I am expecting some number as output from the above code,but I am not getting it out.
I am new to python but started coding with PHP.
Sorry if I go wrong some where.thanks
# By Websten from forums
#
# Given your birthday and the current date, calculate your age in days.
# Compensate for leap days.
# Assume that the birthday and current date are correct dates (and no time travel).
# Simply put, if you were born 1 Jan 2012 and todays date is 2 Jan 2012
# you are 1 day old.
#
# Hint
# A whole year is 365 days, 366 if a leap year.
def nextDay(year, month, day):
"""Simple version: assume every month has 30 days"""
if day < 30:
return year, month, day + 1
else:
if month == 12:
return year + 1, 1, 1
else:
return year, month + 1, 1
def daysBetweenDates(year1, month1, day1, year2, month2, day2):
"""Returns the number of days between year1/month1/day1
and year2/month2/day2. Assumes inputs are valid dates
in Gergorian calendar, and the first date is not after
the second."""
num = 0
# YOUR CODE HERE!
yearx = year1
monthx = month1
dayx = day1
while ((year2 >= year1 ) and ( month2 >= month1 ) and ( day2 >= day1 ) ) :
yearx,monthx,dayx = nextDay(yearx,monthx,dayx)
num = num + 1
num = '5'
return num
print daysBetweenDates(2012,9,30,2012,10,30)
You need to change the line:
while ((year2 >= year1 ) and ( month2 >= month1 ) and ( day2 >= day1 ) ) :
to:
while ((year2 >= yearx ) and ( month2 >= monthx ) and ( day2 >= dayx ) ) :
because you are not changing the value of month1 in your code but that of monthx.
Also, I think your while loop will break when dayx is greather that day2, so your measurement will be off by 1.
I have never mastered the while statement in Python but I think that is your infinite loop it is always true that day2 > day1 etc. So that condition remains true therefore you are stuck with num increasing
What does happen - do you get any error message?
if I were doing this I would set functions to determine
if the years are the same
if the years are the same then calculate the days between them
if the years are not the same calculate the number of days between the first date and the end of the year for that particular year
Calculate the number of days between the beginning of the year of the second date to the second date
Calculate the number of years difference between the end of the first year and the beginning of the second year and convert that into days
It may be clunky but it should get you home
Here is my solution in just one function
def daysBetweenDates(year1, month1, day1, year2, month2, day2):
##
# Your code here.
##
a1=str(month1)+'/'+str(day1)+'/'+str(year1)
date1 = datetime.strptime(a1, '%m/%d/%Y')
a2=str(month2)+'/'+str(day2)+'/'+str(year2)
date2 = datetime.strptime(a2, '%m/%d/%Y')
result= (date2 - date1).days
return result
Related
The goal of the program is to:
validate inputs formats, and the date range should not be more than 2 months
No input/invalid input -> return the first or second half of the current month based on the day of the month
For the first half is from day 1 to day 15 for the second half is from day 16 till the end of the month
The hardcode, which will be the first code snippet, works as expected with the correct output, however the date validation in the second code snippet does not jump to the method indicated in the except block, which is causing me problems and not displaying correct output.
