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weighted average aggregation on multiple columns of df

I'm trying to calculate a weighted average for multiple columns in a dataframe.
This is a sample of my data
Group
Year
Month
Weight(kg)
Nitrogen
Calcium
A
2020
01
10000
10
70
A
2020
01
15000
4
78
A
2021
05
12000
5
66
A
2021
05
10000
8
54
B
2021
08
14000
10
90
C
2021
08
50000
20
92
C
2021
08
40000
10
95
My desired result would look something like this:
What I've tried:
I can get the correct weighted average values for a single column using this function:
(similar to: link)
def wavg(df, value, weight):
d = df[value]
w = df[weight]
try:
return (d * w).sum() / w.sum()
except ZeroDivisionError:
return d.mean()
I can apply this function to a single column of my df:
df2 = df.groupby(["Group", "year", "month"]).apply(wavg, "Calcium", "Weight(kg").to_frame()
(Don't mind the different values, they are correct for the data in my notebook)
The obvious problem is that this function only works for a single column whilst I have a douzens of columns. I therefore tried a for loop:
column_list=[]
for column in df.columns:
column_list.append(df.groupby(["Group", "year", "month"]).apply(wavg, column, "Weight(kg").to_frame())
It calculates the values correctly, but the columns are placed on top of each other instead of next to eachother. They also miss a usefull column name:
How could I adapt my code to return the desired df?

Change function for working by multiple columns and for avoid removing column for grouping are converting to MultiIndex:
def wavg(x, value, weight):
d = x[value]
w = x[weight]
try:
return (d.mul(w, axis=0)).div(w.sum())
except ZeroDivisionError:
return d.mean()
#columns used for groupby
groups = ["Group", "Year", "Month"]
#processing all another columns
cols = df.columns.difference(groups + ["Weight(kg)"], sort=False)
#create index and processing all columns by variable cols
df1 = (df.set_index(groups)
.groupby(level=groups)
.apply(wavg, cols, "Weight(kg)")
.reset_index())
print (df2)
Group Year Month Calcium Nitrogen
0 A 2020 1 28.000000 4.000000
1 A 2020 1 46.800000 2.400000
2 A 2021 5 36.000000 2.727273
3 A 2021 5 24.545455 3.636364
4 B 2021 8 90.000000 10.000000
5 C 2021 8 51.111111 11.111111
6 C 2021 8 42.222222 4.444444

Try via concat() and reset_index():
df=pd.concat(column_list,axis=1).reset_index()
OR
you can make changes here:
column_list=[]
for column in df.columns:
column_list.append(df.groupby(["Group", "year", "month"]).apply(wavg, column, "Weight(kg").reset_index())
#Finally:
df=pd.concat(column_list,axis=1)

How to get weekly averages for column values and week number for the corresponding year based on daily data records with pandas

I'm still learning python and would like to ask your help with the following problem:
I have a csv file with daily data and I'm looking for a solution to sum it per calendar weeks. So for the mockup data below I have rows stretched over 2 weeks (week 14 (current week) and week 13 (past week)). Now I need to find a way to group rows per calendar week, recognize what year they belong to and calculate week sum and week average. In the file input example there are only two different IDs. However, in the actual data file I expect many more.
input.csv
id date activeMembers
1 2020-03-30 10
2 2020-03-30 1
1 2020-03-29 5
2 2020-03-29 6
1 2020-03-28 0
2 2020-03-28 15
1 2020-03-27 32
2 2020-03-27 10
1 2020-03-26 9
2 2020-03-26 3
1 2020-03-25 0
2 2020-03-25 0
1 2020-03-24 0
2 2020-03-24 65
1 2020-03-23 22
2 2020-03-23 12
...
desired output.csv
id week WeeklyActiveMembersSum WeeklyAverageActiveMembers
1 202014 10 1.4
2 202014 1 0.1
1 202013 68 9.7
2 202013 111 15.9
my goal is to:
import pandas as pd
df = pd.read_csv('path/to/my/input.csv')
Here I'd need to group by 'id' + 'date' column (per calendar week - not sure if this is possible) and create a 'week' column with the week number, then sum 'activeMembers' values for the particular week, save as 'WeeklyActiveMembersSum' column in my output file and finally calculate 'weeklyAverageActiveMembers' for the particular week. I was experimenting with groupby and isin parameters but no luck so far... would I have to go with something similar to this:
df.groupby('id', as_index=False).agg({'date':'max',
'activeMembers':'sum'}
and finally save all as output.csv:
df.to_csv('path/to/my/output.csv', index=False)
Thanks in advance!

It seems I'm getting a different week setting than you do:
# should convert datetime column to datetime type
df['date'] = pd.to_datetime(df['date'])
(df.groupby(['id',df.date.dt.strftime('%Y%W')], sort=False)
.activeMembers.agg([('Sum','sum'),('Average','mean')])
.add_prefix('activeMembers')
.reset_index()
)
Output:
id date activeMembersSum activeMembersAverage
0 1 202013 10 10.000000
1 2 202013 1 1.000000
2 1 202012 68 9.714286
3 2 202012 111 15.857143

Python Pandas: how to take only the earliest date in each group

here's a sample dataset i've created for this question:
data1 = pd.DataFrame([['1','303','3/7/2016'],
['4','404','6/23/2011'],
['7','101','3/7/2016'],
['1','303','5/6/2017']],
columns=["code", "ticket #", "CB date"])
data1['CB date'] = pd.to_datetime(data1['CB date'])
data2 = pd.DataFrame([['1','303','2/5/2016'],
['4','404','6/23/2011'],
['7','101','3/17/2016'],
['1','303','4/6/2017']],
columns=["code", "ticket #", "audit date"])
data2['audit date'] = pd.to_datetime(data2['audit date'])
print(data1)
print(data2)
code ticket # CB date
0 1 303 2016-03-07
1 4 404 2011-06-23
2 7 101 2016-03-07
3 1 303 2017-05-06
code ticket # audit date
0 1 303 2016-02-05
1 4 404 2011-06-23
2 7 101 2016-03-17
3 1 303 2017-04-06
I want to merge the two df's, and make sure that the CB dates are always on or after Audit dates:
data_all = data1.merge(data2, how='inner', on=['code', 'ticket #'])
data_all = data_all[data_all['audit date'] <= data_all['CB date']]
print(data_all)
code ticket # CB date audit date
0 1 303 2016-03-07 2016-02-05
2 1 303 2017-05-06 2016-02-05
3 1 303 2017-05-06 2017-04-06
4 4 404 2011-06-23 2011-06-23
However, I only want to keep the rows with earliest CB date after each audit date. So in above output, row 2 shouldn't be there, because row 1 and row 2 both have same audit date 2016/2/5, but I only want to keep row 1 since the CB date is much closer to 2016/2/5 than row 2 CB date does.
Desired output:
code ticket # CB date audit date
0 1 303 2016-03-07 2016-02-05
3 1 303 2017-05-06 2017-04-06
4 4 404 2011-06-23 2011-06-23
I know in SQL I'd have to gorupby code & ticket # & Audit date first, then order CB date in ascending order, then take the item rank = 1 in each group; but how can I do this in Python/Pandas?
I read other posts here but I am still not getting it, so would really appreciate some advice here.
Few posts I read into include:
Pandas Groupy take only the first N Groups
Pandas: select the first couple of rows in each group

I'd do this with an optional sort_values call and a drop_duplicates call.
data_all.sort_values(data_all.columns.tolist())\
.drop_duplicates(subset=['CB date'], keep='first')
code ticket # CB date audit date
0 1 303 2016-03-07 2016-02-05
2 1 303 2017-05-06 2016-02-05
4 4 404 2011-06-23 2011-06-23
I say the sort_values call is optional here, since your data appears to be sorted already. If it isn't, make sure that's part of your solution.

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'.