I have a pandas df that includes two columns: time_in_years (float64) and date (datetime64).
import pandas as pd
df = pd.DataFrame({
'date': ['2009-12-25','2005-01-09','2010-10-31'],
'time_in_years': ['10.3434','5.0977','3.3426']
})
df['date'] = pd.to_datetime(df['date'])
df["time_in_years"] = df.time_in_years.astype(float)
I need to create date2 as a datetime64 column by adding the number of years to the date.
I tried the following but with no luck:
df['date_2'] = df['date'] + datetime.timedelta(years=df['time_in_years'])
I know that with fractions I will not be able to get the exact date, but I want to get the closest new date as possible.
Try package dateutil:
from dateutil.relativedelta import relativedelta
First convert fractional years to number of days, then use lambda function and apply it to dataframe:
df['date_2'] = df.apply(lambda x: x['date'] + relativedelta(days = int(x['time_in_years']*365)), axis = 1)
Result:
date time_in_years date_2
0 2009-12-25 10.3434 2020-04-26
1 2005-01-09 5.0977 2010-02-12
2 2010-10-31 3.3426 2014-03-04
datetime.timedelta also works fine:
df['date_2'] = df.apply(lambda x: x['date'] + datetime.timedelta(days = int(x['time_in_years']*365)), axis = 1)
Please note conversion to int is necessary, because relativedelta and timedelta do not accept fractional values.
Related
I am trying to convert my 15ys worth of daily data into weekly by taking the mean, diff and count of certain features. I tried using .resample but I was not sure if that is the most efficient way.
My sample data:
Date,Product,New Quantity,Price,Refund Flag
8/16/1994,abc,10,0.5,
8/17/1994,abc,11,0.9,1
8/18/1994,abc,15,0.6,
8/19/1994,abc,19,0.4,
8/22/1994,abc,22,0.2,1
8/23/1994,abc,19,0.1,
8/16/1994,xyz,16,0.5,1
8/17/1994,xyz,10,0.9,1
8/18/1994,xyz,12,0.6,1
8/19/1994,xyz,19,0.4,
8/22/1994,xyz,26,0.2,1
8/23/1994,xyz,30,0.1,
8/16/1994,pqr,0,0,
8/17/1994,pqr,0,0,
8/18/1994,pqr,1,1,
8/19/1994,pqr,2,0.6,
8/22/1994,pqr,9,0.1,
8/23/1994,pqr,12,0.2,
This is the output I am looking for:
Date,Product,Net_Quantity_diff,Price_avg,Refund
8/16/1994,abc,9,0.6,1
8/22/1994,abc,-3,0.15,0
8/16/1994,xyz,3,0.6,3
8/22/1994,xyz,4,0.15,1
8/16/1994,pqr,2,0.4,0
8/22/1994,pqr,3,0.15,0
I think the pandas resample method is indeed ideal for this. You can pass a dictionary to the agg method, defining which aggregation function to use for each column. For example:
import numpy as np
import pandas as pd
df = pd.read_csv('sales.txt') # your sample data
df['Date'] = pd.to_datetime(df['Date'])
df = df.set_index(df['Date'])
del df['Date']
df['Refund Flag'] = df['Refund Flag'].fillna(0).astype(bool)
def span(s):
return np.max(s) - np.min(s)
df_weekly = df.resample('w').agg({'New Quantity': span,
'Price': np.mean,
'Refund Flag': np.sum})
df_weekly
New Quantity Price Refund Flag
Date
1994-08-21 19 0.533333 4
1994-08-28 21 0.150000 2
I have a dataframe with dates as seen in the table below. 1st block is what it should look like and the 2nd block is what I get when just adding the BDays. This is an example of what it should look like when completed. I want to use the 1st column and add 5 business days to the dates, but if the 5 Bdays overlaps a holiday (like 15 Feb'21) then I need to add one additional day. It is fairly simple to add the 5Bday using pandas.tseries.offsets import BDay, but i cannot skip the holidays while using the dataframe.
I have tried to use pandas.tseries.holiday import USFederalHolidayCalendar, the workdays and workalendar modules, but cannot figure it out. Anyone have an idea what I can do.
Correct Example
DATE
EXIT DATE +5
2021/02/09
2021/02/17
2021/02/10
2021/02/18
Wrong Example
DATE
EXIT DATE +5
2021/02/09
2021/02/16
2021/02/10
2021/02/17
Here are some examples of code I tried:
import pandas as pd
from workdays import workday
...
df['DATE'] = workday(df['EXIT DATE +5'], days=5, holidays=holidays)
Next Example:
import pandas as pd
from pandas.tseries.holiday import USFederalHolidayCalendar
bday_us = pd.offsets.CustomBusinessDay(calendar=USFederalHolidayCalendar())
dt = df['DATE']
df['EXIT DATE +5'] = dt + bday_us
=========================================
Final code:
Below is the code I finally settled on. I had to define the holidays manually due to the days the NYSE actually trades. Like for instance the day Pres Bush was laid to rest.