import os, sys
import datetime as dt
if __name__ == '__main__':
validate_range('2003/12/23', '2003/12/22')
validate_range('2003/10/23', '2003/11/22')
validate_range('2003/12/23', '2003/10/22')
validate_range('2003/7/23', '2003/10/22')
Code snippet with correct output (does not check if dates are correct format):
def validate_range(first_date, second_date):
start_date = dt.datetime.strptime(first_date, '%Y/%m/%d') # Get first date
end_date = dt.datetime.strptime(second_date, '%Y/%m/%d') # Get second date
num_months = (end_date.year - start_date.year) * 12 + (
end_date.month - start_date.month) # Get year/month within 2 months
if num_months >= 0 and num_months <= 2:
print("In Range")
else:
current_day = dt.datetime.today().day
if current_day >= 1 and current_day <= 15:
print("First half")
if current_day >= 16 and current_day <= 32:
print("Second half")
Correct Output:
In Range
In Range
First half
First half
Code snippet that does not display correct output (with date validation):
def read_args(first_date, second_date):
try:
start_date = dt.datetime.strptime(first_date, '%Y/%m/%d') # Get first date
end_date = dt.datetime.strptime(second_date, '%Y/%m/%d') # Get second date
v_range(start_date, end_date)
except:
w_range()
def v_range(first_date, second_date):
num_months = (second_date.year - first_date.year) * 12 + (
second_date.month - first_date.month) # Get year/month within 2 months
if num_months >= 0 and num_months <= 2:
print("In Range")
def w_range():
current_day = dt.datetime.today().day
if current_day >= 1 and current_day <= 15:
print("First half")
if current_day >= 16 and current_day <= 32:
print("Second half")
Output:
In Range
In Range
Keeping with your use of try / except, might need some improvement, but this example demonstrates it is possible:
import os, sys
import datetime as dt
def read_args(first_date, second_date):
try:
start_date = dt.datetime.strptime(first_date, '%Y/%m/%d') # Get first date
end_date = dt.datetime.strptime(second_date, '%Y/%m/%d') # Get second date
result = v_range(start_date, end_date)
except:
pass
# w_range()
def v_range(first_date, second_date):
num_months = (second_date.year - first_date.year) * 12 + (
second_date.month - first_date.month) # Get year/month within 2 months
if num_months >= 0 and num_months <= 2:
print("In Range")
else:
w_range()
def w_range():
current_day = dt.datetime.today().day
if current_day >= 1 and current_day <= 15:
print("First half")
if current_day >= 16 and current_day <= 32:
print("Second half")
if __name__ == '__main__':
read_args('2003/12/23', '2003/12/22')
read_args('2003/10/23', '2003/11/22')
read_args('2003/12/23', '2003/10/22')
read_args('2003/7/23', '2003/10/22')
def isMagicDate(day,month,year):
if day* month == year %100:
return True
return False
def magic():
for year in range(1900,2000):
for month in range(1,13):
for day in range(1,month,year+1):
if isMagicDate(day,month,year):
print("%02d/%02d/%04d is a magic date "%(day,month,year))
magic()
I need to fix this code he's not giving all magic dates in 1900 to 2000
Using monthRange to adjust for the different days by month (i.e. Feb vs. January)
from calendar import monthrange
def isMagicDate(day, month, year):
return True if day*month == year % 100 else False
def main():
for year in range(1900, 2000):
for month in range(1, 13):
for day in range(1, monthrange(year, month)[1] + 1):
if isMagicDate(day, month, year):
print(f"{day:02d}/{month:02d}/{year:04d} is magic date.")
main()
def isMagicDate(day,month,year):
if day* month == year %100:
return True
else:
return False
def magic():
for year in range(1900,2000):
for month in range(1,13):
for day in range(1,30):
if isMagicDate(day,month,year):
print("%02d/%02d/%04d is a magic date "%(day,month,year))
magic()
TRY THIS!!
U have to put else after the if statement or the second statement will be excuted always,and also every month has 30 or 31 days(my code only take 30 days for every month).
We all know that in the Gregorian calendar that we are currently using, a random day of the week can be one of 7 options.
The intention of my code is based on a segment in Arthur Benjamin's Think Like A Math Genius (2006), whereby you can perform a simple party trick using mathematics. Using codes (basically implementing shortcuts based on the patterns of the Gregorian and finding the remainders after dividing by seven), one can quickly and accurately predict past and future days of the week.
My current attempt results in PyCharm overcomputing and eating up majority of CPU activity. I reckon that this could be because I have not limited the extent of user input 'Year' the same way I limited 'typedMonth'.
