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Pandas: Groupby and sum customer profit, for every 6 months, starting from users first transaction

I have a dataset like this:
Customer ID
Date
Profit
1
4/13/2018
10.00
1
4/26/2018
13.27
1
10/23/2018
15.00
2
1/1/2017
7.39
2
7/5/2017
9.99
2
7/7/2017
10.01
3
5/4/2019
30.30
I'd like to groupby and sum profit, for every 6 months, starting at each users first transaction.
The output ideally should look like this:
Customer ID
Date
Profit
1
4/13/2018
23.27
1
10/13/2018
15.00
2
1/1/2017
7.39
2
7/1/2017
20.00
3
5/4/2019
30.30
The closest I've seem to get on this problem is by using:
df.groupby(['Customer ID',pd.Grouper(key='Date', freq='6M', closed='left')])['Profit'].sum().reset_index()
But, that doesn't seem to sum starting on a users first transaction day.
If the changing of dates is not possible (ex. customer 2 date is 7/1/2017 and not 7/5/2017), then at least summing the profit so that its based on each users own 6 month purchase journey would be extremely helpful. Thank you!

I can get you the first of the month until you find a more perfect solution.
df["Date"] = pd.to_datetime(df["Date"], format="%m/%d/%Y")
df = (
df
.set_index("Date")
.groupby(["Customer ID"])
.Profit
.resample("6MS")
.sum()
.reset_index(name="Profit")
)
print(df)
Customer ID Date Profit
0 1 2018-04-01 23.27
1 1 2018-10-01 15.00
2 2 2017-01-01 7.39
3 2 2017-07-01 20.00
4 3 2019-05-01 30.30

Multiple group counts within data base

I have been presented with a very small dataset that has the date of each time a user logs into a system, I have to use this data set to create a table where I can show for each log-in the cumulative monthly counts of logs and the overall cumulative counts of logs, this is the data set I have:
date
user
1/01/2022
Mark
2/01/2022
Mark
3/02/2022
Mark
4/02/2022
Mark
5/03/2022
Mark
6/03/2022
Mark
7/03/2022
Mark
8/03/2022
Mark
9/03/2022
Mark
and this is my desired output:
row
date
user
monthly_track
acum_track
1
1/01/2022
Mark
1
1
2
2/01/2022
Mark
2
2
3
3/02/2022
Mark
1
3
4
4/02/2022
Mark
2
4
5
5/03/2022
Mark
1
5
6
6/03/2022
Mark
2
6
7
7/03/2022
Mark
3
7
8
8/03/2022
Mark
4
8
9
9/03/2022
Mark
5
9
Why? Let's look at the row number 5. This is the first time the user Mark has logged into the system during the month 3 (March) but it is the 5th overall login in the data set (for the purpose of learning there will only be one year (2022).
I have no idea as to how to get the monthly and overall count together. I can groupby user and sort by date to count how many times in total a user has logged in, but I know that in order to achive my desired output I will have to group by date and user and then make counts based on month but I will have to somehow group the data by user (only) to get the overall count and I dont think I could group twice the data.

First you need to convert date to actual datetime values with to_datetime. The rest is simple with groupby and cumcount:
df['date'] = pd.to_datetime(df['date'], format='%d/%m/%Y')
df['monthly_count'] = df.groupby([df['user'], df['date'].dt.year, df['date'].dt.month]).cumcount() + 1
df['acum_count'] = df.groupby('user').cumcount() + 1
Output:
>>> df
date user monthly_count acum_count
0 2022-01-01 Mark 1 1
1 2022-01-02 Mark 2 2
2 2022-02-03 Mark 1 3
3 2022-02-04 Mark 2 4
4 2022-03-05 Mark 1 5
5 2022-03-06 Mark 2 6
6 2022-03-07 Mark 3 7
7 2022-03-08 Mark 4 8
8 2022-03-09 Mark 5 9

Compute operations within subgroups in pandas

I have a table that has multiple subgroups. For example, person A has a total of three visits and person B has a total of two visits. I also have the time of each visit:
id visit time_of_visit
A 1 2002-01-15
A 2 2003-01-15
A 3 2003-02-15
B 1 1996-08-09
B 2 1998-08-09
I want to compute how long apart each visit is in terms of years for each person. So I want something like this:
id visit time_of_visit difference_in_time
A 1 2002-01-15 na
A 2 2003-01-15 1
A 3 2003-02-15 0.0833
B 1 1996-08-09 na
B 2 1998-08-09 2
Any ideas how to do this in python pandas? Thanks!

