Question

我有一个带有“日期”字段的访问表。它有每个记录的随机日期。我已经构建了一个脚本来将所有记录附加到列表中，然后设置列表以仅过滤掉唯一值：

dateList = []
# cursor search through each record and append all records in the date 
# field to a python list
for row in rows:
   dateList.append(row.getValue("DATE_OBSERVATION").strftime('%m-%d-%Y'))

# Filter unique values to a set
newList = list(set(dateList))

返回（在我的测试表上）：

['07-06-2010'，'06 -24-2010'，'07 -05-2010'，'06 -25-2010']

既然我有“DATE_OBSERVATION”字段的唯一值，我想检测是否：

日期是单一的（即只返回一个唯一日期，因为这是每条记录中的日期）
如果日期是一系列日期（即所有日期都属于连续范围）
如果日期是多个日期，但不在连续日期的范围内

任何建议都将不胜感激！麦克

Answer 1

您可以使用datetime对象的consecutive方法将日期对象简单地转换为整数，而不是滚动自己的.toordinal()函数。序数日期的最大值和最小值之间的差异比集合的长度多一个：

from datetime import datetime

date_strs = ['07-06-2010', '06-24-2010', '07-05-2010', '06-25-2010']
# date_strs = ['02-29-2012', '02-28-2012', '03-01-2012']
# date_strs = ['01-01-2000']
dates = [datetime.strptime(d, "%m-%d-%Y") for d in date_strs]

date_ints = set([d.toordinal() for d in dates])

if len(date_ints) == 1:
    print "unique"
elif max(date_ints) - min(date_ints) == len(date_ints) - 1:
    print "consecutive"
else:
    print "not consecutive"

Answer 2

另一个版本使用与我的其他答案相同的逻辑。

from datetime import date, timedelta

# Definition 1: 1/1/14, 1/2/14, 1/2/14, 1/3/14 is consider consecutive
# Definition 2: 1/1/14, 1/2/14, 1/2/14, 1/3/14 is consider not consecutive

# datelist = [date(2014, 1, 1), date(2014, 1, 3),
#             date(2013, 12, 31), date(2013, 12, 30)]

# datelist = [date(2014, 2, 19), date(2014, 2, 19), date(2014, 2, 20),
#             date(2014, 2, 21), date(2014, 2, 22)]

datelist = [date(2014, 2, 19), date(2014, 2, 21),
            date(2014, 2, 22), date(2014, 2, 20)]

datelist.sort()

previousdate = datelist[0]

for i in range(1, len(datelist)):
    #if (datelist[i] - previousdate).days == 1 or (datelist[i] - previousdate).days == 0:  # for Definition 1
    if (datelist[i] - previousdate).days == 1:    # for Definition 2
        previousdate = datelist[i]
    else:
        previousdate = previousdate + timedelta(days=-1)

if datelist[-1] == previousdate:
    print "dates are consecutive"
else:
    print "dates are not consecutive"

Answer 3

使用您的数据库按升序选择唯一日期：

如果查询返回单个日期，那么这是您的第一个案例

否则查明日期是否连续：

import datetime

def consecutive(a, b, step=datetime.timedelta(days=1)):
    return (a + step) == b

代码布局：

dates = <query database>
if all(consecutive(dates[i], dates[i+1]) for i in xrange(len(dates) - 1)):
   if len(dates) == 1: # unique
      # 1st case: all records have the same date
   else:
      # the dates are a range of dates
else:
   # non-consecutive dates

Answer 4

这是我使用reduce（）函数的版本。

from datetime import date, timedelta


def checked(d1, d2):
    """
    We assume the date list is sorted.
    If d2 & d1 are different by 1, everything up to d2 is consecutive, so d2
    can advance to the next reduction.
    If d2 & d1 are not different by 1, returning d1 - 1 for the next reduction
    will guarantee the result produced by reduce() to be something other than
    the last date in the sorted date list.

    Definition 1: 1/1/14, 1/2/14, 1/2/14, 1/3/14 is consider consecutive
    Definition 2: 1/1/14, 1/2/14, 1/2/14, 1/3/14 is consider not consecutive

    """
    #if (d2 - d1).days == 1 or (d2 - d1).days == 0:  # for Definition 1
    if (d2 - d1).days == 1:                          # for Definition 2
        return d2
    else:
        return d1 + timedelta(days=-1)

# datelist = [date(2014, 1, 1), date(2014, 1, 3),
#             date(2013, 12, 31), date(2013, 12, 30)]

# datelist = [date(2014, 2, 19), date(2014, 2, 19), date(2014, 2, 20),
#             date(2014, 2, 21), date(2014, 2, 22)]

datelist = [date(2014, 2, 19), date(2014, 2, 21),
            date(2014, 2, 22), date(2014, 2, 20)]

datelist.sort()

if datelist[-1] == reduce(checked, datelist):
    print "dates are consecutive"
else:
    print "dates are not consecutive"

如何检测Python中的日期是否连续？

4 个答案: