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How to calculate time difference between two succesive rows with groupby?

I have the following dataset of students taking multiple SAT exams:
df = pd.DataFrame({'student': 'A A A A A B B B C'.split(),
'exam_date':[datetime.datetime(2013,4,1),datetime.datetime(2013,6,1),
datetime.datetime(2013,7,1),datetime.datetime(2013,10,2),
datetime.datetime(2014,1,1),datetime.datetime(2013,11,2),
datetime.datetime(2014,2,2),datetime.datetime(2014,5,2),
datetime.datetime(2014,5,2)]})
print(df)
student exam_date
0 A 2013-04-01
1 A 2013-06-01
2 A 2013-07-01
3 A 2013-10-02
4 A 2014-01-01
5 B 2013-11-02
6 B 2014-02-02
7 B 2014-05-02
8 C 2014-05-02
I want to create a new column diff with the difference of two successive exam dates for each individual student, and then filter the value with a particular threshold, i.e. 75 days. If the student doesn't have two successive dates, we need to drop that student.
I am trying the following script to create the new column:
df['exam_date'] = df.groupby('student')['exam_date'].apply(lambda x: x.sort_values())
df['diff'] = df.groupby('student')['exam_date'].diff() / np.timedelta64(1, 'D')
print(df)
student exam_date diff
0 A 2013-04-01 NaN
1 A 2013-06-01 61.0
2 A 2013-07-01 30.0
3 A 2013-10-02 93.0
4 A 2014-01-01 91.0
5 B 2013-11-02 NaN
6 B 2014-02-02 92.0
7 B 2014-05-02 89.0
8 C 2014-05-02 NaN
Then I'm using query to filter the value and get the output:
df_new = df.query('diff <= 75')
print(df_new)
student exam_date diff
1 A 2013-06-01 61.0
2 A 2013-07-01 30.0
This is correctly selecting the student A and removing the students B and C. However, I'm missing the earliest date for the student A.
Though using df[df['student'].isin(studentList)]I'm getting the desired result, but it's too much of work.
Is there any better way of getting the desired output, maybe using diff() and le()? Any suggestions would be appreciated. Thanks!

What you want is filtering students, but you are filtering exam records.
After you got df_new, just find the students set, and use that to select df:
df[df.student.isin(df_new.student.unique())]
and you'll get:
student exam_date diff
0 A 2013-04-01 NaN
1 A 2013-06-01 61.0
2 A 2013-07-01 30.0
3 A 2013-10-02 93.0
4 A 2014-01-01 91.0

How to create a new column with the last value of the previous year

I have this data frame
import pandas as pd
df = pd.DataFrame({'COTA':['A','A','A','A','A','B','B','B','B'],
'Date':['14/10/2021','19/10/2020','29/10/2019','30/09/2021','20/09/2020','20/10/2021','29/10/2020','15/10/2019','10/09/2020'],
'Mark':[1,2,3,4,5,1,2,3,3]
})
print(df)
based on this data frame I wanted the MARK from the previous year, I managed to acquire the maximum COTA but I wanted the last one, I used .max() and I thought I could get it with .last() but it didn't work.
follow the example of my code.
df['Date'] = pd.to_datetime(df['Date'])
df['LastYear'] = df['Date'] - pd.offsets.YearEnd(0)
s1 = df.groupby(['Found', 'LastYear'])['Mark'].max()
s2 = s1.rename(index=lambda x: x + pd.offsets.DateOffset(years=1), level=1)
df = df.join(s2.rename('Max_MarkLastYear'), on=['Found', 'LastYear'])
print (df)
Found Date Mark LastYear Max_MarkLastYear
0 A 2021-10-14 1 2021-12-31 5.0
1 A 2020-10-19 2 2020-12-31 3.0
2 A 2019-10-29 3 2019-12-31 NaN
3 A 2021-09-30 4 2021-12-31 5.0
4 A 2020-09-20 5 2020-12-31 3.0
5 B 2021-10-20 1 2021-12-31 3.0
6 B 2020-10-29 2 2020-12-31 3.0
7 B 2019-10-15 3 2019-12-31 NaN
8 B 2020-10-09 3 2020-12-31 3.0
How do I create a new column with the last value of the previous year

