Pandas - Subtract min date from max date for each group - python

I want to add a column that is a result of subtraction of min date from max date for each customer_id to this table
Input:
action_date customer_id
2017-08-15 1
2017-08-21 1
2017-08-21 1
2017-09-02 1
2017-08-28 2
2017-09-29 2
2017-10-15 3
2017-10-30 3
2017-12-05 3
And get this table
Output:
action_date customer_id diff
2017-08-15 1 18
2017-08-21 1 18
2017-08-21 1 18
2017-09-02 1 18
2017-08-28 2 32
2017-09-29 2 32
2017-10-15 3 51
2017-10-30 3 51
2017-12-05 3 51
I tried this code, but it puts lots of NaN's
group = df.groupby(by='customer_id')
df['diff'] = (group['action_date'].max() - group['action_date'].min()).dt.days

you can use transform method:
In [23]: df['diff'] = df.groupby('customer_id') \
['action_date'] \
.transform(lambda x: (x.max()-x.min()).days)
In [24]: df
Out[24]:
action_date customer_id diff
0 2017-08-15 1 18
1 2017-08-21 1 18
2 2017-08-21 1 18
3 2017-09-02 1 18
4 2017-08-28 2 32
5 2017-09-29 2 32
6 2017-10-15 3 51
7 2017-10-30 3 51
8 2017-12-05 3 51

Related

How to get Max Value from group of Column Values for each Row [duplicate]

This question already has answers here:
Find the max of two or more columns with pandas
(3 answers)
Closed 2 years ago.
I have a dataframe as below:
F_Time BP BQ BO0 BO1 BO2 BO3 BO4
0 2020-07-10 09:30:00 10780.00 8550 1 28 1 1 2
1 2020-07-10 10:15:00 10788.00 8700 1 5 10 2 1
2 2020-07-10 10:20:00 10780.00 12150 1 1 1 3 76
3 2020-07-10 10:30:00 10770.00 15675 3 2 8 4 94
4 2020-07-10 10:35:00 10760.60 8100 2 1 1 1 29
5 2020-07-10 10:40:00 10750.00 18825 8 9 154 1 1
6 2020-07-10 11:05:00 10725.00 9825 3 4 94 1 1
I want to find the Max Value from group of Column Values (BO0, BO1, BO2, BO3, BO4) for every row:
Expected Output as below:
F_Time BP BQ BO
0 2020-07-10 09:30:00 10780.00 8550 28
1 2020-07-10 10:15:00 10788.00 8700 10
2 2020-07-10 10:20:00 10780.00 12150 76
3 2020-07-10 10:30:00 10770.00 15675 94
4 2020-07-10 10:35:00 10760.60 8100 29
5 2020-07-10 10:40:00 10750.00 18825 154
6 2020-07-10 11:05:00 10725.00 9825 94
Try this.
df["BO"] = df[["BO0", "BO1".....]].max(axis=1)

