Same question as here: group by pandas dataframe and select latest in each group, except instead of latest date, would like to get next upcoming date for each group.
So given a dataframe sorted by date:
id product date
0 220 6647 2020-09-01
1 220 6647 2020-10-03
2 220 6647 2020-12-16
3 826 3380 2020-11-11
4 826 3380 2020-12-09
5 826 3380 2021-05-19
6 901 4555 2020-09-01
7 901 4555 2020-12-01
8 901 4555 2021-11-01
Using todays date (2020-12-01) to determine the next upcoming date, grouping by id or product and selecting the the next upcoming date should give:
id product date
2 220 6647 2020-12-16
5 826 3380 2020-12-09
8 901 4555 2021-11-01
Filter the dates first, then drop duplicates:
df[df['date']>'2020-12-01'].sort_values(['id','date']).drop_duplicates('id')
Output:
id product date
2 220 6647 2020-12-16
4 826 3380 2020-12-09
8 901 4555 2021-11-01
Related
I would like to aggregate my data on week, using pandas grouper, where the week ends at the exact same day of my last date, and not the end of the week.
This is the code I wrote:
fp.groupby(pd.Grouper(key='date',freq='w')).collectionName.nunique().tail(10)
And these are the results:
date
2021-10-03 644
2021-10-10 698
2021-10-17 756
2021-10-24 839
2021-10-31 883
2021-11-07 905
2021-11-14 961
2021-11-21 1028
2021-11-28 990
2021-12-05 726
Freq: W-SUN, Name: collectionName, dtype: int64
The last date I have is 2021-12-02, so I would like that to be the last day of the week aggregate, and it goes back every 7 days, to the end (in this case beginning of the dataset).
I need help with this.
Use pd.DataFrame.resample with rule='1w', on='date' and origin='end_day'
This assumes you can find the last date prior to grouping. See references here: https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html#offset-aliases and here: https://www.programiz.com/python-programming/datetime/strftime
df = pd.DataFrame({'date': pd.date_range(start='2021-01-01 01:04:16', periods=250), 'val':range(1,251)})
date val
0 2021-01-01 01:04:16 1
1 2021-01-02 01:04:16 2
2 2021-01-03 01:04:16 3
3 2021-01-04 01:04:16 4
4 2021-01-05 01:04:16 5
.. ... ...
245 2021-09-03 01:04:16 246
246 2021-09-04 01:04:16 247
247 2021-09-05 01:04:16 248
248 2021-09-06 01:04:16 249
249 2021-09-07 01:04:16 250
[250 rows x 2 columns]
# locate last date and get day of week in correct format
anchor = df['date'].iat[-1].strftime("%a")
df.groupby(pd.Grouper(key='date',freq='w-' + anchor)).nunique().tail(5)
# week ends on the same day as the original dataset
val
date
2021-08-10 7
2021-08-17 7
2021-08-24 7
2021-08-31 7
2021-09-07 7
How to keep the last group within a group using Pandas?
For example, given the following dataset:
id product date
0 220 6647 2014-09-01
1 220 6647 2014-09-03
2 220 6647 2014-10-16
3 826 6647 2014-11-11
4 826 6647 2014-12-09
5 826 6647 2015-05-19
6 901 4555 2014-09-01
7 901 4555 2014-10-05
8 901 4555 2014-11-01
9 401 4555 2015-01-05
10 401 4555 2015-02-01
how to get the last id group from each of the product group succinctly?
id product date
3 826 6647 2014-11-11
4 826 6647 2014-12-09
5 826 6647 2015-05-19
9 401 4555 2015-01-05
10 401 4555 2015-02-01
My question is based on this thread, where we group values of a pandas dataframe and select the latest (by date) from each group:
id product date
0 220 6647 2014-09-01
1 220 6647 2014-09-03
2 220 6647 2014-10-16
3 826 3380 2014-11-11
4 826 3380 2014-12-09
5 826 3380 2015-05-19
6 901 4555 2014-09-01
7 901 4555 2014-10-05
8 901 4555 2014-11-01
using the following
df.loc[df.groupby('id').date.idxmax()]
Say, however, that I want to include the condition that I only want to select the latest (by date) from each group within +/- 5 days. I.e., after grouping I want to find the latest within the following groups:
0 220 6647 2014-09-01 #because only these two are within +/- 5 days of each other
1 220 6647 2014-09-03
2 220 6647 2014-10-16 #spaced more than 5 days apart the above two records
3 826 3380 2014-11-11
.....
which yields
id product date
1 220 6647 2014-09-03
2 220 6647 2014-10-16
3 826 3380 2014-11-11
4 826 3380 2014-12-09
5 826 3380 2015-05-19
5 826 3380 2015-05-19
6 901 4555 2014-09-01
7 901 4555 2014-10-05
8 901 4555 2014-11-01
Dataset with price:
id product date price
0 220 6647 2014-09-01 100 #group 1
1 220 6647 2014-09-03 120 #group 1 --> pick this
2 220 6647 2014-09-05 0 #group 1
3 826 3380 2014-11-11 150 #group 2 --> pick this
4 826 3380 2014-12-09 23 #group 3 --> pick this
5 826 3380 2015-05-12 88 #group 4 --> pick this
6 901 4555 2015-05-15 32 #group 4
7 901 4555 2015-10-05 542 #group 5 --> pick this
8 901 4555 2015-11-01 98 #group 6 --> pick this
I think you need create groups by apply with list comprehension and between, then convert to numeric groups by factorize, last use your solution with loc + idxmax:
df['date'] = pd.to_datetime(df['date'])
df = df.reset_index(drop=True)
td = pd.Timedelta('5 days')
def f(x):
x['g'] = [tuple((x.index[x['date'].between(i - td, i + td)])) for i in x['date']]
return x
df2 = df.groupby('id').apply(f)
df2['g'] = pd.factorize(df2['g'])[0]
print (df2)
id product date price g
0 220 6647 2014-09-01 100 0
1 220 6647 2014-09-03 120 0
2 220 6647 2014-09-05 0 0
3 826 3380 2014-11-11 150 1
4 826 3380 2014-12-09 23 2
5 826 3380 2015-05-12 88 3
6 901 4555 2015-05-15 32 4
7 901 4555 2015-10-05 542 5
8 901 4555 2015-11-01 98 6
df3 = df2.loc[df2.groupby('g')['price'].idxmax()]
print (df3)
id product date price g
1 220 6647 2014-09-03 120 0
3 826 3380 2014-11-11 150 1
4 826 3380 2014-12-09 23 2
5 826 3380 2015-05-12 88 3
6 901 4555 2015-05-15 32 4
7 901 4555 2015-10-05 542 5
8 901 4555 2015-11-01 98 6
Or use a two-liner:
df2=pd.to_numeric(df.groupby('id')['date'].diff(-1).astype(str).str[:-25]).abs().fillna(6)
print(df.loc[df2.index[df2>5].tolist()])
Output:
id product date
1 220 6647 2014-09-03
2 220 6647 2014-10-16
3 826 3380 2014-11-11
4 826 3380 2014-12-09
5 826 3380 2015-05-19
6 901 4555 2014-09-01
7 901 4555 2014-10-05
8 901 4555 2014-11-01
So use diff and slice using string slice, and absolute all the values, then drop the ones less than 5, get those indexes, then get the indexes in the in df.
A B C D yearweek
0 245 95 60 30 2014-48
1 245 15 70 25 2014-49
2 150 275 385 175 2014-50
3 100 260 170 335 2014-51
4 580 925 535 2590 2015-02
5 630 126 485 2115 2015-03
6 425 90 905 1085 2015-04
7 210 670 655 945 2015-05
The last column contains the the year along with the weeknumber. Is it possible to convert this to a datetime column with pd.to_datetime?
