我有兩個月的時間，從開始到結束。我想迭代之間的所有月份，即我想將所有月份提取為循環中的字符串。start = '1905' # May 2019end = '2003' # March 2020for month in range(start, end): # I want to get string of each month in same format here有人可以幫助我干凈有效地獲得這個嗎？我的一年總是在（2000-2099）之間
4 回答
慕容708150
TA貢獻1831條經驗 獲得超4個贊
period_range與將日期字符串轉換為日期時間一起使用：
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自定義格式：
print (p.strftime('%B'))
Index(['May', 'June', 'July', 'August', 'September', 'October', 'November',
? ? ? ?'December', 'January', 'February', 'March'],
? ? ? dtype='object')
for val in p:
? ? print (val)
回首憶惘然
TA貢獻1847條經驗 獲得超11個贊
一個簡單的解決方案是這樣的：
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}")
輸出：
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
滄海一幻覺
TA貢獻1824條經驗 獲得超5個贊
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)
?
這將產生字符串
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對象"%y%m"將返回您想要的格式。
繁星淼淼
TA貢獻1775條經驗 獲得超11個贊
像這樣的生成器函數應該可以解決問題。
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))
輸出是
(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
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