Say for the year of 2020, how do I iterate through the days in the months so that my outcome would be in the following format:
Jan1
Jan2
Jan3
....
Jan31
Feb1
I've tried so many things online but I couldnt find an answer. Please help :(
Both of these methods will handle leap years correctly out of the box.
Using a simple while loop:
from datetime import datetime, timedelta
def iter_days(year):
dt = datetime(year, 1, 1)
while dt.year == year:
yield dt
dt += timedelta(days=1)
Using date rules:
from datetime import datetime
from dateutil.rrule import rrule, DAILY
def iter_days(year):
first_date = datetime(year, 1, 1)
last_date = datetime(year, 12, 31)
return rrule(DAILY, dtstart=first_date, until=last_date)
Both would be used the same:
for dt in iter_days(2020):
print(dt.strftime('%b%-d'))
The format string '%b%-d' will give you the format you specified in your question. I don't know if that was a requirement or not.
This is crude but gets what you want for 2020. You'll need to change 366 to 365 for non-leap-years.
#!/usr/bin/python3
import datetime
startDate = '2020-01-01'
start = datetime.datetime.strptime(startDate, '%Y-%m-%d')
for dayNum in range(0,366):
dayOfYear = start + datetime.timedelta(days=dayNum)
print(dayOfYear.strftime('%b %d, %Y'))
The calendar module offers quite a bit of functionality.
Here is a solution that works for any given year
import calendar as cal
for mi in range(1,13):
_, days = cal.monthrange(2020, mi)
for d in range(1, days+1):
print(cal.month_name[mi], d)
Related
What I am trying to figure out is how to add "Cases" and "Deaths" for each day, so that it starts with: "1/19/2020 Cases" and "1/19/2020 Deaths" then "1/20/2020 Cases" etc. It seems the append function does not work for this, and I don't know how else to add this. It doesn't seem like python has a way to do this task. My eventual goal is to make this a pandas dataframe.
import pandas as pd
dates = pd.date_range(start = '1/19/2020', end = '12/31/2021')
lst = dates.repeat(repeats = 2)
print(lst)
Thanks
If I am not mistaken, I don't think there's a way to do it with purely pandas. However with python and datetime, you can do so:
import pandas as pd
from datetime import timedelta, date
def daterange(start_date, end_date):
# Credit: https://stackoverflow.com/a/1060330/10640517
for n in range(int((end_date - start_date).days)):
yield start_date + timedelta(n)
dates = []
start_date = date(2020, 1, 19) # Start date here
end_date = date(2021, 12, 31) # End date here
for single_date in daterange(start_date, end_date):
dates.append(single_date.strftime("%m/%d/%Y") + " Cases")
dates.append(single_date.strftime("%m/%d/%Y") + " Deaths")
pdates = pd.DataFrame(dates)
print (pdates)
Is this what you want? If not, I can delete it.
I have two tables:
-event dates
-return dates
Some event dates are not at a trading day.
How can I change the event date to the next trading day?
So if event date is not in return dates, take the next day in return dates.
The approach to change weekend days to working days does not work because of days like Christmas.
The best would be to look up the next day in the return table.
for i in event['date']:
if i is not in return ['date'].values:
event ['date']=i+datetime.timedelta(days=1)
but this doenst work
I am working with dataframes and dates have the format datetime64[ns]. If the event date does not exist in return date than event date plus one day
Edit
After the clarifications concerning the desired logic, here is the new solution
from datetime import datetime, timedelta
import numpy as np
import pandas as pd
# Create two df
event_date = datetime.now()
event_dates = pd.DataFrame([datetime(2020, 2, _) for _ in range(1, 29)], columns=['date'])
print(event_dates.date[0])
# 2020-02-01 00:00:00
return_dates = pd.DataFrame([datetime(2020, 1, _) for _ in range(1, 32)], columns=['date'])
# Apply logic
event_dates.date = [_ if _ in return_dates.date else _ + timedelta(days=1) for _ in event_dates.date]
print(event_dates.date[0])
# 2020-02-02 00:00:00
Base Python
Here is a solution using the standard datetime library
from datetime import datetime
from typing import List
def get_next_trade_date(date: datetime, date_list: List[datetime]) -> datetime: # The annotations here are just to specify the types of the objects
if date in date_list: # Check if the date is contained in the list
return date
delta, res = None, None # Initialize both to None
for _ in date_list:
tmp = abs((date - _).days) # Time difference in current iteration
if not delta or tmp < delta: # See bullet point 1.
delta, res = tmp, _
return res
if __name__ == '__main__':
event_date = datetime.now()
return_dates = [datetime(2020, 1, _) for _ in range(1, 32)]
print(get_next_trade_date(event_date, return_dates))
# 2020-01-01 00:00:00
Notice that
The condition not delta or tmp < delta is twofold: in the first iteration delta, res are both None so we will overwrite them with tmp, _. We catch this by using not delta. The other part (tmp < delta) is more obvious: if we have a new minimal delta then we overwrite delta, res.