How to copy paste values from another dataset conditional on a column

I have df1
Id Data Group_Id
0 1 A 1
1 2 B 2
2 3 B 3
...
100 4 A 101
101 5 A 102
...
and df2
Timestamp Group_Id
2012-01-01 00:00:05.523 1
2013-07-01 00:00:10.757 2
2014-01-12 00:00:15.507. 3
...
2016-03-05 00:00:05.743 101
2017-12-24 00:00:10.407 102
...
I want to match the 2 datasets by Group_Id, then copy only date from Timestamp in df2 and paste to a new column in df1 based on corresponding Group_Id, name the column day1.
Then I want to add 6 more columns next to day1, name them day2, ..., day7 with the next six days based on day1. So it looks like:
Id Data Group_Id day1 day2 day3 ... day7
0 1 A 1 2012-01-01 2012-01-02 2012-01-03 ...
1 2 B 2 2013-07-01 2013-07-02 2013-07-03 ...
2 3 B 3 2014-01-12 2014-01-13 2014-01-14 ...
...
100 4 A 101 2016-03-05 2016-03-06 2016-03-07 ...
101 5 A 102 2017-12-24 2017-12-25 2017-12-26 ...
...
Thanks.
First we need merge here
df1=df1.merge(df2,how='left')
s=pd.DataFrame([pd.date_range(x,periods=6,freq ='D') for x in df1.Timestamp],index=df1.index)
s.columns+=1
df1.join(s.add_prefix('Day'))
another approach here, basically just merges the dfs, grabs the date from the timestamp and makes 6 new columns adding a day each time:
import pandas as pd
df1 = pd.read_csv('df1.csv')
df2 = pd.read_csv('df2.csv')
df3 = df1.merge(df2, on='Group_Id')
df3['Timestamp'] = pd.to_datetime(df3['Timestamp']) #only necessary if not already timestamp
df3['day1'] = df3['Timestamp'].dt.date
for i in (range(1,7)):
df3['day'+str(i+1)] = df3['day1'] + pd.Timedelta(i,unit='d')
output:
Id Data Group_Id Timestamp day1 day2 day3 day4 day5 day6 day7
0 1 A 1 2012-01-01 00:00:05.523 2012-01-01 2012-01-02 2012-01-03 2012-01-04 2012-01-05 2012-01-06 2012-01-07
1 2 B 2 2013-07-01 00:00:10.757 2013-07-01 2013-07-02 2013-07-03 2013-07-04 2013-07-05 2013-07-06 2013-07-07
2 3 B 3 2014-01-12 00:00:15.507 2014-01-12 2014-01-13 2014-01-14 2014-01-15 2014-01-16 2014-01-17 2014-01-18
3 4 A 101 2016-03-05 00:00:05.743 2016-03-05 2016-03-06 2016-03-07 2016-03-08 2016-03-09 2016-03-10 2016-03-11
4 5 A 102 2017-12-24 00:00:10.407 2017-12-24 2017-12-25 2017-12-26 2017-12-27 2017-12-28 2017-12-29 2017-12-30
note that I copied your data frame into a csv and only had the 5 entires so the index is not the same as your example (i.e. 100, 101)
you can delete the timestamp col if not needed

Upsample each pandas dateindex row including previous dates within group

My data looks something like this:
ID1 ID2 Date Values
1 1 2018-01-05 75
1 1 2018-01-06 83
1 1 2018-01-07 17
1 1 2018-01-08 15
1 2 2018-02-01 85
1 2 2018-02-02 98
2 1 2018-02-15 54
2 1 2018-02-16 17
2 1 2018-02-17 83
2 1 2018-02-18 94
2 2 2017-12-18 16
2 2 2017-12-19 84
2 2 2017-12-20 47
2 2 2017-12-21 28
2 2 2017-12-22 38
All the operations must be done within groups of ['ID1', 'ID2'].
What I want to do is upsample the dataframe in a way such that I end up with a sub-dataframe for each 'Date' index which includes all previous dates including the current one from it's own ['ID1', 'ID2'] group. The resulting dataframe should look like this:
ID1 ID2 DateGroup Date Values
1 1 2018-01-05 2018-01-05 75
1 1 2018-01-06 2018-01-05 75
1 1 2018-01-06 2018-01-06 83
1 1 2018-01-07 2018-01-05 75
1 1 2018-01-07 2018-01-06 83
1 1 2018-01-07 2018-01-07 17
1 1 2018-01-08 2018-01-05 75
1 1 2018-01-08 2018-01-06 83
1 1 2018-01-08 2018-01-07 17
1 1 2018-01-08 2018-01-08 15
1 2 2018-02-01 2018-02-01 85
1 2 2018-02-02 2018-02-01 85
1 2 2018-02-02 2018-02-02 98
2 1 2018-02-15 2018-02-15 54
2 1 2018-02-16 2018-02-15 54
2 1 2018-02-16 2018-02-16 17
2 1 2018-02-17 2018-02-15 54
2 1 2018-02-17 2018-02-16 17
2 1 2018-02-17 2018-02-17 83
2 1 2018-02-18 2018-02-15 54
2 1 2018-02-18 2018-02-16 17
2 1 2018-02-18 2018-02-17 83
2 1 2018-02-18 2018-02-18 94
2 2 2017-12-18 2017-12-18 16
2 2 2017-12-19 2017-12-18 16
2 2 2017-12-19 2017-12-19 84
2 2 2017-12-20 2017-12-18 16
2 2 2017-12-20 2017-12-19 84
2 2 2017-12-20 2017-12-20 47
2 2 2017-12-21 2017-12-18 16
2 2 2017-12-21 2017-12-19 84
2 2 2017-12-21 2017-12-20 47
2 2 2017-12-21 2017-12-21 28
2 2 2017-12-22 2017-12-18 16
2 2 2017-12-22 2017-12-19 84
2 2 2017-12-22 2017-12-20 47
2 2 2017-12-22 2017-12-21 28
2 2 2017-12-22 2017-12-22 38
The dataframe I'm working with is quite big (~20 million rows), thus I would like to avoid iterating through each row.
Is it possible to use a function or combination of pandas functions like resample/apply/reindex to achieve what I need?
Assuming ID1 and ID2 is your original Index. You should reset the index, set Date as Index, reset the index back to [ID1, ID2]:
df = df.reset_index().set_index(['Date']).resample('d').ffill().reset_index().set_index(['ID1','ID2'])
If your 'Date' field is string, then you should be converting it into datetime before resampling on that field. You can use the below for that:
df['Date'] = pd.to_datetime(df['Date'], format='%d/%m/%Y')

