use dt.strftime to re-create your object (string) columns.
s = pd.to_datetime(
df[0].astype(str) + df[1].astype(str).str.zfill(4), format="%Y%m%d%H%M"
)
0 2016-01-01 01:00:00
1 2016-01-01 02:00:00
2 2016-01-01 03:00:00
8780 2016-12-31 21:00:00
8781 2016-12-31 22:00:00
8782 2016-12-31 23:00:00
8783 2017-01-01 00:00:00
dtype: datetime64[ns]
df[1] = (s - pd.DateOffset(hours=1)).dt.strftime("%H%M").str.lstrip("0").str.zfill(3)
df[0] = (s - pd.DateOffset(hours=1)).dt.strftime("%Y%d%m")
print(df)
0 1 2 3 4 5 6 7
0 20160101 000 7.977169 109404.0 20160101 100 4.028678 814.0
1 20160101 100 8.420204 128546.0 20160101 200 4.673662 2152.0
2 20160101 200 9.515370 165931.0 20160101 300 8.019863 8100.0
8780 20163112 2000 4.198906 11371.0 20161231 2100 0.995571 131.0
8781 20163112 2100 4.787433 19083.0 20161231 2200 1.029809 NaN
8782 20163112 2200 3.987506 9354.0 20161231 2300 0.900942 NaN
8783 20163112 2300 3.284947 1815.0 20170101 0 0.899262 NaN
Use, DataFrame.shift to shift the columns 0, 1, then use Series.bfill on column 0 of df2 to fill the missing values, then use .fillna on column 1 of df2 to fill the NaN values, finally use Dataframe.join to join the dataframe df2 with the dataframe df1:
df2 = df1[['0', '1']].shift()
df2['0'] = df2['0'].bfill()
df2['1'] = df2['1'].fillna('000')
df2 = df2.join(df1.loc[:, '2':])
# print(df2)
0 1 2 3 4 5 6 7
0 20160101 000 7.977169 109404.0 20160101 100 4.028678 814.0
1 20160101 100 8.420204 128546.0 20160101 200 4.673662 2152.0
2 20160101 200 9.515370 165931.0 20160101 300 8.019863 8100.0
...
8780 20160101 300 4.198906 11371.0 20161231 2100 0.995571 131.0
8781 20161231 2100 4.787433 19083.0 20161231 2200 1.029809 NaN
8782 20161231 2200 3.987506 9354.0 20161231 2300 0.900942 NaN
8783 20161231 2300 3.284947 1815.0 20170101 0 0.899262 NaN
You can do subtraction in pandas (considering that the data in your dataframe are not string type)
I will show you an example on how it can be done
import pandas as pd
df = pd.DataFrame()
df['time'] = [0,100,500,2100,2300,0] #creating dataframe
df['time'] = df['time']-100 #This is what you want to do
Now your data will be subtracted by 100.
There is a case when subtracting 0 you will get -100 as time. For that you can do this:
for i in range(len(df['time'])):
if df['time'].iloc[i]== -100:
df['time'].iloc[i]=2300
I have a running balance sheet showing customer balances after inflows and (outflows) by date. It looks something like this:
ID DATE AMOUNT RUNNING AMOUNT
-- ---------------- ------- --------------
10 27/06/2019 14:30 100 100
10 29/06/2019 15:26 -100 0
10 03/07/2019 01:56 83 83
10 04/07/2019 17:53 15 98
10 05/07/2019 15:09 -98 0
10 05/07/2019 15:53 98.98 98.98
10 05/07/2019 19:54 -98.98 0
10 07/07/2019 01:36 90.97 90.97
10 07/07/2019 13:02 -90.97 0
10 07/07/2019 16:32 39.88 39.88
10 08/07/2019 13:41 50 89.88
20 08/01/2019 09:03 890.97 890.97
20 09/01/2019 14:47 -91.09 799.88
20 09/01/2019 14:53 100 899.88
20 09/01/2019 14:59 -399 500.88
20 09/01/2019 18:24 311 811.88
20 09/01/2019 23:25 50 861.88
20 10/01/2019 16:18 -861.88 0
20 12/01/2019 16:46 894.49 894.49
20 25/01/2019 05:40 -871.05 23.44
I have attempted using lag() but I seem not to understand how to use it yet.
SELECT ID, MEDIAN(DIFF) MEDIAN_AGE
FROM
(
SELECT *, DATEDIFF(day, Lag(DATE, 1) OVER(ORDER BY ID), DATE
)AS DIFF
FROM TABLE 1
WHERE RUNNING AMOUNT = 0
)
GROUP BY ID;
The expected result would be:
ID MEDIAN_AGE
-- ----------
10 1
20 2
Please help in writing out the query that gives the expected result.
As already pointed out, you are using syntax that isn't valid for Oracle, including functions that don't exist and column names that aren't allowed.
You seem to want to calculate the number of days between a zero running-amount and the following non-zero running-amount; lead() is probably easier than lag() here, and you can use a case expression to only calculate it when needed:
select id, date_, amount, running_amount,
case when running_amount = 0 then
lead(date_) over (partition by id order by date_) - date_
end as diff
from your_table;
ID DATE_ AMOUNT RUNNING_AMOUNT DIFF
---------- -------------------- ---------- -------------- ----------
10 2019-06-27 14:30:00 100 100
10 2019-06-29 15:26:00 -100 0 3.4375
10 2019-07-03 01:56:00 83 83
10 2019-07-04 17:53:00 15 98
10 2019-07-05 15:09:00 -98 0 .0305555556
10 2019-07-05 15:53:00 98.98 98.98
10 2019-07-05 19:54:00 -98.98 0 1.2375
10 2019-07-07 01:36:00 90.97 90.97
10 2019-07-07 13:02:00 -90.97 0 .145833333
10 2019-07-07 16:32:00 39.88 39.88
10 2019-07-08 13:41:00 50 89.88
20 2019-01-08 09:03:00 890.97 890.97
20 2019-01-09 14:47:00 -91.09 799.88
20 2019-01-09 14:53:00 100 899.88
20 2019-01-09 14:59:00 -399 500.88
20 2019-01-09 18:24:00 311 811.88
20 2019-01-09 23:25:00 50 861.88
20 2019-01-10 16:18:00 -861.88 0 2.01944444
20 2019-01-12 16:46:00 894.49 894.49
20 2019-01-25 05:40:00 -871.05 23.44
Then use the median() function, rounding if desired to get your expected result:
select id, median(diff) as median_age, round(median(diff)) as median_age_rounded
from (
select id, date_, amount, running_amount,
case when running_amount = 0 then
lead(date_) over (partition by id order by date_) - date_
end as diff
from your_table
)
group by id;
ID MEDIAN_AGE MEDIAN_AGE_ROUNDED
---------- ---------- ------------------
10 .691666667 1
20 2.01944444 2
db<>fiddle