I'm quite new to python and programming in general and I made this program that lets you add days to today's current date and it would output what that date would be. Not very useful, but I got quite annoyed at how much space it was taking and was wondering if it could be made smaller and more condensed, since it's quite simple. I dont actually have a clue. So if someone could show me that would be nice.
import datetime
calendar = {1:"Sunday", 2:"Monday", 3:"Tuesday", 4:"Wednesday", 5:"Thursday", 6:"Friday", 0:"Saturday"}
Day = eval(input("Enter how many days you want to look into the future: "))
todayDate = datetime.date.today()
x = todayDate.strftime("%A")
if x == "Sunday" :
y = ((1+ Day) % 7)
print(calendar[y])
exit()
if x == "Monday" :
y = ((2+ Day) % 7)
print(calendar[y])
exit()
if x == "Tuesday" :
y = ((3+ Day) % 7)
print(calendar[y])
exit()
if x == "Wednesday" :
y = ((4+ Day) % 7)
print(calendar[y])
exit()
if x == "Thursday" :
y = ((5+ Day) % 7)
print(calendar[y])
exit()
if x == "Friday" :
y = ((6+ Day) % 7)
print(calendar[y])
exit()
if x == "Saturday" :
y = ((7+ Day) % 7)
print(calendar[y])
exit()
offset = {d: o for o, d in calendar.items()}
y = (offset[x] + day) % 7
print(calendar[y])
You can use timedelta to add any amount of time to a date.
from datetime import date, timedelta
x=5#or any amount of days you want
date = date.today()
new_date=date+timedelta(days=x)
print(new_date.strftime('%A'))
If you didn't wish to create an inverse dictionary there are multiple things you can do with your own code.
dont repeat yourself (the printing and exiting can be done after the if statement
use elif to avoid checking if statements unnecessarily (wouldn't currently happen with seperate exits though)
x = todayDate.strftime("%A")
if x == "Sunday" :
y = ((1+ Day) % 7)
elif x == "Monday" :
y = ((2+ Day) % 7)
elif x == "Tuesday" :
y = ((3+ Day) % 7)
elif x == "Wednesday" :
y = ((4+ Day) % 7)
elif x == "Thursday" :
y = ((5+ Day) % 7)
elif x == "Friday" :
y = ((6+ Day) % 7)
elif x == "Saturday" :
y = ((7+ Day) % 7)
print(calendar[y])
exit()
Python 3.10 will bring along the match statement that will make this slightly more efficient
x = todayDate.strftime("%A")
match x:
case "Sunday" :
y = ((1+ Day) % 7)
case "Monday" :
y = ((2+ Day) % 7)
case "Tuesday" :
y = ((3+ Day) % 7)
case "Wednesday" :
y = ((4+ Day) % 7)
case "Thursday" :
y = ((5+ Day) % 7)
case "Friday" :
y = ((6+ Day) % 7)
case "Saturday" :
y = ((7+ Day) % 7)
Usually when condensing/deduplicating code the first thing you should do is identify what is it that you're repeating a lot.
On your code, for instance, we see this "structure" many times:
if x == <SOME DAY>:
y = ((<SOME INTEGER> + Day) % 7)
print(calendar[y])
exit()
So we'd like to abstract that away, so we only write it once (see DRY principle).