better grouping of label frequency by month from dataframe

I have a dataframe with a date+time and a label, which I want to reshape into date (/month) columns with label frequencies for that month:
date_time label
1 2017-09-26 17:08:00 0
3 2017-10-03 13:27:00 2
4 2017-10-04 19:04:00 0
11 2017-10-11 18:28:00 1
27 2017-10-13 11:22:00 0
28 2017-10-13 21:43:00 0
39 2017-10-16 14:43:00 0
40 2017-10-16 21:39:00 0
65 2017-10-21 21:53:00 2
...
98 2017-11-01 20:08:00 3
99 2017-11-02 12:00:00 3
100 2017-11-02 12:01:00 2
109 2017-11-02 12:03:00 3
110 2017-11-03 22:24:00 0
111 2017-11-04 09:05:00 3
112 2017-11-06 12:36:00 3
113 2017-11-06 12:48:00 2
128 2017-11-07 15:20:00 2
143 2017-11-10 16:36:00 3
144 2017-11-10 20:00:00 0
145 2017-11-10 20:02:00 0
I group the label frequency by month with this line (thanks partially to this post):
df2 = df.groupby([pd.Grouper(key='date_time', freq='M'), 'label'])['label'].count()
which outputs
date_time label
2017-09-30 0 1
2017-10-31 0 6
1 1
2 8
3 2
2017-11-30 0 25
4 2
5 1
2 4
3 11
2017-12-31 0 14
5 3
2 5
3 7
2018-01-31 0 8
4 1
5 1
2 2
3 3
but, as mentioned before, I would like to get the data by month/date columns:
2017-09-30 2017-10-31 2017-11-30 2017-12-31 2018-01-31
0 1 6 25 14 8
1 0 1 0 0 0
2 0 8 4 5 2
3 0 2 11 7 3
4 0 0 2 0 1
5 0 0 1 3 1
currently I can do sort of divide the data with
pd.concat([df2[m] for m in df2.index.levels[0]], axis=1).fillna(0)
but I lose the column names:
label label label label label
0 1.0 6.0 25.0 14.0 8.0
1 0.0 1.0 0.0 0.0 0.0
2 0.0 8.0 4.0 5.0 2.0
3 0.0 2.0 11.0 7.0 3.0
4 0.0 0.0 2.0 0.0 1.0
5 0.0 0.0 1.0 3.0 1.0
So I have to do a longer version where I generate a series, rename it, concatenate and then fill in the blanks:
m_list = []
for m in df2.index.levels[0]:
m_labels = df2[m]
m_labels = m_labels.rename(m)
m_list.append(m_labels)
pd.concat(m_list, axis=1).fillna(0)
resulting in
2017-09-30 2017-10-31 2017-11-30 2017-12-31 2018-01-31
0 1.0 6.0 25.0 14.0 8.0
1 0.0 1.0 0.0 0.0 0.0
2 0.0 8.0 4.0 5.0 2.0
3 0.0 2.0 11.0 7.0 3.0
4 0.0 0.0 2.0 0.0 1.0
5 0.0 0.0 1.0 3.0 1.0
Is there a shorter/more elegant way to get to this last datagrame from my original one?
You just need unstack here
df.groupby([pd.Grouper(key='date_time', freq='M'), 'label'])['label'].count().unstack(0,fill_value=0)
Out[235]:
date_time 2017-09-30 2017-10-31 2017-11-30
label
0 1 5 3
1 0 1 0
2 0 2 3
3 0 0 6
Base on your groupby output
s.unstack(0,fill_value=0)
Out[240]:
date_time 2017-09-30 2017-10-31 2017-11-30 2017-12-31 2018-01-31
label
0 1 6 25 14 8
1 0 1 0 0 0
2 0 8 4 5 2
3 0 2 11 7 3
4 0 0 2 0 1
5 0 0 1 3 1