groupby.diff on a datetime column will give you
df['time_of_visit'] = pd.to_datetime(df['time_of_visit'])
df.groupby('id')['time_of_visit'].diff()
Out:
0 NaT
1 365 days
2 31 days
3 NaT
4 730 days
Name: time_of_visit, dtype: timedelta64[ns]
However, timedeltas cannot give you years as it is not a standard measure. You can always convert by your own rules of course (for example divide by 365).
df.groupby('id')['time_of_visit'].diff().dt.days / 365
Out:
0 NaN
1 1.000000
2 0.084932
3 NaN
4 2.000000
Name: time_of_visit, dtype: float64

Python selecting row from second dataframe based on complex criteria

I have two dataframes, one with some purchasing data, and one with a weekly calendar, e.g.
df1:
purchased_at product_id cost
01-01-2017 1 £10
01-01-2017 2 £8
09-01-2017 1 £10
18-01-2017 3 £12
df2:
week_no week_start week_end
1 31-12-2016 06-01-2017
2 07-01-2017 13-01-2017
3 14-01-2017 20-01-2017
I want to use data from the two to add a 'week_no' column to df1, which is selected from df2 based on where the 'purchased_at' date in df1 falls between the 'week_start' and 'week_end' dates in df2, i.e.
df1:
purchased_at product_id cost week_no
01-01-2017 1 £10 1
01-01-2017 2 £8 1
09-01-2017 1 £10 2
18-01-2017 3 £12 3
I've searched but I've not been able to find an example where the data is being pulled from a second dataframe using comparisons between the two, and I've been unable to correctly apply any examples I've found, e.g.
df1.loc[(df1['purchased_at'] < df2['week_end']) &
(df1['purchased_at'] > df2['week_start']), df2['week_no']
was unsuccessful, with the ValueError 'can only compare identically-labeled Series objects'
Could anyone help with this problem, or I'm open to suggestions if there is a better way to achieve the same outcome.
edit to add further detail of df1
df1 full dataframe headers
purchased_at purchase_id product_id product_name transaction_id account_number cost
01-01-2017 1 1 A 1 AA001 £10
01-01-2017 2 2 B 1 AA001 £8
02-01-2017 3 1 A 2 AA008 £10
03-01-2017 4 3 C 3 AB040 £12
...
09-01-2017 12 1 A 10 AB102 £10
09-01-2017 13 2 B 11 AB102 £8
...
18-01-2017 20 3 C 15 AA001 £12
So the purchase_id increases incrementally with each row, the product_id and product_name have a 1:1 relationship, the transaction_id also increases incrementally, but there can be multiple purchases within a transaction.

If your dataframes are to big you can use this trick.
Do a full cartisian product join of all records to all records:
df_out = pd.merge(df1.assign(key=1),df2.assign(key=1),on='key')
Next filter out those records that do not match criteria in this case, where purchased_at is not between week_start and week_end
(df_out.query('week_start < purchased_at < week_end')
.drop(['key','week_start','week_end'], axis=1))
Output:
purchased_at product_id cost week_no
0 2017-01-01 1 £10 1
3 2017-01-01 2 £8 1
7 2017-01-09 1 £10 2
11 2017-01-18 3 £12 3
If you do have large dataframes then you can use this numpy method as proposed by PiRSquared.
a = df1.purchased_at.values
bh = df2.week_end.values
bl = df2.week_start.values
i, j = np.where((a[:, None] >= bl) & (a[:, None] <= bh))
pd.DataFrame(
np.column_stack([df1.values[i], df2.values[j]]),
columns=df1.columns.append(df2.columns)
).drop(['week_start','week_end'],axis=1)
Output:
purchased_at product_id cost week_no
0 2017-01-01 00:00:00 1 £10 1
1 2017-01-01 00:00:00 2 £8 1
2 2017-01-09 00:00:00 1 £10 2
3 2017-01-18 00:00:00 3 £12 3

You could just use time.strftime() to extract the week number from the date. If you want to keep counting the weeks upwards, you need to define a "zero year" as the start of your time-series and offset the week_no accordingly:
import pandas as pd
data = {'purchased_at': ['01-01-2017', '01-01-2017', '09-01-2017', '18-01-2017'], 'product_id': [1,2,1,3], 'cost':['£10', '£8', '£10', '£12']}
df = pd.DataFrame(data, columns=['purchased_at', 'product_id', 'cost'])
def getWeekNo(date, year0):
datetime = pd.to_datetime(date, dayfirst=True)
year = int(datetime.strftime('%Y'))
weekNo = int(datetime.strftime('%U'))
return weekNo + 52*(year-year0)
df['week_no'] = df.purchased_at.apply(lambda x: getWeekNo(x, 2017))
Here, I use pd.to_dateime() to convert the datestring from df into a datetime-object. strftime('%Y') returns the year and strftime('%U') the week (with the first week of a year starting on it's first Sunday. If weeks should start on Monday, use '%W' instead).
This way, you don't need to maintain a seperate DataFrame only for week numbers.

