I developing a course app. In my app, if a student register for a new course, the teacher must input the starting course date for the student.
And in this app, the student can pick up the day that they want to learn to the teacher.
e.g: The student wants to learn 3 days for each week, Monday, Tuesday and Friday.
And for the day that the student taking, they must pay to the teacher on that day.
So, I want my app can automatically display on what day a student must pay the tuition to the teacher.
The thing that I need here is: if the teacher inputted the starting course date on 20-10-2019 (%d-m-Y%), and the days that the student taking are Monday, Tuesday and Friday, I want to print: print('please, pay it') every week on the selected day.
So, I want something like this:
import datetime
starting_course_date = datetime.datetime.strptime("20-10-2019", "%d-%m-%Y")
student_taking_days = [{'day': 'Monday'}, {'day': 'Tuesday'}, {'day': 'Friday'}]
today = datetime.datetime.utcnow()
if today > starting_course_date and today is one of day on the student_taking_days:
print('please, pay it')
Please, any answer, source or refer tutorial will be very appreciated :)
You can get weekday of your today. More info here
import datetime
weekday = datetime.datetime.today().weekday()
if (weekday in [0, 1, 4]): // 0 is monday, 1 is tuesday and 4 is friday
print('please, pay it')
You event can get day name like this using strftime:
weekday_name = weekday.strftime("%A")
if (weekday_name in ['Monday', 'Tuesday', 'Friday']):
print('please, pay it')
weekday = today.weekday()
This gives the result as an integer starting from Monday = 0.
You could directly use this variable in your 'if' statement.
if weekday == 0 or weekday == 1 or weekday == 4:
print('please, pay it')
Related
I'm new to python and new to pandas. Of course, if my project used exact dates, I could easily do this, but unfortunately, the date type is a little different, and as you can see, the sign 08 is after the year 1401, which means it is the eighth month of the year 1401.
I currently know that these 3 customers have bought from me this month. But I want to know if these 3 customers bought from me in the previous month or two months ago? If they do, I will give them a discount.
Of course, I should also say that the number 08 is not always fixed, but it could be 09 in the next month. I just want to know if they bought from me 1 month ago or not?
According to the picture, now only Sara should get a discount
You could convert the purchase date to an integer and calculate the number of months from there.
For instance, you have the purchase month 1901/07 and you want to know in 1901/08 how many months the last purchase took place. So you convert both values to integers and subtract them (190108 - 190107 = 1).
import pandas as pd
df = pd.DataFrame({'customer': ['david', 'sara'], 'date': ['1901/03', '1901/07']})
# Manually setting the reference month (190108 for Year 1901 and Month 08)
df['months'] = 190108 - df['date'].replace('/', '', regex=True).astype(int)
# Check if eligible for discount
df['discount'] = df['months'].isin([1, 2])
customer
date
months
discount
0
david
1901/03
5
False
1
sara
1901/07
1
True
To compare with today's month you could to the following:
df['months'] = int(pd.Timestamp.now().strftime('%Y%m'))\
- df['date'].replace('/', '', regex=True).astype(int)
Problem Statement:
Am developing a custom job scheduler that needs to be run on given days. It takes start date and end date as string and third param is list of week days on which job should run.
Start day can be different with given days but first job should run on next valid day
Let suppose Start date is 2022-09-07 (so day name is Wednesday) but given frequency days are ["Monday", "Friday", "Saturday"] so i need to run my first job on coming Friday and for this i need to calculate difference between start date and first valid day (in this case it's Friday)
So how can i do this python to run my first job on valid day (that can be in any position of given frequency days list) and also after one job complete i need to also get next valid day. I did some work but unfortunately its not working. Here is what i did
sorted_week_days_list = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
start_date = "2022-09-07"
valid_frequency_days = ["Monday", "Tuesday", "Friday"] # It can be any days in sorted order
start_date_object = datetime.datetime.strptime(start_date, "%Y-%m-%d")
given_start_day = start_date_object.strftime("%A")
if given_start_day not in valid_frequency_days:
# Need help to implement logic to get date for valid day
You should use the datetime.weekday() method to pull out the day of the week for days of interest. Assuming that you have dates similar to the format you show above, it is easy to convert, and also just use the day index for your "allowable start days" (Monday=0).
Then you can jig up a little function to look for the next start date in your sorted list and figure out how many days you need to wait.
Example below does that and also "rolls over" the weekend as needed.
Code:
from datetime import datetime, timedelta
from bisect import bisect_left
start_date = "2022-09-09"
valid_start_dates = [1, 4] # It can be any days in sorted order
start_date_object = datetime.strptime(start_date, "%Y-%m-%d")
d=start_date_object.weekday()
print(f'the numbered day of the week is: {d}')
def days_till_start(day, valid_start_days):
idx = bisect_left(valid_start_days, day)
if idx >= len(valid_start_days): # wrap around to next start
return valid_start_days[0] + 7 - day
elif day == valid_start_days[idx]:
return 0
else:
return valid_start_days[idx] - day
print(days_till_start(d, valid_start_dates))
start_dates = ['2022-09-05', '2022-09-06', '2022-09-07', '2022-09-08', '2022-09-09', '2022-09-10']
start_wkdys = [datetime.strptime(d, "%Y-%m-%d").weekday() for d in start_dates]
for d in start_wkdys:
print(f'day index is: {d}')
print(f'next start date is {days_till_start(d, valid_start_dates)} away')
print()
Output:
the numbered day of the week is: 4
0
day index is: 0
next start date is 1 away
day index is: 1
next start date is 0 away
day index is: 2
next start date is 2 away
day index is: 3
next start date is 1 away
day index is: 4
next start date is 0 away
day index is: 5
next start date is 3 away
I want to make the date of startOf("week") its start from Monday on current week
Is it possible for make this? My code:
moment().startOf("week").toDate();
This startOf start from saturday proof startOf start from saturday
Take a look at the First Day of Week and First Week of Year section
From the docs (emphasis is mine)
Locale#week.dow should be an integer representing the first day of the week, 0 is Sunday, 1 is Monday, ..., 6 is Saturday.