So we'd like to write something like this:
def increment_days_and_print(day, increment):
y = ((day + increment) % 7)
print(calendar[y])
exit()
Then we can replace all that repeating code with a call to the function we defined:
def increment_days_and_print(day, increment):
y = ((day + increment) % 7)
print(calendar[y])
exit()
calendar = {1:"Sunday", 2:"Monday", 3:"Tuesday", 4:"Wednesday", 5:"Thursday", 6:"Friday", 0:"Saturday"}
Day = eval(input("Enter how many days you want to look into the future: "))
todayDate = datetime.date.today()
x = todayDate.strftime("%A")
if x == "Sunday":
increment_days_and_print(1, Day)
if x == "Monday":
increment_days_and_print(2, Day)
if x == "Tuesday":
increment_days_and_print(3, Day)
if x == "Wednesday":
increment_days_and_print(4, Day)
if x == "Thursday":
increment_days_and_print(5, Day)
if x == "Friday":
increment_days_and_print(6, Day)
if x == "Saturday":
increment_days_and_print(7, Day)
That's much better already, isn't it? We could also refactor it like this:
def increment_days_and_print(day, increment):
y = ((day + increment) % 7)
print(calendar[y])
exit()
calendar = {1:"Sunday", 2:"Monday", 3:"Tuesday", 4:"Wednesday", 5:"Thursday", 6:"Friday", 0:"Saturday"}
Day = eval(input("Enter how many days you want to look into the future: "))
todayDate = datetime.date.today()
x = todayDate.strftime("%A")
if x == "Sunday":
start = 1
if x == "Monday":
start = 2
if x == "Tuesday":
start = 3
if x == "Wednesday":
start = 4
if x == "Thursday":
start = 5
if x == "Friday":
start = 6
if x == "Saturday":
start = 7
increment_days_and_print(start, Day)
Finally, notice that we're essentially setting start to a svalue, given a value of x. This is exactly what a dict is for:
def increment_days_and_print(day, increment):
y = ((day + increment) % 7)
print(calendar[y])
exit()
calendar = {1:"Sunday", 2:"Monday", 3:"Tuesday", 4:"Wednesday", 5:"Thursday", 6:"Friday", 0:"Saturday"}
Day = eval(input("Enter how many days you want to look into the future: "))
todayDate = datetime.date.today()
x = todayDate.strftime("%A")
weekday_numbers = {
"Sunday": 1,
"Monday": 2,
"Tuesday": 3,
"Wednesday": 4,
"Thursday": 5,
"Friday": 6,
"Saturday": 0
}
start = weekday_numbers[x]
increment_days_and_print(start, Day)
Finally, doing some cleanup:
weekday_numbers = [
(1, "Sunday"),
(2, "Monday"),
(3, "Tuesday"),
(4, "Wednesday"),
(5, "Thursday"),
(6, "Friday"),
(0, "Saturday")
]
num_to_weekday = dict((d, w) for (d, w) in weekday_numbers)
weekday_to_num = dict((w, d) for (d, w) in weekday_numbers)
increment = int(input("Enter how many days you want to look into the future: "))
today = datetime.date.today()
weekday = today.strftime("%A")
start = weekday_to_num[weekday]
next_day = ((start + increment) % 7)
print(num_to_weekday[next_day])
Since we only called the function once, we didn't need it anymore; it worked only as a tool to help us see what simplifications we could make along the way.
Related
i'm working on making a program in python that uses arrays since it's the new chapter i'm learning. The goal of my program will ask the user for month day and day and I seem to have it almost done except that i have an issue with 2 lines in this whole code. I will write a comment in where i'm getting the errors.
def calc_zellers_congruence(get_day: int, get_year: int, get_month: int) -> str:
# Calculates the month day and year
days = [""] * 7
day_Of_Week = 0
day_of_month = 0
century = 0
year_of_the_century = get_year % 100
days[1] = "Sunday"
days[2] = "Monday"
days[3] = "Tuesday"
days[4] = "Wednessday"
days[5] = "Thursday"
days[6] = "Friday"
days[0] = "Saturday"
dayOfWeek = day_of_month + 13 * get_month + int(1) / 5 + year_of_the_century + int(year_of_the_century) / 4 + int(
century) / 4 + 5 * century % 7
name_of_day = days[day_Of_Week]
return name_of_day
def displayResults():
print("Result" + day)
def get_day():
print("Enter a Day")
get_day = int(input())
return get_day
def get_month():
print("Enter a Month")
month = int(input())
return month
def get_year():
print("Enter a Year")
get_year: int(input())
return year
# Main
# This program will calculate birthdays using arrays and zellers congruence
# References...Programming Fundamentals WiKi, J Robinson, GeeksForGeek.com
year = get_year()
month = get_month()
day = get_day()
calc_Zeller_Congruence: calc_zellers_congruence(get_year, get_month, get_day)
displayResults(calc_zellers_congruence)
I've tried renaming them but have gotten more errors after switching them from uppercase to lowercase and also adding snake cases.
I need to check if a given date is exactly a month ago from today, for example, if today is 01-Nov-2021 then exactly a month ago will be 01-Oct-2021 (not exactly 30 days.)
I wrote a code and it works fine
today = fields.Date.from_string(fields.Date.today())
if today.month == 1:
one_month_ago = today.replace(year=today.year - 1, month=12)
else:
extra_days = 0
while True:
try:
one_month_ago = today.replace(month=today.month - 1, day=today.day -
extra_days)
break
except ValueError:
extra_days += 1
if one_month_ago == given_date:
# do something
else:
# do something
It handles well mostly all the cases but mishandles some cases. For example, the given date is 31-March-2021 and today date is 30-April-2021 and 31-April-2021 will not come to compare. I need my code to run daily and check something. It also mishandles cases of 29-31 January because 29-31 February will not come to compare.
