“wie man durch Daten in Python schaufelt” Code-Antworten

Python iteriert durch Date Range

from datetime import timedelta, date

def daterange(start_date, end_date):
    for n in range(int ((end_date - start_date).days)):
        yield start_date + timedelta(n)

start_date = date(2013, 1, 1)
end_date = date(2015, 6, 2)
for single_date in daterange(start_date, end_date):
    print(single_date.strftime("%Y-%m-%d"))
Testy Teira

wie man durch Daten in Python schaufelt

import datetime

# The size of each step in days
day_delta = datetime.timedelta(days=1)

start_date = datetime.date.today()
end_date = start_date + 7*day_delta

for i in range((end_date - start_date).days):
    print(start_date + i*day_delta)
Modern Manx

Wie man alle Daten in Python überschreitet

# this DOES account for leap years
years = ['2023', '2024', '2028']
months = ['01', '02', '03', '04', '05', '06',
              '07', '08', '09', '10', '11', '12']
days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

for i in range(len(years)):
    for j in range(len(months)):
        if (((int(years[i]) % 4 == 0 and int(years[i]) % 100 != 0) or (int(years[i])% 400 == 0)) and months[j] == '02'):
            for k in range(1, days[j] + 2):
                print(years[i] + '-' + months[j] + '-' + str(k))
        else:
            for k in range(1, days[j] + 1):
                print(years[i] + '-' + months[j] + '-' + str(k))
            

#2023-01-1
#2023-01-2
#2023-01-3
#...
liamxaro

Wie man alle Daten in Python überschreitet

#this does not account for leap years
years = [str(i) for i in range(2010, 2022)]
months = ['01', '02', '03', '04', '05', '06',
              '07', '08', '09', '10', '11', '12']
days = [31, 28 , 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

for i in range(len(years)):
  for j in range(len(months)):
    for k in range(1, days[j] + 1):
      print("The date is: " + years[i] + months[j] + str(k))
      



liamxaro

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