So just by running this code before doing any calls to moment
// From 2.12.0 onward
moment.updateLocale('en', {
week : {
dow : 1
}
});
It will set the first day of the week to Monday.
After that calling moment().startOf("week").toDate() will give you a Monday.
What I want to do is figure out what X of the month this is, and returning the relative delta constant for it (Su Mo Tu...). I have found many examples of jumping to a specific day of the month (1). For instance today is the 3rd Tuesday of December and I can get to it by doing this: + relativedelta(month=12, day=1, weekday=TU(3))) but what I want to do is the opposite:
Put in today's date and subtract the first of the month and get something like TU(3) or if it were the 4th wednesday to get: WE(4)
My ultimate goal is to then be able to transfer this constant to a different month or timedelta object and find the equivalent 3rd Tuesday, or 4th Wednesday, etc...
This is a solution that I have come up with, maybe you'll find it less complicated.
It also seems to be about 4 times faster, which if you process a lot of dates can make a difference.
from datetime import *
from dateutil.relativedelta import *
def weekDayOfTheMonth(xdate):
daylist = [MO,TU,WE,TH,FR,SA,SU]
weekday = xdate.weekday()
firstDayOfTheMonth = datetime(xdate.year, xdate.month, 1)
interval = (weekday + 7 - firstDayOfTheMonth.weekday() ) % 7
firstOfThisWeekDay = datetime(xdate.year, xdate.month, 1 + interval)
n = ((xdate.day - firstOfThisWeekDay.day) / 7) + 1
return daylist[weekday](n)
print(weekDayOfTheMonth(datetime.today()))
print(weekDayOfTheMonth(datetime(2018,11,24)))
Basically what happens is that:
I find what day of the week is the first day of given month.
Based on that information I can easily calculate first day of any given weekday in given month.
Then I can even more easily calculate that for example 18th of December 2018 is third Tuesday of this month.
Ok I found a way using rrule to create a list of days in the month that share the current weekday up until today, then length of this list becomes the Nth. Than I use a list as a lookup table for the weekday constants. Not tested to see if this will work for every day of the month but this is a start.
from datetime import *; from dateutil.relativedelta import *
from dateutil.rrule import rrule, WEEKLY
import calendar
def dayofthemonth(xdate):
daylist = [MO,TU,WE,TH,FR,SA,SU]
thisweekday = daylist[xdate.weekday()]
thisdaylist = list(rrule(freq=WEEKLY, dtstart=xdate+relativedelta(day=1), until=xdate, byweekday=xdate.weekday()))
return thisweekday(len(thisdaylist))
print(dayofthemonth(datetime.today())) #correctly returns TU(+3) for 2018, 12, 18
I would like to calculte the number of the week in a year. I see this post
In this post the acepted answer use this code:
public static int GetIso8601WeekOfYear(DateTime time)
{
// Seriously cheat. If its Monday, Tuesday or Wednesday, then it'll
// be the same week# as whatever Thursday, Friday or Saturday are,
// and we always get those right
DayOfWeek day = CultureInfo.InvariantCulture.Calendar.GetDayOfWeek(time);
if (day >= DayOfWeek.Monday && day <= DayOfWeek.Wednesday)
{
time = time.AddDays(3);
}
// Return the week of our adjusted day
return CultureInfo.InvariantCulture.Calendar.GetWeekOfYear(time, CalendarWeekRule.FirstFourDayWeek, DayOfWeek.Monday);
}
However I see a problem. The number of the week is repeated if the last day of a month is in Sunday.
For example, the last day of March of 2013 year is in sunday. This week is the number 13th, is correct. But in April, how C# use always 6 weeks in a month to calculate the number of week, the first week of april has not any day of april, because all the days belong to march because the last day of the week is 30th March. So C# says that the first week of april is th 15th week, but this is incorrect, it has to be 14th.
So I would like to know if there are any way to calculate the number of a week in a right way.
EDIT:
I mean this:
In march, the last week is this:
25 26 27 28 29 30 31
This is the 13th, is correct.
In april, the first week is:
1 2 3 4 5 6 7
And this week is calculated as 15th.
So if I see the march calendar the last week is calculated as 13th and if I see the april calendar the last week of march is caluclated as 14th. This is incorrect.
SOLUTION:
DateTime dtCalendar = Calendar.DisplayDate;
int gridRow = (int)GetValue(Grid.RowProperty);
// Return the week of our adjusted day
int wueekNumber= System.Globalization.CultureInfo.InvariantCulture.Calendar.GetWeekOfYear(dtCalendar, System.Globalization.CalendarWeekRule.FirstDay, DayOfWeek.Monday);
if (dtCalendar.DayOfWeek == DayOfWeek.Monday)
{
gridRow = gridRow - 1;
}
Text = (weekNumbe r+ gridRow - 1).ToString();
Thanks.
The problem is that you are using the wrong CalendarWeekRule. To get the result you want you should use FirstDay. I have seen various codes in internet saying that you should use FirstFourDayWeek but, after some testing, I realised that the "right one" is FirstDay. I have tested it with your example and it delivers the right result: 14th week.
int targetYear = 2013;
DateTime targetDate = new DateTime(targetYear, 4, 1);
int week = System.Globalization.CultureInfo.CurrentCulture.Calendar.GetWeekOfYear(targetDate, System.Globalization.CalendarWeekRule.FirstDay, DayOfWeek.Monday);