import datetime as dt
import pandas as pd
from pandas.tseries.holiday import USFederalHolidayCalendar
from pandas.tseries.offsets import BDay
from pandas.tseries.holiday import AbstractHolidayCalendar, Holiday, nearest_workday, \
USMartinLutherKingJr, USPresidentsDay, GoodFriday, USMemorialDay, \
USLaborDay, USThanksgivingDay
class USTradingCalendar(AbstractHolidayCalendar):
rules = [
Holiday('NewYearsDay', month=1, day=1, observance=nearest_workday),
USMartinLutherKingJr,
USPresidentsDay,
GoodFriday,
USMemorialDay,
Holiday('USIndependenceDay', month=7, day=4, observance=nearest_workday),
Holiday('BushDay', year=2018, month=12, day=5),
USLaborDay,
USThanksgivingDay,
Holiday('Christmas', month=12, day=25, observance=nearest_workday)
]
offset = 5
df = pd.DataFrame(['2019-10-11', '2019-10-14', '2017-04-13', '2018-11-28', '2021-07-02'], columns=['DATE'])
df['DATE'] = pd.to_datetime(df['DATE'])
def offset_date(start, offset):
return start + pd.offsets.CustomBusinessDay(n=offset, calendar=USTradingCalendar())
df['END'] = df.apply(lambda x: offset_date(x['DATE'], offset), axis=1)
print(df)
Input data
df = pd.DataFrame(['2021-02-09', '2021-02-10', '2021-06-28', '2021-06-29', '2021-07-02'], columns=['DATE'])
df['DATE'] = pd.to_datetime(df['DATE'])
Suggested solution using apply
from pandas.tseries.holiday import USFederalHolidayCalendar
from pandas.tseries.offsets import BDay
def offset_date(start, offset):
return start + pd.offsets.CustomBusinessDay(n=offset, calendar=USFederalHolidayCalendar())
offset = 5
df['END'] = df.apply(lambda x: offset_date(x['DATE'], offset), axis=1)
DATE END
2021-02-09 2021-02-17
2021-02-10 2021-02-18
2021-06-28 2021-07-06
2021-06-29 2021-07-07
2021-07-02 2021-07-12
PS: If you want to use a particular calendar such as the NYSE, instead of the default USFederalHolidayCalendar, I recommend following the instructions on this answer, about creating a custom calendar.
Alternative solution which I do not recommend
Currently, to the best of my knowledge, pandas do not support a vectorized approach to your problem. But if you want to follow a similar approach to the one you mentioned, here is what you should do.
First, you will have to define an arbitrary far away end date that includes all the periods you might need and use it to create a list of holidays.
holidays = USFederalHolidayCalendar().holidays(start='2021-02-09', end='2030-02-09')
Then, you pass the holidays list to CustomBusinessDay through the holidays parameter instead of the calendar to generate the desired offset.
offset = 5
bday_us = pd.offsets.CustomBusinessDay(n=offset, holidays=holidays)
df['END'] = df['DATE'] + bday_us
However, this type of approach is not a true vectorized solution, even though it might seem like it. See the following SO answer for further clarification. Under the hood, this approach is probably doing a conversion that is not efficient. This why it yields the following warning.
PerformanceWarning: Non-vectorized DateOffset being applied to Series
or DatetimeIndex
Here's one way to do it
import pandas as pd
from pandas.tseries.holiday import USFederalHolidayCalendar
from datetime import timedelta as td
def get_exit_date(date):
holiday_list = cals.holidays(start=date, end=date + td(weeks=2)).tolist()
# 6 periods since start date is included in set
n_bdays = pd.bdate_range(start=date, periods=6, freq='C', holidays=holiday_list)
return n_bdays[-1]
df = pd.read_clipboard()
cals = USFederalHolidayCalendar()
# I would convert this to datetime
df['DATE'] = pd.to_datetime(df['DATE'])
df['EXIT DATE +5'] = df['DATE'].apply(get_exit_date)
this is using bdate_range which returns a datetime index
Results:
DATE EXIT DATE +5
0 2021-02-09 2021-02-17
1 2021-02-10 2021-02-18
Another option is instead of dynamically creating the holiday list. You could also just choose a start date and leave it outside the function like so:
def get_exit_date(date):
# 6 periods since start date is included in set
n_bdays = pd.bdate_range(start=date, periods=6, freq='C', holidays=holiday_list)
return n_bdays[-1]
df = pd.read_clipboard()
cals = USFederalHolidayCalendar()
holiday_list = cals.holidays(start='2021-01-01').tolist()
# I would convert this to datetime
df['DATE'] = pd.to_datetime(df['DATE'])
df['EXIT DATE +5'] = df['DATE'].apply(get_exit_date)
I have a question regarding resampling of DataFrames.
import pandas as pd
df = pd.DataFrame([['2005-01-20', 10], ['2005-01-21', 20],
['2005-01-27', 40], ['2005-01-28', 50]],
columns=['date', 'num'])
# Convert the column to datetime
df['date'] = pd.to_datetime(df['date'])
# Resample and aggregate results by week
df = df.resample('W', on='date')['num'].sum().reset_index()
print(df.head())
# OUTPUT:
# date num
# 0 2005-01-23 30
# 1 2005-01-30 90
Everything works as expected, but I would like to better understand what exactly resample(),['num'] and sum() do here.
