Question

我已经给出了这样的日期字符串：

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

计算天数差异的方便python方法是什么？假设时区是相同的。

字符串由linux命令返回。

编辑：谢谢，这么多好的答案

Answer 1

使用strptime。

样本用法：

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

修改

您还可以用人类可读的形式打印时差，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)

Original source获取时间片段。

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

这与其他答案不同，但对于想要展示更具人类可读性（且更不精确）的人来说，它可能会有所帮助。我很快就这样做了，欢迎提出建议。

（请注意，它假设until_seconds是更晚的时间戳。）

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

试试这个：

>>> (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)

有关如何更好地打印timedelta对象的详细信息，您可以看到this previous post。

Python人类可读的日期差异

6 个答案: