前一个月的Python日期

我试图用python获取前一个月的日期。以下是我的尝试:

str( time.strftime('%Y') ) + str( int(time.strftime('%m'))-1 )

然而，这种方式有两个原因:首先，它将返回2012年2月的20122(而不是201202);其次，它将返回0而不是1月的12。

我一下子就解决了这个麻烦

echo $(date -d"3 month ago" "+%G%m%d")

我认为，如果bash有一种内置的方式来实现这一目的，那么python应该提供更好的东西，而不是强迫自己编写脚本来实现这一目标。当然我可以这样做:

if int(time.strftime('%m')) == 1:
    return '12'
else:
    if int(time.strftime('%m')) < 10:
        return '0'+str(time.strftime('%m')-1)
    else:
        return str(time.strftime('%m') -1)

我还没有测试这段代码，我不想使用它(除非我找不到任何其他方法:/)

谢谢你的帮助!

当前回答

基于bgporter的回答。

def prev_month_range(when = None): 
    """Return (previous month's start date, previous month's end date)."""
    if not when:
        # Default to today.
        when = datetime.datetime.today()
    # Find previous month: https://stackoverflow.com/a/9725093/564514
    # Find today.
    first = datetime.date(day=1, month=when.month, year=when.year)
    # Use that to find the first day of this month.
    prev_month_end = first - datetime.timedelta(days=1)
    prev_month_start = datetime.date(day=1, month= prev_month_end.month, year= prev_month_end.year)
    # Return previous month's start and end dates in YY-MM-DD format.
    return (prev_month_start.strftime('%Y-%m-%d'), prev_month_end.strftime('%Y-%m-%d'))

2012-05-09 14:41:02

其他回答


from datetime import datetime, timedelta
from dateutil.relativedelta import relativedelta

#Months(0-12) (1 for Previous month)

#last day of (n) previous month (n=months)
day = datetime(2023, 1, 13)
n = 1
lastDayMonth = ((day - relativedelta(months=n) + relativedelta(day=31)).date());

#First day of previous month (n=months=1)
firstDayMonth = ((day - relativedelta(months=n) + relativedelta(day=1)).date());

print("Last Day of Month - "+ str(lastDayMonth))
print("First Day of Month - "+ str(firstDayMonth))

#Last business day (Friday) of prev (n) month if last day on weekend
lastBusDay = (lastDayMonth - timedelta(max(1,(lastDayMonth.weekday() + 6) % 7 - 3))) if lastDayMonth.weekday() in (5,6) else lastDayMonth

print("Last Business Day of Month - " + str(lastBusinessDay))

print()

2023-02-01 17:53:31

这很简单。这样做

from dateutil.relativedelta import relativedelta
from datetime import datetime

today_date = datetime.today()
print "todays date time: %s" %today_date

one_month_ago = today_date - relativedelta(months=1)
print "one month ago date time: %s" % one_month_ago
print "one month ago date: %s" % one_month_ago.date()

输出如下: 美元python2.7 main.py

todays date time: 2016-09-06 02:13:01.937121
one month ago date time: 2016-08-06 02:13:01.937121
one month ago date: 2016-08-06

2016-09-02 06:40:04

Datetime和Datetime。Timedelta类是您的朋友。

今天发现。用这个找到这个月的第一天。使用timedelta将某一天备份到上个月的最后一天。打印您正在寻找的YYYYMM字符串。

是这样的:

 import datetime
 today = datetime.date.today()
 first = today.replace(day=1)
 last_month = first - datetime.timedelta(days=1)
 print(last_month.strftime("%Y%m"))

201202打印出来了。

2012-03-15 17:35:12

import pandas as pd

lastmonth = int(pd.to_datetime("today").strftime("%Y%m"))-1

print(lastmonth)

202101

from datetime import date, timedelta
YYYYMM = (date.today().replace(day=1)-timedelta(days=1)).strftime("%Y%m")

2021-02-05 18:44:18

明确的方式:

import datetime
result = (datetime.datetime.today().month - 2) % 12 + 1

问题是如何将月[1,2,3，…]， 12]到[12,1,2，…]11)。

Step1: month = month - 1转移[1,2,3，…]， 12] to[0,1,2，…]11)。

Step2: month = (month - 1) % 12 transfer[0,1,2，…]， 11] to[11, 0,1，…], 10]。

Step3: month = month + 1 transfer[11, 0, 1，…]， 10]到[12,1,2，…]11)。

因此，结果是result =(月- 2)% 12 + 1

2022-02-22 04:17:13

前一个月的Python日期

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