QUESTION #1
Why the following happens:
The result of df.resample('W', on='date') is DatetimeIndexResampler.
The result of df.resample('W', on='date')['num'] is pandas.core.groupby.SeriesGroupBy.
The result of df.resample('W', on='date')['num'].sum() is
date
2005-01-23 30
2005-01-30 90
Freq: W-SUN, Name: num, dtype: int64
QUESTION #2
Is there a way to produce the same results without resampling? For example, using groupby.
Answer1
As the docs says, .resample returns a Resampler Object. Hence you get DatetimeIndexResampler because date is a datetime object.
Now, you get <pandas.core.groupby.SeriesGroupBy because you are looking for Series from the dataframe based of off the Resampler object.
Oh by the way,
df.groupby([pd.Grouper(key='date', freq='W-SUN')])['num']
Would return
<pandas.core.groupby.SeriesGroupBy as well.
Now when you do .sum(), you are getting the sum over the requested axis of the dataframe. You get a Series because you are doing sum over the pandas.core.series.Series.
Answer2
You can achieve results using groupby with the help from Grouper as follow:
df.groupby([pd.Grouper(key='date', freq='W-SUN')])['num'].sum()
Output:
date
2005-01-23 30
2005-01-30 90
Name: num, dtype: int64
I asked (and answered) a question here Pandas ffill resampled data grouped by column where I wanted to know how to ffill a date range for each unique entry for a column (my assets column).
My solution requires that the asset "id" is a column. However, the data makes more sense to me as a multiindex. Furthermore I would like more fields in the multiindex. Is the only way of filling forward to drop the non-date fields from the multiiindex before ffilling?
A modified version of my example (to work on a df with multiindex) here:
from datetime import datetime, timedelta
import pytz
some_time = datetime(2018,4,2,20,20,42)
start_date = datetime(some_time.year,some_time.month,some_time.day).astimezone(pytz.timezone('Europe/London'))
end_date = start_date + timedelta(days=1)
start_date = start_date + timedelta(hours=some_time.hour,minutes=(0 if some_time.minute < 30 else 30 ))
df = pd.DataFrame(['A','B'],columns=['asset_id'])
df2=df.copy()
df['datetime'] = start_date
df2['datetime'] = end_date
df['some_property']=0
df.loc[df['asset_id']=='B','some_property']=2
df = df.append(df2).set_index(['asset_id','datetime'])
With what is arguably my crazy solution here:
df = df.reset_index()
df = df.set_index('datetime').groupby('asset_id').resample('30T').ffill().drop('asset_id',axis=1)
df = df.reset_index().set_index(['asset_id','datetime'])
Can I avoid all that re-indexing?
I'm still fairly new to pandas and the script i wrote to accomplish a seemily easy task seems needlessly complicated. If you guys know of an easier way to accomplish this I would be extremely grateful.
task:
I hate two spreadsheets (df1&df2), each with an identifier (mrn) and a date. my task is to retrieve an value from df2 for each row in df1 if the following conditions are met:
the identifier for a given row in df1 exists in df2
if above is true, then retrieve the value in df2 if the associated date is within a +/-5 day range from the date in df1.
I have written the following code which accomplishes this:
#%%housekeeping
import numpy as np
import pandas as pd
import csv
import datetime
from datetime import datetime, timedelta
import sys
from io import StringIO
#%%dataframe import
df1=',mrn,date,foo\n0,1,2015-03-06,n/a\n1,11,2009-08-14,n/a\n2,14,2009-05-18,n/a\n3,20,2010-06-19,n/a\n'
df2=',mrn,collection Date,Report\n0,1,2015-03-06,report to import1\n1,11,2009-08-12,report to import11\n2,14,2009-05-21,report to import14\n3,20,2010-06-25,report to import20\n'
df1 = pd.read_csv(StringIO(df1))
df2 = pd.read_csv(StringIO(df2))
#converting to date-time format
df1['date']=pd.to_datetime(df1['date'])
df2['collection Date']=pd.to_datetime(df2['collection Date'])
#%%mask()
def mask(df2, rangeTime):
mask= (df2> rangeTime -timedelta(days=5)) & (df2 <= rangeTime + timedelta(days=5))
return mask
#%% detailLoop()
i=0
for element in df1["mrn"]:
df1DateIter = df1.ix[i, 'date']
df2MRNmatch= df2.loc[df2['mrn']==element, ['collection Date', 'Report']]
df2Date= df2MRNmatch['collection Date']
df2Report= df2MRNmatch['Report']
maskOut= mask(df2Date, df1DateIter)
dateBoolean= maskOut.iloc[0]
if dateBoolean==True:
df1.ix[i, 'foo'] = df2Report.iloc[0]
i+=1
#: once the script has been run the df1 looks like:
Out[824]:
mrn date foo
0 1 2015-03-06 report to import1
1 11 2009-08-14 report to import11
2 14 2009-05-18 report to import14
3 20 2010-06-19 NaN