Develop Reference

Cognitive Complexity of functions should not be too high

When I use SonarLint to check code, it notifies Critical that Cognitive Complexity is a measure of how hard the control flow of a function is to understand. Functions with high Cognitive Complexity will be difficult to maintain.
I use a lot of if else statement, but can not use switch case.
This is my code:
str_time = str_time.lower()
if (bool(re.search(r'\d', str_time)) == True) and ('tối' or 'chiều' in str_time):
day = re.findall(r'\d+', str_time)[0]
if int(day) < datetime.date.today().day:
month = datetime.date.today().month + 1
else:
month = datetime.date.today().month
year = datetime.date.today().year
day = f'{year}-{month}-{day}'
return format_datetime(day)
elif 'hôm nay' in str_time or 'hn' in str_time or 'chiều nay' in str_time or 'tối nay' in str_time:
return format_datetime(datetime.date.today())
elif 'ngày mai' in str_time or 'mai' in str_time:
day = datetime.date.today() + datetime.timedelta(days=1)
elif 'ngày mốt' in str_time or 'mốt' in str_time:
day = datetime.date.today() + datetime.timedelta(days=2)
elif 'thứ 2 tuần sau' in str_time:
num = 7 - datetime.date.today().weekday() + 0
day = datetime.date.today() + datetime.timedelta(days=num)
elif 'thứ 3 tuần sau' in str_time:
num = 7 - datetime.date.today().weekday() + 1
day = datetime.date.today() + datetime.timedelta(days=num)

Sonar lint is right. It seems your code complexity is high. You should create smaller methods or change the logic. But if that is not possible, just ignore the linter.

Finding weekday with specific date

I am learning python and going through some interactive exercises. Specifically, I'm working on Friday the 13th.
I have rewritten several iterations of this but can never seem to lock it down. With this version, it seems to get hung up when run with the simulated start date of 2025-06-12 which means there's a problem with the "this month" section. Since it returns an accurate Friday the 13th except not 2025-06-13, I suspect it's a problem with the elif statement, particularly the
and date.fromisoformat(current_year + '-' + current_month + '-13').weekday == 4:
Here's the most recent iteration of this.
def friday_the_13th():
from datetime import date
current_year = str(date.today().year)
current_month = str(date.today().month)
if len(current_month) == 1: current_month = '0' + current_month
#Function to increment to the 13th of next month
def NextMonth13(startdate):
lst_date = str(startdate)
lst_date = lst_date.split('-')
month = int(lst_date[1])
if month == 12:
year = str(int(lst_date[0]) + 1)
month = '01'
return str(year + '-' + month + '-' + '13')
else:
year = lst_date[0]
month = str(month + 1)
if len(month) == 1: month = '0' + month
return str(year + '-' + month + '-' + '13')
# Return today if today is Friday the 13th
if date.today().weekday() == 4 and date.today().day == 13:
return date.today()
# Check if this month's 13th is in the future and if it's a Friday
elif date.today().day < 13 and date.fromisoformat(current_year + '-' + current_month + '-13').weekday == 4:
return str(date.fromisoformat(current_year + '-' + current_month + '-13'))
#Check next month and return result if Friday 13
else:
result = NextMonth13(date.today())
while not (date.fromisoformat(result).weekday() == 4):
result = NextMonth13(result)
if date.fromisoformat(result).weekday() == 4:
return result
Would someone mind giving me some guidance on what I might be doing wrong?

First, your error is that you forgot the parenthesis after the weekday method call: date.fromisoformat(current_year + '-' + current_month + '-13').weekday() == 4 (FYI, date.fromisoformat(current_year + '-' + current_month + '-13').weekday returns the memory address of the method, something like this <built-in method weekday of datetime.date object at 0x7fa4e36058f0>. As you can see, it is nowhere near the result you were expecting, so it was normal for your program to behave this way.)
Second, you are needlessly complicating yourself by doing str conversions all the time:
def friday_the_13th():
from datetime import datetime, timedelta
days_passed = 0
today = datetime.today()
while True:
curr = today + timedelta(days=days_passed)
if curr.day == 13 and datetime.weekday(curr) == 4:
return str(datetime.date(curr))
days += 1
This is more readable and less prone to error as you only convert to string at the end, after you've handled all your calculations.

Not sure if it helps but to calculate the future Friday 13th you can do something like:
import datetime
def get_13th_future(startdate, months):
result=[]
year=startdate.year
month=startdate.month
checkdate = datetime.date(year=year, month=month, day=13)
for i in range(month):
if checkdate.weekday()==4:
result.append(checkdate.isoformat())
month+=1
if month==13:
year+=1
month=1
checkdate=datetime.date(year=year,month=month,day=13)
return result
startdate=datetime.datetime.now()
print(get_13th_future(startdate,1000))
If you like to search for a specific date you might construct a set instead of the list.

Update date in Python

so here is my problem, I am trying to do a little program that gives the user the next date when he will have to pay rent.
Here is my code:
curdate = datetime.date(2015, 01, 01)
rent_date = datetime.date(curdate.year, curdate.month+1, 01)
one_day = datetime.timedelta(days = 1)
one_week = datetime.timedelta(weeks = 1)
one_month = datetime.timedelta(weeks = 4)
def rent_date_calc(cd, rd):
if cd.month == 12:
rd.replace(cd.year+1, 01, 01)
else:
rd.replace(cd.year, cd.month+1, 01)
def time_pass(rd, cd, a, al):
if rd < cd:
for a in al:
a.finances -= a.rent
move_fwd = raw_input("Would you like to move forward one day(1), one week (2) or one month (3)?")
if move_fwd == "1":
curdate += one_day
elif move_fwd == "2":
curdate += one_week
else:
curdate += one_month
time_pass(rent_date, curdate, prodcomp, prodcomps)
rent_date_calc(curdate, rent_date)
print "Rent date: " + str(rent_date)
The problem I have is that rent_date always stays the same (2015-02-01)
Any idea why?

