如何获取日期之间的周期？

Question

我有两个月的时间，从开始到结束。我想迭代之间的所有月份，即我想将所有月份提取为循环中的字符串。start = '1905' # May 2019end = '2003' # March 2020for month in range(start, end):     # I want to get string of each month in same format here有人可以帮助我干净有效地获得这个吗？我的一年总是在（2000-2099）之间

Answer 1

period_range与将日期字符串转换为日期时间一起使用：

start = '1905' # May 2019

end = '2003' # March 2020

p = pd.period_range(pd.to_datetime(start, format="%y%m"), 

                    pd.to_datetime(end, format="%y%m"), freq='M')

print (p)

PeriodIndex(['2019-05', '2019-06', '2019-07', '2019-08', '2019-09', '2019-10',

             '2019-11', '2019-12', '2020-01', '2020-02', '2020-03'],

            dtype='period[M]', freq='M')

然后用于PeriodIndex.strftime自定义格式：

print (p.strftime('%B'))

Index(['May', 'June', 'July', 'August', 'September', 'October', 'November',

       'December', 'January', 'February', 'March'],

      dtype='object')

for val in p:

    print (val)

Answer 2

一个简单的解决方案是这样的：

months = {

            '01': 'Jan',

            '02': 'Feb',

            '03': 'Mar',

            '04': 'Apr',

            '05': 'May',

            '06': 'Jun',

            '07': 'Jul',

            '08': 'Aug',

            '09': 'Sep',

            '10': 'Oct',

            '11': 'Nov',

            '12': 'Dec'

        }

start = '1905'

end = '2003'

for i, j in enumerate(range(int(start[0:2]), int(end[0:2])+1)):

  if i == 0 and start[0:2] != end[0:2]:

    s = int(start[2:4])

    e = 12

  if i > 0 and start[0:2] != end[0:2] and j != int(end[0:2]):

    s = 1

    e = 12

  if i > 0 and start[0:2] != end[0:2] and j == int(end[0:2]):

    s = 1

    e = int(end[2:4]) + 1

    

  if i == 0 and start[0:2] == end[0:2]:

    s = int(start[2:4])

    e = int(end[2:4]) + 1

  for p in range(s, e):

    index = '0' + str(p) if p < 10 else str(p)

    print(f"Year: {j}, month {months[index]}")

    print(f"{j}{index}")

输出：

Year: 19, month May

1905

Year: 19, month Jun

1906

Year: 19, month Jul

1907

Year: 19, month Aug

1908

Year: 19, month Sep

1909

Year: 19, month Oct

1910

Year: 19, month Nov

1911

Year: 20, month Jan

2001

Year: 20, month Feb

2002

Year: 20, month Mar

2003

Answer 3

您可以使用date及其格式：

from datetime import date, datetime

def parse(strg):

    return datetime.strptime(strg, "%y%m").date()

start = parse("1905")

end = parse("2003")

dt = start

while dt <= end:

    print(f"{dt:%y%m}  -  {dt:%B %Y}")

    dt = date(year=dt.year + dt.month // 12, month=(dt.month % 12) + 1, day=1)

 

这将产生字符串

1905  -  May 2019

1906  -  June 2019

1907  -  July 2019

1908  -  August 2019

1909  -  September 2019

1910  -  October 2019

1911  -  November 2019

1912  -  December 2019

2001  -  January 2020

2002  -  February 2020

2003  -  March 2020

date使用格式字符串格式化对象"%y%m"将返回您想要的格式。

Answer 4

像这样的生成器函数应该可以解决问题。

def parse_yymm_string(yymm_string):

    # Parse an `YYMM` string into a tuple of (Year, Month)

    return (2000 + int(yymm_string[:2], 10), int(yymm_string[2:], 10))

def ym_pair_to_yymm(ym_pair):

    # Convert a (Year, Month) tuple into an `YYMM` string

    return f"{ym_pair[0] - 2000:02d}{ym_pair[1]:02d}"

def generate_yms(start, end): 

    parsed_end = parse_yymm_string(end)

    curr_year, curr_month = parse_yymm_string(start)

    while (curr_year, curr_month) <= parsed_end:

        yield (curr_year, curr_month)

        curr_month += 1

        if curr_month > 12:

            curr_year += 1

            curr_month = 1

start = '1905' # May 2019

end = '2003' # March 2020

for ym in generate_yms('1905', '2003'):

    print(ym, ym_pair_to_yymm(ym))

输出是

(2019, 5) 1905

(2019, 6) 1906

(2019, 7) 1907

(2019, 8) 1908

(2019, 9) 1909

(2019, 10) 1910

(2019, 11) 1911

(2019, 12) 1912

(2020, 1) 2001

(2020, 2) 2002

(2020, 3) 2003