"Python人類可讀的日期差異"

Question

我給出了這樣的日期字符串：

Mon Jun 28 10:51:07 2010
Fri Jun 18 10:18:43 2010
Wed Dec 15 09:18:43 2010

Answer 1

使用strptime 。

樣品用法：

from datetime import datetime

my_date = datetime.strptime('Mon Jun 28 10:51:07 2010', '%a %b %d %H:%M:%S %Y')
print my_date

編輯：

您還可以用人類可讀的形式打印時差，如下所示：

from time import strptime
from datetime import datetime

def date_diff(older, newer):
    """
    Returns a humanized string representing time difference

    The output rounds up to days, hours, minutes, or seconds.
    4 days 5 hours returns '4 days'
    0 days 4 hours 3 minutes returns '4 hours', etc...
    """

    timeDiff = newer - older
    days = timeDiff.days
    hours = timeDiff.seconds/3600
    minutes = timeDiff.seconds%3600/60
    seconds = timeDiff.seconds%3600%60

    str = ""
    tStr = ""
    if days > 0:
        if days == 1:   tStr = "day"
        else:           tStr = "days"
        str = str + "%s %s" %(days, tStr)
        return str
    elif hours > 0:
        if hours == 1:  tStr = "hour"
        else:           tStr = "hours"
        str = str + "%s %s" %(hours, tStr)
        return str
    elif minutes > 0:
        if minutes == 1:tStr = "min"
        else:           tStr = "mins"           
        str = str + "%s %s" %(minutes, tStr)
        return str
    elif seconds > 0:
        if seconds == 1:tStr = "sec"
        else:           tStr = "secs"
        str = str + "%s %s" %(seconds, tStr)
        return str
    else:
        return None

older = datetime.strptime('Mon Jun 28 10:51:07 2010', '%a %b %d %H:%M:%S %Y')
newer = datetime.strptime('Tue Jun 28 10:52:07 2010', '%a %b %d %H:%M:%S %Y')
print date_diff(older, newer)

時間片段的原始來源 。

Answer 2

>>> import datetime
>>> a = datetime.datetime.strptime("Mon Jun 28 10:51:07 2010", "%a %b %d %H:%M:%S %Y")
>>> b = datetime.datetime.strptime("Fri Jun 18 10:18:43 2010", "%a %b %d %H:%M:%S %Y")
>>> c = a-b
>>> c.days
10

Answer 3

#!/usr/bin/env python

import datetime

def hrdd(d1, d2):
    """
    Human-readable date difference.
    """
    _d1 = datetime.datetime.strptime(d1, "%a %b %d %H:%M:%S %Y")
    _d2 = datetime.datetime.strptime(d2, "%a %b %d %H:%M:%S %Y")
    diff = _d2 - _d1
    return diff.days # <-- alternatively: diff.seconds 

if __name__ == '__main__':
    d1 = "Mon Jun 28 10:51:07 2010"
    d2 = "Fri Jun 18 10:18:43 2010"
    d3 = "Wed Dec 15 09:18:43 2010"

    print hrdd(d1, d2)
    # ==> -11
    print hrdd(d2, d1)
    # ==> 10
    print hrdd(d1, d3)
    # ==> 169
    # ...

Answer 4

這與其他答案不同，但對於想要展示更具人類可讀性（且更不精確）的人來說，它可能會有所幫助。 我很快就這樣做了，歡迎提出建議。

（請注意，它假設until_seconds是更晚的時間戳。）

def readable_delta(from_seconds, until_seconds=None):
    '''Returns a nice readable delta.

    readable_delta(1, 2)           # 1 second ago
    readable_delta(1000, 2000)     # 16 minutes ago
    readable_delta(1000, 9000)     # 2 hours, 133 minutes ago
    readable_delta(1000, 987650)   # 11 days ago
    readable_delta(1000)           # 15049 days ago (relative to now)
    '''

    if not until_seconds:
        until_seconds = time.time()

    seconds = until_seconds - from_seconds
    delta = datetime.timedelta(seconds=seconds)

    # deltas store time as seconds and days, we have to get hours and minutes ourselves
    delta_minutes = delta.seconds // 60
    delta_hours = delta_minutes // 60

    ## show a fuzzy but useful approximation of the time delta
    if delta.days:
        return '%d day%s ago' % (delta.days, plur(delta.days))
    elif delta_hours:
        return '%d hour%s, %d minute%s ago' % (delta_hours, plur(delta_hours), delta_minutes, plur(delta_minutes))
    elif delta_minutes:
        return '%d minute%s ago' % (delta_minutes, plur(delta_minutes))
    else:
        return '%d second%s ago' % (delta.seconds, plur(delta.seconds))

def plur(it):
    '''Quick way to know when you should pluralize something.'''
    try:
        size = len(it)
    except TypeError:
        size = int(it)
    return '' if size==1 else 's'

Answer 5

試試這個：

>>> (datetime.datetime.strptime("Mon Jun 28 10:51:07 2010", "%a %b %d %H:%M:%S %Y") - datetime.datetime.strptime("Fri Jun 18 10:18:43 2010", "%a %b %d %H:%M:%S %Y")).days
10

Answer 6

from datetime import datetime

resp = raw_input("What is the first date ?")
date1 = datetime.strptime(resp,"%a %b %d %H:%M:%S %Y")
resp2 = raw_input("What is the second date ?")
date2 = datetime.strptime(resp2,"%a %b %d %H:%M:%S %Y")
res = date2-date1
print str(res)

有關如何更好地打印timedelta對象的詳細信息，您可以看到此前一篇文章 。

Answer 7

這是@the_void 改進的 date_diff() 例程。 它將以以下方式打印差異，具體取決於差異的大小：

1d 
12 sec
30 min 12 sec
3h 30m 
5d 3h 30m