From the given date, you can find the previous month and year and using these two obtained values, find the length of the previous month. The last thing to be done will be to compare the day of the given date with the length of the previous month and accordingly, return the desired date.
Demo:
from datetime import date
from calendar import monthrange
def date_a_month_ago(today):
x = today.month - 1
previous_month = 12 if x == 0 else x
year = today.year - 1 if x == 0 else today.year
last_day_of_previous_month = monthrange(year, previous_month)[1]
day = last_day_of_previous_month if today.day > last_day_of_previous_month else today.day
return date(year, previous_month, day)
# Tests
print(date_a_month_ago(date(2021, 11, 1)))
print(date_a_month_ago(date(2021, 1, 31)))
print(date_a_month_ago(date(2021, 12, 31)))
print(date_a_month_ago(date(2021, 3, 29)))
print(date_a_month_ago(date(2020, 3, 29)))
print(date_a_month_ago(date(2021, 3, 30)))
print(date_a_month_ago(date(2020, 3, 30)))
Output:
2021-10-01
2020-12-31
2021-11-30
2021-02-28
2020-02-29
2021-02-28
2020-02-29
ONLINE DEMO
May be like this:
from typing import Tuple
def last_month(year: int, month: int) -> Tuple[int, int]:
y, m = year, month - 1
if m == 0:
y, m = y - 1, 12
return y, m
def one_month_ago(today: datetime) -> datetime:
y, m = last_month(today.year, today.month)
dt = today.replace(year=y, month=m)
for day in range(today.day, 0, -1):
try:
return dt.replace(day=day)
except ValueError:
...
I did this and it's giving me the required output:
from datetime import date
from calendar import monthrange
def is_one_month(given_date, today):
x = today.month - 1
previous_month = 12 if x == 0 else x
year = today.year - 1 if x == 0 else today.year
last_day_of_previous_month = monthrange(year, previous_month)[1]
day = last_day_of_previous_month if today.day > last_day_of_previous_month else today.day
one_month_ago = date(year, previous_month, day)
if today.month == 2:
if given_date.month == today.month-1 and given_date.year == today.year and given_date.day >= 28:
return 'it is one month before'
if today.month == 4 or today.month == 6 or today.month == 9 or today.month == 11:
if given_date.month == today.month-1 and given_date.day == 31:
return 'it is one month before'
if one_month_ago == given_date:
return 'it is one month before'
else:
return 'it is NOT one month before'
print(is_one_month(date(2021, 1, 30), date(2021, 2, 28)))
Output:
it is one month before
I have an assignment in my computer science class and need help fixing a function but I have no clue what's wrong! The function is called 'days_left'. This function takes in three variables, Day, Month, and Year. It is supposed to output how many total days are left in that year. I have tried my best but cannot figure it out! Any help will be greatly appreciated. Here's the script:
def leap_year(year):
if (year % 4) == 0:
if (year % 100) == 0:
if (year % 400) == 0:
return True
else:
return False
else:
return True
else:
return False
def number_of_days(month, year):
days31 = [1, 3, 5, 7, 8, 10, 12]
days30 = [4, 6, 9, 11]
if month in days31:
return 31
elif month in days30:
return 30
else:
if not leap_year(year):
return 28
else:
return 29
def days_left(month, day, year):
days = 0
for i in range(1, month):
days += number_of_days(i, year)
days += day
for x in range(year):
if leap_year(year):
return 366 - days
else:
return 365 - days
if __name__ == '__main__':
print("Please enter a date: \n")
d = int(input("Day: "))
m = int(input("Month: "))
y = int(input("Year: "))
print("\nMenu:")
print("1) Calculate the number of days in the given month.")
print("2) Calculate the number of days left in the given year.")
selection = int(input())
if selection == 1:
print(number_of_days(m, y))
elif selection == 2:
print(days_left(d, m, y))
else:
print('')
def days_left(day, month, year):
It should be 'day, month' instead 'month, day' since you are calling function with days_left(d, m, y)
This is my first question here so please forgive and educate on any formatting errors. I am new to Python and going through automate the boring stuff. Decided to expand the date detection project by using the clipboard and formatting some things also. The problem I have is in any operation taken on the year part of the REGEX.
I have commented out my last attempt to validate the year and gave up and changed the REGEX to only find dates from 1000 to 2999 and skipped code validation of the dates.
I now need to validate leap years but I'm back to having to work with the year variable, but once again no operation has any effect.
Basically the problem is I can extract the year value and display it but I cannot modify it or do checks against it.
#! python3
#! detect dates in a block of text
import pyperclip
import re
#!import numpy as np
text = str(pyperclip.paste())
def datedetection(text):
dateRegex = re.compile(
r"""(
(\d|\d\d) #! match day
(/{1}) #! match /
(\d|\d\d) #! match month
(/{1}) #! match /
([1|2][0-9][0-9][0-9]) #! match year
)""",
re.VERBOSE,
)
matches = []
for groups in dateRegex.findall(text):
day = str(groups[1])
slash1 = str(groups[2])
month = str(groups[3])
slash2 = str(groups[4])
year = str(groups[5])
month_range_30 = ["04", "06", "09", "11"]
month_range_31 = ["01", "03", "05", "07", "08", "10", "12"]
month_range_Feb = ["02"]
#!year_range = np.arange(1000, 3000, 1).tolist()
if len(day) == 1:
day = "0" + day
else:
day = day
if len(month) == 1:
month = "0" + month
else:
month = month
if month in month_range_31:
if int(day) > 31:
day = "Too many days in a month with only 31 days."
slash1 = month = slash2 = year = ""
elif month in month_range_30:
if int(day) > 30:
day = "Too many days in a month with only 30 days."
slash1 = month = slash2 = year = ""
elif month in month_range_Feb:
if int(day) > 29:
day = "Too many days in February."
slash1 = month = slash2 = year = ""
elif int(month) > 12:
day = "Found an invalid month."
slash1 = month = slash2 = year = ""
elif month in month_range_Feb:
if (
int(day) == 29
and (int(year) % 4 == 0)
and (int(year) % 400 == 0)
and (int(year) % 100 == 0)
):
day = day
elif month in month_range_Feb:
if (
int(day) == 29
and (int(year) % 4 == 0)
and (int(year) % 100 != 0)
):
day = "Found an invalid leap year."
slash1 = month = slash2 = year = ""
#!elif year not in year_range:
#!day = "Year is out of range."
#!slash1 = month = slash2 = year = ""
dates = "".join([day, slash1, month, slash2, year])
matches.append(dates)
if len(matches) > 0:
pyperclip.copy("\n".join(matches))
print("Copied to clipboard:")
print("\n".join(matches))
else:
print("No dates found.")
datedetection(text)
In my approach to this project, I considered validating the days, months, and year ranges as part of the regular expression. I then defined functions to check for the leap year, and validate the number of days according to the months.
I found that way simpler and easier to understand and follow. As below:
dateRegex = re.compile(r'([0-3][0-9])/([0-1][0-9])/([1-2][0-9]{3})')
def is_leap_year(year):
year = int(year)
if year % 4 == 0:
if year % 100 == 0:
return year % 400 == 0
else:
return True
else:
return False
def is_valid_date(day, month, year):
if month == '02':
if is_leap_year(year):
return int(day) <= 29
else:
return int(day) <= 28
elif month in ['04', '06', '09', '11']:
return int(day) <= 30
else:
return int(day) <= 31
You can find the rest of my code below.
https://gist.github.com/alialbusaidi/f56e4c9342f622434f8bff0549f94884
The problem was the operations before the year operations. The day and month operations were overwriting the year values. Not entirely sure how or why at this point, but moving the year code above the day and month code has started to fix the issue.
hay guys i need to write a program that i give a date and then i give numbers of days and then i get a new date ( the numbers of days can be + or - ). i will be happy if someone help me. (about the first and second line , i just need them in the output)
specific_date =input('Enter a date:')
numofdays=input("Enter number of days:")
from datetime import datetime, timedelta
specific_date = datetime(day/month/year)
new_date = specific_date + timedelta(numofdays)
if (year % 400 == 0):
leap_year = True
elif (year % 100 == 0):
leap_year = False
elif (year % 4 == 0):
leap_year = True
else:
leap_year = False
month=['1','2','3','4','5','6','7','8','9','10','11','12']
if month in (1, 3, 5, 7, 8, 10, 12):
day in [1,31]
elif month == 2:
if leap_year:
day in[1,29]
else:
day in[1,28]
if month in (4,6,9,11):
day in [1,30]