Aggregate to 15min based timestamp to hour and find sum, avg and max for multiple columns in pandas

I have a dataframe with period_start_time by every 15 minutes and now I need to aggregate to 1 hour and calculate sum and avg for almost every column in dataframe (it has about 20 columns) and
PERIOD_START_TIME ID val1 val2
06.21.2017 22:15:00 12 3 0
06.21.2017 22:30:00 12 5 6
06.21.2017 22:45:00 12 0 3
06.21.2017 23:00:00 12 5 2
...
06.21.2017 22:15:00 15 9 2
06.21.2017 22:30:00 15 0 2
06.21.2017 22:45:00 15 1 5
06.21.2017 23:00:00 15 0 1
...
Desired output:
PERIOD_START_TIME ID val1(avg) val1(sum) val1(max) ...
06.21.2017 22:00:00 12 3.25 13 5
...
06.21.2017 23:00:00 15 2.25 10 9 ...
And for columns val2 too, and for every other column in dataframe.
I have no idea how to group by period start time for every hour, not for the whole day, no idea how to start.
I believe you need Series.dt.floor for Hours and then aggregate by agg:
df = df.groupby([df['PERIOD_START_TIME'].dt.floor('H'),'ID']).agg(['mean','sum', 'max'])
#for columns from MultiIndex
df.columns = df.columns.map('_'.join)
print (df)
val1_mean val1_sum val1_max val2_mean val2_sum \
PERIOD_START_TIME ID
2017-06-21 22:00:00 12 2.666667 8 5 3 9
15 3.333333 10 9 3 9
2017-06-21 23:00:00 12 5.000000 5 5 2 2
15 0.000000 0 0 1 1
val2_max
PERIOD_START_TIME ID
2017-06-21 22:00:00 12 6
15 5
2017-06-21 23:00:00 12 2
15 1
df = df.reset_index()
print (df)
PERIOD_START_TIME ID val1_mean val1_sum val1_max val2_mean val2_sum \
0 2017-06-21 22:00 12 2.666667 8 5 3 9
1 2017-06-21 22:00 15 3.333333 10 9 3 9
2 2017-06-21 23:00 12 5.000000 5 5 2 2
3 2017-06-21 23:00 15 0.000000 0 0 1 1
val2_max
0 6
1 5
2 2
3 1
Very similarly you can convert PERIOD_START_TIME to a pandas Period.
df['PERIOD_START_TIME'] = df['PERIOD_START_TIME'].dt.to_period('H')
df.groupby(['PERIOD_START_TIME', 'ID']).agg(['max', 'min', 'mean']).reset_index()

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