Time series: Mean per hour per day per Id number

I am a somewhat beginner programmer and learning python (+pandas) and hope I can explain this well enough. I have a large time series pd dataframe of over 3 million rows and initially 12 columns spanning a number of years. This covers people taking a ticket from different locations denoted by Id numbers(350 of them). Each row is one instance (one ticket taken).
I have searched many questions like counting records per hour per day and getting average per hour over several years. However, I run into the trouble of including the 'Id' variable.
I'm looking to get the mean value of people taking a ticket for each hour, for each day of the week (mon-fri) and per station.
I have the following, setting datetime to index:
Id Start_date Count Day_name_no
149 2011-12-31 21:30:00 1 5
150 2011-12-31 20:51:00 1 0
259 2011-12-31 20:48:00 1 1
3015 2011-12-31 19:38:00 1 4
28 2011-12-31 19:37:00 1 4
Using groupby and Start_date.index.hour, I cant seem to include the 'Id'.
My alternative approach is to split the hour out of the date and have the following:
Id Count Day_name_no Trip_hour
149 1 2 5
150 1 4 10
153 1 2 15
1867 1 4 11
2387 1 2 7
I then get the count first with:
Count_Item = TestFreq.groupby([TestFreq['Id'], TestFreq['Day_name_no'], TestFreq['Hour']]).count().reset_index()
Id Day_name_no Trip_hour Count
1 0 7 24
1 0 8 48
1 0 9 31
1 0 10 28
1 0 11 26
1 0 12 25
Then use groupby and mean:
Mean_Count = Count_Item.groupby(Count_Item['Id'], Count_Item['Day_name_no'], Count_Item['Hour']).mean().reset_index()
However, this does not give the desired result as the mean values are incorrect.
I hope I have explained this issue in a clear way. I looking for the mean per hour per day per Id as I plan to do clustering to separate my dataset into groups before applying a predictive model on these groups.
Any help would be grateful and if possible an explanation of what I am doing wrong either code wise or my approach.
Thanks in advance.
I have edited this to try make it a little clearer. Writing a question with a lack of sleep is probably not advisable.
A toy dataset that i start with:
Date Id Dow Hour Count
12/12/2014 1234 0 9 1
12/12/2014 1234 0 9 1
12/12/2014 1234 0 9 1
12/12/2014 1234 0 9 1
12/12/2014 1234 0 9 1
19/12/2014 1234 0 9 1
19/12/2014 1234 0 9 1
19/12/2014 1234 0 9 1
26/12/2014 1234 0 10 1
27/12/2014 1234 1 11 1
27/12/2014 1234 1 11 1
27/12/2014 1234 1 11 1
27/12/2014 1234 1 11 1
04/01/2015 1234 1 11 1
I now realise I would have to use the date first and get something like:
Date Id Dow Hour Count
12/12/2014 1234 0 9 5
19/12/2014 1234 0 9 3
26/12/2014 1234 0 10 1
27/12/2014 1234 1 11 4
04/01/2015 1234 1 11 1
And then calculate the mean per Id, per Dow, per hour. And want to get this:
Id Dow Hour Mean
1234 0 9 4
1234 0 10 1
1234 1 11 2.5
I hope this makes it a bit clearer. My real dataset spans 3 years with 3 million rows, contains 350 Id numbers.

Your question is not very clear, but I hope this helps:
df.reset_index(inplace=True)
# helper columns with date, hour and dow
df['date'] = df['Start_date'].dt.date
df['hour'] = df['Start_date'].dt.hour
df['dow'] = df['Start_date'].dt.dayofweek
# sum of counts for all combinations
df = df.groupby(['Id', 'date', 'dow', 'hour']).sum()
# take the mean over all dates
df = df.reset_index().groupby(['Id', 'dow', 'hour']).mean()

You can use the groupby function using the 'Id' column and then use the resample function with how='sum'.