This is the original code:
print("Enter the full date: in dd/Month/yyyy")
Date = int(input("dd:"))
typedMonth = str(input("Month:"))
Year = int(input("yyyy: "))
while not int(Year) in range(0,3000):
Year = int(input("Please enter year (yyyy) limited to [0 - 3000] : "))
if typedMonth in ['January']:
Month = 3
while int(Year) % 4 == 0:
Month = 2
elif typedMonth in ['February']:
Month = 6
while int(Year) % 4 == 0:
Month = 5
elif typedMonth in ['March', 'November']:
Month = 6
elif typedMonth in ['April', 'July']:
Month = 2
elif typedMonth in ['May']:
Month = 4
elif typedMonth in ['August']:
Month = 5
elif typedMonth in ['September', 'December']:
Month = 1
elif typedMonth in ['October']:
Month = 3
elif typedMonth in ['June']:
Month = 0
else:
Month = 0
remYear = Year%400
if 300 <= remYear <= 400:
Year = Year + 1
elif 200 <= remYear <= 300:
Year = Year + 3
elif 100 <= remYear <= 200:
Year = Year + 5
else:
Year = Year + 0
print(remYear)
Day = (Date + int(Month) + remYear) % 7
print(Day)
Isolated testing shows that the first four lines of code work as intended (if I just print typedMonth), for instance.
In silo, the segment near the end from remYear= Year%400 also works as intended.
Hence, to my untrained eye, the 'overcomputation' is likely due to the stretch of code that I use to fix the Month variable from user input string typedMonth. To emphasise, the initial input for Month starts with the string typedMonth. I then use while statements to convert user input of string to an integer value. This integer value is attributed to variable 'Month' so that the final computation of integers 'Date', 'Month' and 'Year' can be made. Remainder will correspond to day of the week (e.g. 0 = Sunday, 1 = Monday, 2 = Tuesday, etc).
This is the updated code (third iteration). Thanks to the comments below, I am now able to enter most years as input and the program will run. Logic errors still persist, especially with the codes assigned to the months. I will fix them in future iterations.
print("Enter the full date: in dd/Month/yyyy")
Date = int(input("dd:"))
typedMonth = str(input("Month:"))
Year = int(input("yyyy: "))
while not int(Year) in range(1,3000):
Year = int(input("Please enter year (yyyy) limited to [0 - 3000] : "))
if typedMonth in ['January']:
Month = 3
while int(Year) % 4 == 0:
Month = 2
elif typedMonth in ['February']:
Month = 6
while int(Year) % 4 == 0:
Month = 5
elif typedMonth in ['March', 'November']:
Month = 6
elif typedMonth in ['April', 'July']:
Month = 2
elif typedMonth in ['May']:
Month = 4
elif typedMonth in ['June']:
Month = 0
elif typedMonth in ['August']:
Month = 5
elif typedMonth in ['September', 'December']:
Month = 1
elif typedMonth in ['October']:
Month = 3
remYear = Year%400
if 300 <= remYear <= 400:
Year = Year + 1
elif 200 <= remYear <= 300:
Year = Year + 3
elif 100 <= remYear <= 200:
Year = Year + 5
else:
Year = Year + 0
calcYear = int(Year % 100)
Day=(Date + Month + calcYear) % 7
print("The date is %d"%Date)
print("The month is %d"%Month)
print("The calculated year is %d"%calcYear)
print("The code for day is %d"%Day)
if Day in [0]:
print("You were born on a Sunday")
elif Day in [1]:
print("You were born on a Monday")
elif Day in [2]:
print("You were born on a Tuesday")
elif Day in [3]:
print("You were born on a Wednesday")
elif Day in [4]:
print("You were born on a Thursday")
elif Day in [5]:
print("You were born on a Friday")
elif Day in [6]:
print("You were born on a Saturday")
print("~~End Of Program~~")
Recommendations for restricting the range of computation of Year (limiting user input for Year does not fix the above problems) or restructuring the value for variable month from initial string input 'typedMonth' would be greatly appreciated.
P.S.1 Perhaps I should use another software for this project? I just started out with python so i don't know what applications would be more suited for such programs. Do recommend if you think that this could be a viable solution!
P.S.2 Made progress! The program kind of works for millennial babies (aka enter year after 2001), i still need to fix the logic!