Your code is not altering anything because datetime is an immutable object, and when you call replace on it, it returns a new datetime, instead of modifying the first one.
You should return the new object from the function and assign it back to rent_date:
def rent_date_calc(cd, rd):
if cd.month == 12:
return rd.replace(cd.year+1, 01, 01)
else:
return rd.replace(cd.year, cd.month+1, 01)
...
rent_date = rent_date_calc(curdate, rent_date)

Your functions have to return a new rent date. You just need to add the following lines in your code:
return cd
new_rent_date = rent_date_calc(curdate, rent_date)
====================================================================
curdate = datetime.date(2015, 1, 1)
rent_date = datetime.date(curdate.year, curdate.month+1, 1)
one_day = datetime.timedelta(days = 1)
one_week = datetime.timedelta(weeks = 1)
one_month = datetime.timedelta(weeks = 4)
def rent_date_calc(cd, rd):
if cd.month == 12:
new_date = rd.replace(cd.year+1, 1, 1)
else:
new_date = rd.replace(cd.year, cd.month+1, 1)
return new_date
def time_pass(rd, cd, a, al):
if rd < cd:
for a in al:
a.finances -= a.rent
# Not sure what this function should return...
move_fwd = raw_input("Would you like to move forward one day(1), one week (2) or one month (3)?")
if move_fwd == "1":
curdate += one_day
elif move_fwd == "2":
curdate += one_week
else:
curdate += one_month
# time_pass(rent_date, curdate, prodcomp, prodcomps)
new_rent_date = rent_date_calc(curdate, rent_date)
print "Rent date: " + str(new_rent_date)

Is there a best way to change given number of days to years months weeks days in Python?

I am giving number of days to convert them to years, months, weeks and days, but I am taking default days to 365 and month days to 30. How do I do it in an effective way?
def get_year_month_week_day(days):
year = days / 365
days = days % 365
month = days / 30
days = days % 30
week = days / 7
day = days % 7
return year,month,week,day
def add_s(num):
if num > 1:
return 's '
return ' '
#register.filter
def daysleft(fdate):
cdate = datetime.datetime.now().date()
days = (fdate.date() - cdate).days
if days == 0:
return "Today"
elif days == 1:
return "Tomorrow"
elif days > 0:
year, month, week, day = get_year_month_week_day(days)
print year, month, week, day
days_left = ""
if year > 0:
days_left += str(year) + " year" + add_s(year)
if month > 0:
days_left += str(month) + " month" + add_s(month)
if week > 0:
days_left += str(week) + " week" + add_s(week)
if day > 0:
days_left += str(day) + " day" + add_s(day)
return days_left + " left"
else:
return "No time left"

It is much easier if you use a third-party library named python-dateutil:
>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> now = datetime.datetime.now()
>>> td = datetime.timedelta(days=500)
>>> five_hundred_days_ago = now - td
>>> print relativedelta(now, five_hundred_days_ago)
relativedelta(years=+1, months=+4, days=+13)

Checking the format/contents of a string

This program is intended to ask for the date as dd/mm/yyyy. It should then check to see if the user inputted the date in the correct format (dd/mm/yyyy). My program is not able to recognize the format correctly. This is my program:
date = (input("enter the date as dd/mm/yyyy: "))
date = day, month, year = date.split("/")
if date == (day + '/' + month + '/' + year):
print (date)
if len(day) == 1 or len(day) == 2:
print("1")
if len(month) == 1 or len(month) == 2:
print("2")
if len(year) == 4:
print ("3")
else:
if len(day) == 1 or len(day) == 2:
print("4")
if len(month) == 1 or len(month) == 2:
print("5")
if len(year) == 4:
print ("6")
The numbers being printed currently have no other purpose than to just check the validity of the date. So far, only 4,5, and 6 are being printed, meaning my program is not recognizing the formatting of the date.

Your solution doesn't work because date=day, month, year = date.split("/") sets date to a list, then you're comparing it to a string (day + '/' + month + '/' + year). However, your solution is a solved problem, do instead:
import datetime
date = (input("enter the date as dd/mm/yyyy: "))
try: datetime.datetime.strptime(date,"%d/%m/%Y")
except ValueError: # incorrect format
In addition, you probably are turning this into a datetime object later on anyway, so you can do so in the try block!
As a further optimization, be aware that many users won't WANT to enter their dates using / as a datesep! Do some introspection on your input, and adjust your datesep appropriately.
date = input("enter the date: ")
if "-" in date: datesep = "-"
elif "/" in date: datesep = "/"
elif "." in date: datesep = "."
else: datesep = ""
if len(date) < 6: yeartype = "%y"
elif date[-4:-2] not in ("19","20"): yeartype = "%y"
else: yeartype = "%Y"
try: date = datetime.datetime.strptime(date,"%d{0}%m{0}{1}".format(datesep,yeartype))
except ValueError: # invalid date
Now your code will end up with a valid datetime object of Feb 2nd 2014 for:
02022014
222014
0222014
222014
020214
02214
2214
02-02-2014
02/02/2014
2-2-14
2/2/2014
2/2/14
etc etc etc

You can use the datetime module:
import datetime
def checkdate(date):
try:
datelist = date.split('/')
datetime.datetime(year=int(datelist[2]), month=int(datelist[1]),day=int(datelist[0]))
return True
except:
return False