python - 计算当前月份的星期日数

Question

如何在 Python 中获取当月的星期日数？

有人对此有任何想法吗？

Answer 1

这为您提供了您想要的当前月份的星期日数：

import calendar
from datetime import datetime

In [367]: len([1 for i in calendar.monthcalendar(datetime.now().year,
                                  datetime.now().month) if i[6] != 0])
Out[367]: 4

Answer 2

我碰巧需要一个解决方案，但对这里的解决方案不满意，所以我想出了自己的：

import calendar

year = 2016
month = 3
day_to_count = calendar.SUNDAY

matrix = calendar.monthcalendar(year,month)

num_days = sum(1 for x in matrix if x[day_to_count] != 0)

Answer 3

我的看法：（不必担心在正确的月份等......）

from calendar import weekday, monthrange, SUNDAY

y, m = 2012, 10

days = [weekday(y, m, d+1) for d in range(*monthrange(y, m))]
print days.count(SUNDAY)

或者，正如@mgilson所指出的，您可以取消 list-comp，并将其全部包装为生成器：

sum(1 for d in range(*monthrange(y,m)) if weekday(y,m,d+1)==SUNDAY)

我想，你可以输入一个：

from collections import Counter
days = Counter(weekday(y, m, d + 1) for d in range(*monthrange(y, m)))
print days[SUNDAY]

Answer 4

我会这样做：

import datetime

today = datetime.date.today()
day = datetime.date(today.year, today.month, 1)
single_day = datetime.timedelta(days=1)

sundays = 0
while day.month == today.month:
    if day.weekday() == 6:
        sundays += 1
    day += single_day

print 'Sundays:', sundays

Answer 5

另一个使用calendarand 的例子datetime：

import datetime
import calendar
today = datetime.date.today()
m = today.month
y = today.year
sum(1 for week in calendar.monthcalendar(y,m) if week[-1])

也许一个稍微快一点的方法是：

first_day,month_len = monthrange(y,m)
date_of_first_sun = 1+6-first_day
print sum(1 for x in range(date_of_first_sun,month_len+1,7))

Answer 6

import calendar

MONTH = 10    
sundays = 0
cal = calendar.Calendar()

for day in cal.itermonthdates(2012, MONTH):
    if day.weekday() == 6 and day.month == MONTH:
        sundays += 1

PAY ATTENTION: Here are the Calendar.itermonthdates's docs:

Return an iterator for one month. The iterator will yield datetime.date values and will always iterate through complete weeks, so it will yield dates outside the specified month.

That's why day.month == MONTH is needed

If you want the weekdays to be in range 0-6, use day.weekday(), if you want them to be in range 1-7, use day.isoweekday()

Answer 7

您可以使用 ISO 周数执行此操作：

from datetime import date
bom = date.today().replace(day=1)                 # first day of current month
eom = (date(bom.year, 12, 31) if bom.month == 12 else
       (bom.replace(month=bom.month + 1) - 1))    # last day of current month
_, b_week, _ = bom.isocalendar()
_, e_week, e_weekday = eom.isocalendar()
num_sundays = (e_week - b_week) + (1 if e_weekday == 7 else 0)

一般来说，对于一周中的某一天（1 = 星期一，7 = 星期日），计算如下：

num_days = ((e_week - b_week) +
            (-1 if b_weekday > day else 0) +
            ( 1 if e_weekday >= day else 0))

Answer 8

我的解决方案。

以下内容受到@Lingson 回答的启发，但我认为它的循环较少。

import calendar

def get_number_of_weekdays(year: int, month: int) -> list:
    main_calendar = calendar.monthcalendar(year, month)
    number_of_weeks = len(main_calendar)
    number_of_weekdays = []

    for i in range(7):
        number_of_weekday = number_of_weeks

        if main_calendar[0][i] == 0:
            number_of_weekday -= 1
        
        if main_calendar[-1][i] == 0:
            number_of_weekday -= 1

        number_of_weekdays.append(number_of_weekday)

    return sum(number_of_weekdays)  # In my application I needed the number of each weekday, so you could return just the list to do that.