I've tried:
pd.to_datetime(df.yearweek, format='%Y-%U')
0 2014-01-01
1 2014-01-01
2 2014-01-01
3 2014-01-01
4 2015-01-01
5 2015-01-01
6 2015-01-01
7 2015-01-01
Name: yearweek, dtype: datetime64[ns]
But that output is incorrect, while I believe %U should be the format string for week number. What am I missing here?
You need another parameter for specify day - check this:
df = pd.to_datetime(df.yearweek.add('-0'), format='%Y-%W-%w')
print (df)
0 2014-12-07
1 2014-12-14
2 2014-12-21
3 2014-12-28
4 2015-01-18
5 2015-01-25
6 2015-02-01
7 2015-02-08
Name: yearweek, dtype: datetime64[ns]
I have the table below in a Pandas dataframe:
date user_id whole_cost cost1
02/10/2012 00:00:00 1 1790 12
07/10/2012 00:00:00 1 364 15
30/01/2013 00:00:00 1 280 10
02/02/2013 00:00:00 1 259 24
05/03/2013 00:00:00 1 201 39
02/10/2012 00:00:00 3 623 1
07/12/2012 00:00:00 3 90 0
30/01/2013 00:00:00 3 312 90
02/02/2013 00:00:00 5 359 45
05/03/2013 00:00:00 5 301 34
02/02/2013 00:00:00 5 359 1
05/03/2013 00:00:00 5 801 12
..
The table was extracted from a csv file using the following query :
import pandas as pd
newnames = ['date','user_id', 'whole_cost', 'cost1']
df = pd.read_csv('expenses.csv', names = newnames, index_col = 'date')
I have to analyse the profile of my users and for this purpose:
I would like to group (for each user - they are thousands) queries by month summing the query whole_cost for the entire month e.g. if user_id=1 was has a whole cost of 1790 on 02/10/2012 with cost1 12 and on the 07/10/2012 with whole cost 364, then it should have an entry in the new table of 2154 (as the whole cost) on 31/10/2012 (end of the month end-point representing the month - all dates in the transformed table will be month ends representing the whole month to which they relate).
In 0.14 you'll be able to groupby monthly and another column at the same time:
In [11]: df
Out[11]:
user_id whole_cost cost1
2012-10-02 1 1790 12
2012-10-07 1 364 15
2013-01-30 1 280 10
2013-02-02 1 259 24
2013-03-05 1 201 39
2012-10-02 3 623 1
2012-12-07 3 90 0
2013-01-30 3 312 90
2013-02-02 5 359 45
2013-03-05 5 301 34
2013-02-02 5 359 1
2013-03-05 5 801 12
In [12]: df1 = df.sort_index() # requires sorted DatetimeIndex
In [13]: df1.groupby([pd.TimeGrouper(freq='M'), 'user_id'])['whole_cost'].sum()
Out[13]:
user_id
2012-10-31 1 2154
3 623
2012-12-31 3 90
2013-01-31 1 280
3 312
2013-02-28 1 259
5 718
2013-03-31 1 201
5 1102
Name: whole_cost, dtype: int64
until 0.14 I think you're stuck with doing two groupbys:
In [14]: g = df.groupby('user_id')['whole_cost']
In [15]: g.resample('M', how='sum').dropna()
Out[15]:
user_id
1 2012-10-31 2154
2013-01-31 280
2013-02-28 259
2013-03-31 201
3 2012-10-31 623
2012-12-31 90
2013-01-31 312
5 2013-02-28 718
2013-03-31 1102
dtype: float64
With timegrouper getting deprecated, you can replace it with Grouper to get the same results
df.groupby(['user_id', pd.Grouper(key='date', freq='M')]).agg({'whole_cost':sum})
df.groupby(['user_id', df['date'].dt.dayofweek]).agg({'whole_cost':sum})