I only considered days intervals ((date - _).days), you could go further into details (see datetime.timedelta for more info)
coming from R I believe there must be a simpler solution using numpy - see below
Numpy
This solution uses numpy. (date_list - date) is an array of timedeltas, (date_list - date).argmin() returns the index of the minimal value.
from datetime import datetime
import numpy as np
def get_next_trade_date(date: datetime, date_list: np.ndarray) -> datetime:
return date_list[(date_list - date).argmin()]
if __name__ == '__main__':
event_date = datetime.now()
return_dates = np.array([datetime(2020, 1, _) for _ in range(1, 32)])
print(get_next_trade_date(event_date, return_dates))
# 2020-01-01 00:00:00
I have a pandas dataframe that looks like:
import pandas as pd
df1 = pd.DataFrame({'Counterparty':['Bank','Client','Bank','Bank','Bank','Bank'],
'Date':['4Q18','1Q19','2Q19','4Q21','FY22','H123']
})
I want to convert the 'Date' column from a string to a date such that the date is the last date for that particular period. ie 'FQ18'= 31st Dec 2018, '1Q19' = 31st Mar 2019, 'FY22' = 31st Dec 2022,'H123'= 30th June 2023
Any suggestions how to achieve this ?
As mentioned by #jpp, you're going to have to do some customization. There isn't existing functionality to map "FY22" to 2022-12-31, to my knowledge. Here's something to get you started, based on the limited example you've shown:
import re
import pandas as pd
from pandas.core.tools.datetimes import DateParseError
from pandas.tseries import offsets
halfyr = re.compile(r'H(?P<half>\d)(?P<year>\d{2})')
fiscalyr = re.compile(r'FY(?P<year>\d{2})')
def try_qend(date):
try:
return pd.to_datetime(date) + offsets.QuarterEnd()
except (DateParseError, ValueError):
halfyr_match = halfyr.match(date)
if halfyr_match:
half, year = [int(i) for i in halfyr_match.groups()]
month = 6 if half == 1 else 12
return pd.datetime(2000 + year, month, 1) + offsets.MonthEnd()
else:
fiscalyr_match = fiscalyr.match(date)
if fiscalyr_match:
year = int(fiscalyr_match.group('year'))
return pd.datetime(2000 + year, 12, 31)
else:
# You're SOL
return pd.NaT
def parse_dates(dates):
return pd.to_datetime([try_qend(date) for date in dates])
Assumptions:
All years are 20yy, not 19xx.
The regex patterns here completely describe the year-half/fiscal-year syntax set.
Example:
dates = ['4Q18','1Q19','2Q19','4Q21','FY22','H123']
parse_dates(dates)
DatetimeIndex(['2018-12-31', '2019-03-31', '2019-06-30', '2021-12-31',
'2022-12-31', '2023-06-30'],
dtype='datetime64[ns]', freq=None)
Need help in Pulling stock value of last business day of a month in a time series/dataframe
I am executing the fol code:
import pandas as pd
import datetime
import matplotlib
import warnings
warnings.filterwarnings('ignore')
start = datetime.datetime(2014, 3, 31)
end = datetime.datetime(2017, 9, 30)
stocks = ['AAPL', 'GOOG']
col = 'Adj Close'
df = web.get_data_yahoo(stocks,start,end)
data = df.ix[col]
dataframe = pd.DataFrame(data)
This gives me a dataframe with all the close values.
I want to get the values only from the last business day of the month
Please ignore the question, I've managed to find the answer
Title says it all. I have a string in date/time format and I'm not sure how to convert it to decimal. Thanks
import datetime
import re
timestr = '4 days, 23:56:46.063070'
matches = re.match('(\d+) days, (.*)', timestr)
assert matches
days = int(matches.group(1))
time = datetime.datetime.strptime(matches.group(2), '%H:%M:%S.%f')
timediff = time - datetime.datetime(1900, 1, 1) # timedelta
hours = days*24 + timediff.total_seconds()/60/60
By the way that gives 119.946 not 119.997.
import pandas as pd
pd.Timedelta("4 days, 23:56:46.063070") / pd.Timedelta(1, 'h')
119.94612863055555