如何檢測Python中的日期是否連續？

Question

我有一個訪問表與'日期'字段。它有每個記錄的隨機日期。我已經建立了一個腳本追加的所有記錄到一個列表，然後將列表設置過濾出獨特的價值：

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）：

[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（即只有一個唯一的日期是r因爲那是每個記錄中的日期）
• 如果日期是日期範圍（即，所有日期的陷入連續的範圍內）
• 如果日期是多個日期，但不是在一個範圍內的連續日期

任何建議，將不勝感激的！ Mike

Answer 1

而不是滾動自己的consecutive函數，您可以簡單地使用datetime對象的.toordinal()方法將日期對象轉換爲整數。設定順序日期的最大值與最小值之差小於設定的長度多了一個：

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

使用數據庫升序選擇獨特日期：

• 如果該查詢返回一個日期這是你第一種情況

• 否則找出日期是否是連續的：

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

代碼佈局：

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 3

這裏的使用減少（）函數，我的版本。

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

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"