P.S.3 Thanks to the comments below, I am now able to run the program. There are still glaring logic errors within the code. I endeavor to commence work on this project when time allows. I will update this post accordingly. Thanks for joining me in my learning journey!
You asked if Python was a suitable tool for this sort of project. It is, but it is much more suitable if you use it properly. In other words, don't make the problem unnecessarily difficult.
I understand your impulse to implement a particular algorithm to get comfortable with programming, but I have to say that this isn't a good choice. Date arithmetic is usually a pain. If you want to use computational shortcuts (and you should), then take a look at the shortcuts that the Python programming environment offers.
For example:
>>> import datetime
>>> from dateutil import parser
>>> date = "25/December/1999"
>>> datetime.datetime.strftime(parser.parse(date), "%A")
'Saturday'
The dateutil module isn't part of the standard library but it is well worth the trouble of installing.
When ever I ran this the output is always 0 where i was expecting like 29 or 30.
import datetime
def days_in_month(year, month):
"""
Inputs:
year - an integer between datetime.MINYEAR and datetime.MAXYEAR
representing the year
month - an integer between 1 and 12 representing the month
Returns:
The number of days in the input month.
"""
#if month is december, we proceed to next year
def month_december(month):
if month == 12:
return 1
else:
return month
#if month is december, we proceed to next year
def year_december(year, month):
if month == 12:
return new_year + 1
else:
return year
#verify if month/year is valid
if (month < 1) or (month > 12):
print ("please enter a valid month")
elif (year < 1) or (year > 9999):
print ("please enter a valid year between 1 - 9999")
else:
#subtract current month from next month then get days
date1 = (datetime.date(year_december(year, month), month_december(month), 1) - datetime.date(year, month, 1)).days
print (date1)
days_in_month(1997, 1)
As daniel told there is a standard library method. It is always better to reuse than reinvent.
import calender
def days_in_month(year, month):
if (month < 1) or (month > 12):
print ("please enter a valid month")
elif (year < 1) or (year > 9999):
print ("please enter a valid year between 1 - 9999")
else:
return calendar.monthrange(year, month)[1]
You forgot to add one month at first parameter.
Your codes:
datetime.date(year_december(year, month), month_december(month), 1)
= datetime.date(year, month, 1)
New codes:
date1 = (datetime.date(year_december(year, month+1), month_december(month+1), 1) - datetime.date(year, month, 1)).days
But pay an attention on the result will be wrong if December, you need to improve it like below:
Full codes:
import datetime
def days_in_month(year, month):
"""
Inputs:
year - an integer between datetime.MINYEAR and datetime.MAXYEAR
representing the year
month - an integer between 1 and 12 representing the month
Returns:
The number of days in the input month.
"""
#if month is december, we proceed to next year
def month_december(month):
if month > 12:
return month-12 #minus 12 if cross year.
else:
return month
#if month is december, we proceed to next year
def year_december(year, month):
if month > 12:
return year + 1
else:
return year
#verify if month/year is valid
if (month < 1) or (month > 12):
print ("please enter a valid month")
elif (year < 1) or (year > 9999):
print ("please enter a valid year between 1 - 9999")
else:
#subtract current month from next month then get days
date1 = (datetime.date(year_december(year, month+1), month_december(month+1), 1) - datetime.date(year, month, 1)).days
print (date1)
days_in_month(1997, 12)
days_in_month(1998, 1)
days_in_month(1998, 2)
Test Case
days_in_month(1997, 12)
days_in_month(1998, 1)
days_in_month(1998, 2)
Output
31
31
28
[Finished in 0.187s]
There is a really cool method I like, and you don't have to import anything. You just need to have two one-line functions. For input you need the year (as integer) and month (as integer) - which both comes from the date it self.
So:
function: check if it's a leap year or not.
function: create a list of the days (corrected with leap year) and call the item of it with the month.
def is_leap_year(year):
return ((year % 4 == 0) and (year % 100 != 0)) or (year % 400 == 0)
def get_days_in_month(year, month):
return [31, (29 if is_leap_year(year) else 28), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31][month - 1]
Check it out.
