如何檢測Python中的日期是否連續？

Question

我有一個帶有“日期”字段的訪問表。 它有每個記錄的隨機日期。 我已經構建了一個腳本來將所有記錄附加到列表中，然后設置列表以僅過濾掉唯一值：

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))

這將返回（在我的測試表上）：

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

既然我有“DATE_OBSERVATION”字段的唯一值，我想檢測是否：

• 日期是單一的（即只返回一個唯一的日期，因為這是每條記錄中的日期）
• 如果日期是一系列日期（即所有日期都屬於連續范圍）
• 如果日期是多個日期，但不在連續日期的范圍內

我們歡迎所有的建議！ 麥克風

Answer 1

您可以使用datetime對象的.toordinal()方法將日期對象簡單地轉換為整數，而不是滾動自己的consecutive函數。 序數日期的最大值和最小值之間的差異比集合的長度多一個：

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

另一個版本使用與我的其他答案相同的邏輯。

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

這是我使用reduce（）函數的版本。

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"

Answer 4

使用數據庫按升序選擇唯一日期：

• 如果查詢返回單個日期，那么這是您的第一個案例

• 否則查明日期是否連續：

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

代碼布局：

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