Dynamic Dates difference calculation Pandas

customer_id Order_date
1 2015-01-16
1 2015-01-19
2 2014-12-21
2 2015-01-10
1 2015-01-10
3 2018-01-18
3 2017-03-04
4 2019-11-05
4 2010-01-01
3 2019-02-03
Lets say I have data like this
Basically for an ecommerce firm some people buy regularly, some buy once every year, some buy monthly once etc. I need to find the difference between frequency of each transaction for each customer.
This will be a dynamic list, since some people will have transacted thousand times, some would have transacted once, some ten times etc. Any ideas on how to achieve this.
Output needed:
customer_id Order_date_Difference_in_days
1 6,3 #Difference b/w first 2 dates 2015-01-10 and 2015-01-16
#is 6 days and diff b/w next 2 consecutive dates is
#2015-01-16 and 2015-01-19 is #3 days
2 20
3 320,381
4 3596
Basically these are the differences between dates after sorting them first for each customer id

You can also use the below for the current output:
m=(df.assign(Diff=df.sort_values(['customer_id','Order_date'])
.groupby('customer_id')['Order_date'].diff().dt.days).dropna())
m=m.assign(Diff=m['Diff'].astype(str)).groupby('customer_id')['Diff'].agg(','.join)
customer_id
1 6.0,3.0
2 20.0
3 320.0,381.0
4 3595.0
Name: Diff, dtype: object

First we need to sort the data by customer id and the order date
ensure your datetime is a proper date time call df['Order_date'] = pd.to_datetime(df['Order_date'])
df.sort_values(['customer_id','Order_date'],inplace=True)
df["days"] = df.groupby("customer_id")["Order_date"].apply(
lambda x: (x - x.shift()) / np.timedelta64(1, "D")
)
print(df)
customer_id Order_date days
4 1 2015-01-10 NaN
0 1 2015-01-16 6.0
1 1 2015-01-19 3.0
2 2 2014-12-21 NaN
3 2 2015-01-10 20.0
6 3 2017-03-04 NaN
5 3 2018-01-18 320.0
9 3 2019-02-03 381.0
8 4 2010-01-01 NaN
7 4 2019-11-05 3595.0
then you can do a simple agg but you'll need to conver the value into a string.
df.dropna().groupby("customer_id")["days"].agg(
lambda x: ",".join(x.astype(str))
).to_frame()
days
customer_id
1 6.0,3.0
2 20.0
3 320.0,381.0
4 3595.0

Year to date average in dataframe

I have a dataframe that I am trying to calculate the year-to-date average for my value columns. Below is a sample dataframe.
date name values values2
0 2019-01-01 a 1 1
1 2019-02-01 a 3 3
2 2019-03-01 a 2 2
3 2019-04-01 a 6 2
I want to create new columns (values_ytd & values2_ytd) that will average the values from January to the latest period within the same year (April in sample data). I will need to group the data by year & name when calculating the averages. I am looking for an output similar to this.
date name values values2 values2_ytd values_ytd
0 2019-01-01 a 1 1 1 1
1 2019-02-01 a 3 3 2 2
2 2019-03-01 a 2 2 2 2
3 2019-04-01 a 6 2 2 3
I have tried unsuccesfully to using expanding().mean(), but most likely I was doing it wrong. My main dataframe has numerous name categories and many more columns. Here is the code I was attempting to use
df1.groupby([df1['name'], df1['date'].dt.year], as_index=False).expanding().mean().loc[:, 'values':'values2'].add_suffix('_ytd').reset_index(drop=True,level=0)
but am receiving the following error.
NotImplementedError: ops for Expanding for this dtype datetime64[ns] are not implemented
Note: This code below works perfectly when substituting cumsum() for .expanding().mean()to create a year-to-date sum of the values, but I cant figure it out for averages
df1.groupby([df1['name'], df1['date'].dt.year], as_index=False).cumsum().loc[:, 'values':'values2'].add_suffix('_ytd').reset_index(drop=True,level=0)
Any help is greatly appreciated.

Try this:
df['date'] = pd.to_datetime(df['date'])
df.set_index('date', inplace=True)
df[['values2_ytd', 'values_ytd']] = df.groupby([df.index.year, 'name'])['values','values2'].expanding().mean().reset_index(level=[0,1], drop=True)
df
name values values2 values2_ytd values_ytd
date
2019-01-01 a 1 1 1.0 1.0
2019-02-01 a 3 3 2.0 2.0
2019-03-01 a 2 2 2.0 2.0
2019-04-01 a 6 2 3.0 2.0
Example using multiple names and years:
date name values values2
0 2019-01-01 a 1 1
1 2019-02-01 a 3 3
2 2019-03-01 a 2 2
3 2019-04-01 a 6 2
4 2019-01-01 b 1 4
5 2019-02-01 b 3 4
6 2020-01-01 a 1 1
7 2020-02-01 a 3 3
8 2020-03-01 a 2 2
9 2020-04-01 a 6 2
Output:
name values values2 values2_ytd values_ytd
date
2019-01-01 a 1 1 1.0 1.0
2019-02-01 a 3 3 2.0 2.0
2019-03-01 a 2 2 2.0 2.0
2019-04-01 a 6 2 3.0 2.0
2019-01-01 b 1 4 1.0 4.0
2019-02-01 b 3 4 2.0 4.0
2020-01-01 a 1 1 1.0 1.0
2020-02-01 a 3 3 2.0 2.0
2020-03-01 a 2 2 2.0 2.0
2020-04-01 a 6 2 3.0 2.0

You should set date column as index: df.set_index('date', inplace=True) and then use df.resample('AS').groupby('name').mean()

Elegant resample for groups in Pandas

For a given pandas data frame called full_df which looks like
index id timestamp data
------- ---- ------------ ------
1 1 2017-01-01 10.0
2 1 2017-02-01 11.0
3 1 2017-04-01 13.0
4 2 2017-02-01 1.0
5 2 2017-03-01 2.0
6 2 2017-05-01 9.0
The start and end dates (and the time delta between start and end) are varying.
But I need a id wise resampled version (added rows marked with *)
index id timestamp data
------- ---- ------------ ------ ----
1 1 2017-01-01 10.0
2 1 2017-02-01 11.0
3 1 2017-03-01 NaN *
4 1 2017-04-01 13.0
5 2 2017-02-01 1.0
6 2 2017-03-01 2.0
7 2 2017-04-01 NaN *
8 2 2017-05-01 9.0
Because the dataset is very large I was wondering if there is more efficient way of doing so than
Do full_df.groupby('id')
Do for each group df
df.index = pd.DatetimeIndex(df['timestamp'])
all_days = pd.date_range(df.index.min(), df.index.max(), freq='MS')
df = df.reindex(all_days)
Combine all groups again with a new index
That's time consuming and not very elegant. Any ideas?

Using resample
In [1175]: (df.set_index('timestamp').groupby('id').resample('MS').asfreq()
.drop(['id', 'index'], 1).reset_index())
Out[1175]:
id timestamp data
0 1 2017-01-01 10.0
1 1 2017-02-01 11.0
2 1 2017-03-01 NaN
3 1 2017-04-01 13.0
4 2 2017-02-01 1.0
5 2 2017-03-01 2.0
6 2 2017-04-01 NaN
7 2 2017-05-01 9.0
Details
In [1176]: df
Out[1176]:
index id timestamp data
0 1 1 2017-01-01 10.0
1 2 1 2017-02-01 11.0
2 3 1 2017-04-01 13.0
3 4 2 2017-02-01 1.0
4 5 2 2017-03-01 2.0
5 6 2 2017-05-01 9.0
In [1177]: df.dtypes
Out[1177]:
index int64
id int64
timestamp datetime64[ns]
data float64
dtype: object

Edit to add: this way does the min/max of dates for full_df, not df. If there wide variation in start/end dates between IDs this will unfortunately inflate the dataframe and #JohnGalt method is better. Nevertheless I'll leave this here as an alternate approach as it ought to be faster than groupby/resample for cases where it is appropriate.
I think the most efficient approach is likely going to be with stack/unstack or melt/pivot.
You could do something like this, for example:
full_df.set_index(['timestamp','id']).unstack('id').stack('id',dropna=False)
index data
timestamp id
2017-01-01 1 1.0 10.0
2 NaN NaN
2017-02-01 1 2.0 11.0
2 4.0 1.0
2017-03-01 1 NaN NaN
2 5.0 2.0
2017-04-01 1 3.0 13.0
2 NaN NaN
2017-05-01 1 NaN NaN
2 6.0 9.0
Just add reset_index().set_index('id') if you want it to display more like how you have it above. Note in particular the use of dropna=False with stack which preserves the NaN placeholders. Without that, the stack/unstack method just leaves you back where you started.
This method automatically includes the min & max dates, and all dates present for at least one timestamp. If there are interior timestamps missing for everyone, then you need to add a resample like this:
full_df.set_index(['timestamp','id']).unstack('id')\
.resample('MS').mean()\
.stack('id',dropna=False)