Python：如何計算日期列表中的日期範圍？

Question

我有日期的列表，例如：

['2011-02-27', '2011-02-28', '2011-03-01', '2011-04-12', '2011-04-13', '2011-06-08'] 

如何找到包含這些日期內連續日期範圍？在上面的示例中，範圍應爲：

[{"start_date": '2011-02-27', "end_date": '2011-03-01'}, 
{"start_date": '2011-04-12', "end_date": '2011-04-13'}, 
{"start_date": '2011-06-08', "end_date": '2011-06-08'} 
] 

謝謝。

Answer 1

這個作品，，但我不滿意它，將工作在一個更清潔的解決方案編輯答案。做完後，這裏是一個乾淨，工作液：

import datetime 
import pprint 

def parse(date): 
    return datetime.date(*[int(i) for i in d.split('-')]) 

def get_ranges(dates): 
    while dates: 
     end = 1 
     try: 
      while dates[end] - dates[end - 1] == datetime.timedelta(days=1): 
       end += 1 
     except IndexError: 
      pass 

     yield { 
      'start-date': dates[0], 
      'end-date': dates[end-1] 
     } 
     dates = dates[end:] 

dates = [ 
    '2011-02-27', '2011-02-28', '2011-03-01', 
    '2011-04-12', '2011-04-13', 
    '2011-06-08' 
] 

# Parse each date and convert it to a date object. Also ensure the dates 
# are sorted, you can remove 'sorted' if you don't need it 
dates = sorted([parse(d) for d in dates]) 

pprint.pprint(list(get_ranges(dates))) 

，相對輸出：

[{'end-date': datetime.date(2011, 3, 1), 
    'start-date': datetime.date(2011, 2, 27)}, 
{'end-date': datetime.date(2011, 4, 13), 
    'start-date': datetime.date(2011, 4, 12)}, 
{'end-date': datetime.date(2011, 6, 8), 
    'start-date': datetime.date(2011, 6, 8)}] 

Answer 2

試圖忍者GaretJax的編輯：;）在

def date_to_number(date): 
    return datetime.date(*[int(i) for i in date.split('-')]).toordinal() 

def number_to_date(number): 
    return datetime.date.fromordinal(number).strftime('%Y-%m-%d') 

def day_ranges(dates): 
    day_numbers = set(date_to_number(d) for d in dates) 
    start = None 
    # We loop including one element guaranteed not to be in the set, to force the 
    # closing of any range that's currently open. 
    for n in xrange(min(day_numbers), max(day_numbers) + 2): 
    if start == None: 
     if n in day_numbers: start = n 
    else: 
     if n not in day_numbers: 
     yield { 
      'start_date': number_to_date(start), 
      'end_date': number_to_date(n - 1) 
     } 
     start = None 

list(
    day_ranges([ 
    '2011-02-27', '2011-02-28', '2011-03-01', 
    '2011-04-12', '2011-04-13', '2011-06-08' 
    ]) 
) 

Answer 3

from datetime import datetime, timedelta 

dates = ['2011-02-27', '2011-02-28', '2011-03-01', '2011-04-12', '2011-04-13', '2011-06-08'] 
d = [datetime.strptime(date, '%Y-%m-%d') for date in dates] 
test = lambda x: x[1] - x[0] != timedelta(1) 
slices = [0] + [i+1 for i, x in enumerate(zip(d, d[1:])) if test(x)] + [len(dates)] 
ranges = [{"start_date": dates[s], "end_date": dates[e-1]} for s, e in zip(slices, slices[1:])] 

結果以下：

>>> pprint.pprint(ranges) 
[{'end_date': '2011-03-01', 'start_date': '2011-02-27'}, 
{'end_date': '2011-04-13', 'start_date': '2011-04-12'}, 
{'end_date': '2011-06-08', 'start_date': '2011-06-08'}] 

slices列表理解獲取所有指數，其中上一個日期不是當前日期前一天。將0添加到前面，並將len(dates)添加到最後，並且每個日期範圍可以描述爲dates[slices[i]:slices[i+1]-1]。

Answer 4

我的主題是輕微的變化（我最初建的開始/結束列表，並拉上他們返回記錄，但我更喜歡@Karl Knechtel的產生辦法）：

from datetime import date, timedelta 

ONE_DAY = timedelta(days=1) 

def find_date_windows(dates): 
    # guard against getting empty list 
    if not dates: 
     return 

    # convert strings to sorted list of datetime.dates 
    dates = sorted(date(*map(int,d.split('-'))) for d in dates) 

    # build list of window starts and matching ends 
    lastStart = lastEnd = dates[0] 
    for d in dates[1:]: 
     if d-lastEnd > ONE_DAY: 
      yield {'start_date':lastStart, 'end_date':lastEnd} 
      lastStart = d 
     lastEnd = d 
    yield {'start_date':lastStart, 'end_date':lastEnd} 

下面是測試情況：

tests = [ 
    ['2011-02-27', '2011-02-28', '2011-03-01', '2011-04-12', '2011-04-13', '2011-06-08'], 
    ['2011-06-08'], 
    [], 
    ['2011-02-27', '2011-02-28', '2011-03-01', '2011-04-12', '2011-04-13', '2011-06-08', '2011-06-10'], 
] 
for dates in tests: 
    print dates 
    for window in find_date_windows(dates): 
     print window 
    print 

打印：

['2011-02-27', '2011-02-28', '2011-03-01', '2011-04-12', '2011-04-13', '2011-06-08'] 
{'start_date': datetime.date(2011, 2, 27), 'end_date': datetime.date(2011, 3, 1)} 
{'start_date': datetime.date(2011, 4, 12), 'end_date': datetime.date(2011, 4, 13)} 
{'start_date': datetime.date(2011, 6, 8), 'end_date': datetime.date(2011, 6, 8)} 

['2011-06-08'] 
{'start_date': datetime.date(2011, 6, 8), 'end_date': datetime.date(2011, 6, 8)} 

[] 

['2011-02-27', '2011-02-28', '2011-03-01', '2011-04-12', '2011-04-13', '2011-06-08', '2011-06-10'] 
{'start_date': datetime.date(2011, 2, 27), 'end_date': datetime.date(2011, 3, 1)} 
{'start_date': datetime.date(2011, 4, 12), 'end_date': datetime.date(2011, 4, 13)} 
{'start_date': datetime.date(2011, 6, 8), 'end_date': datetime.date(2011, 6, 8)} 
{'start_date': datetime.date(2011, 6, 10), 'end_date': datetime.date(2011, 6, 10)} 

Answer 5

下面是一個替代的解決方案：它RET甕恩列表（開始，完成）的元組，因爲這是我所需要的;）。

這使列表發生變化，所以我需要複製一份。顯然，這會增加內存使用量。我懷疑list.pop（）不是超高效的，但這可能取決於python中list的實現。

def collapse_dates(date_list): 
    if not date_list: 
     return date_list 
    result = [] 
    # We are going to alter the list, so create a (sorted) copy. 
    date_list = sorted(date_list) 
    while len(date_list): 
     # Grab the first item: this is both the start and end of the range. 
     start = current = date_list.pop(0) 
     # While the first item in the list is the next day, pop that and 
     # set it to the end of the range. 
     while len(date_list) and date_list[0] == current + datetime.timedelta(1): 
      current = date_list.pop(0) 
     # That's a completed range. 
     result.append((start,current)) 

    return result 

你可以很容易地改變附加行來追加一個字典，或者yield而不是附加到列表。

哦，我的假設他們已經是日期。