I have a procedure that calculates age in days. I am thinking of using it to help me calculate my pay days.
I want it to ask for date 1 and then date 2, which would also be the current date.
However, I want it to be fancier and not even ask for date 2.
I want to use (time.strftime('%Y,%-m,%-d')) for the input of the second set of dates when I run this code.
I already have the code built for calculating my total net income from my 3 jobs.
y1,m1,d1 = 1994,4,1 #Birth_Date
y2,m2,d2 = 2014,1,1 #Current_Date
print daysBetweenDates(y1,m1,d1,y2,m2,d2)
My problem is that if I made y2,m2,d2 = (time.strftime('%Y,%-m,%-d'))
it will print it out with quotes at end so --> '2014,2,3'
and then everything just falls apart. I have been looking everywhere to find the solution
and everyone keeps talking about the datetime stuff but my problem with that is that is returns
datetime.strptime('2014-12-04', '%Y-%m-%d').date()
datetime.date(2014, 2, 3)
not only the date but also the datetime.date
I then tried
s = datetime.strptime('2014-12-04', '%Y-%m-%d').date()
s = string [16:-1] #
to remove all excess stuff but that didn't work also.
Please help me figure this one out guys.
I want y2,m2,d2 = current date without quotations.
You could calculate age in day(days between date) with the following:
import datetime
start_date = datetime.datetime.strptime('1975-10-03', '%Y-%m-%d').date()
end_date = datetime.datetime.strptime('2014-12-05', '%Y-%m-%d').date()
day_between_dates = (end_date - start_date).days
>> 14307
Intro to Computer Science Problem: Age in Days
I should have informed everyone of the constraints of: No Imports
I apologize for that, it was my first question I ever posed and had no idea how to pose them correctly.
My solution below without any imports:
def isLeapYear(year):
if year % 4 == 0:
if year % 100 != 0:
return True
else:
if year % 400 == 0:
return True
return False
def daysInMonth(year,month):
if month == 1:
return 31
if month == 2:
if isLeapYear(year) == True:
return 29
return 28
if month == 3:
return 31
if month == 4:
return 30
if month == 5:
return 31
if month == 6:
return 30
if month == 7:
return 31
if month == 8:
return 31
if month == 9:
return 30
if month == 10:
return 31
if month == 11:
return 30
if month == 12:
return 31
def nextDay(year, month, day):
"""Simple version: assume every month has 30 days"""
if day < daysInMonth(year,month):
return year, month, day + 1
else:
if month == 12:
return year + 1, 1, 1
else:
return year, month + 1, 1
def dateIsAfter(year1, month1, day1, year2, month2, day2):
"""Returns True if year1-month1-day1 is after year2-month2-day2. Otherwise, returns False."""
if year1 > year2:
return True
if year1 == year2:
if month1 > month2:
return True
if month1 == month2:
return day1 > day2
return False
def daysBetweenDates(year1, month1, day1, year2, month2, day2):
"""Returns the number of days between year1/month1/day1
and year2/month2/day2. Assumes inputs are valid dates
in Gregorian calendar."""
# program defensively! Add an assertion if the input is not valid!
assert dateIsAfter(year2, month2, day2, year1, month1, day1)
days = 0
while dateIsAfter(year2, month2, day2, year1, month1, day1):
days += 1
(year1, month1, day1) = nextDay(year1, month1, day1)
return (days)
y1,m1,d1 = 1994,7,14 #Birth_Date
y2,m2,d2 = 2014,1,1 #Current_Date
print daysBetweenDates(y1,m1,d1,y2,m2,d2)
My answer: 1 Year & 7 Months later :).
I had solved it, but just forgot to post it as an answer in the community.
I am leaving this here if anyone else might find this useful to learn from.
P.S However, thank you #user3277225 for your answer. It is a very elegant and simple solution that I might use later when I need to.