使用Python将连续日期分组

给定：

dates = [
  datetime(2014, 10, 11), 

  datetime(2014, 10, 1), 
  datetime(2014, 10, 2),
  datetime(2014, 10, 3),

  datetime(2014, 10, 5), 
  datetime(2014, 10, 5), 
  datetime(2014, 10, 6), 

  datetime(2014, 10, 22), 
  datetime(2014, 10, 20),
  datetime(2014, 10, 21),

  datetime(2014, 10, 9), 

  datetime(2014, 10, 7),
  datetime(2014, 10, 6)
]

expect = [
  [datetime(2014, 10, 1), datetime(2014, 10, 3)],
  [datetime(2014, 10, 5), datetime(2014, 10, 7)],
   datetime(2014, 10, 9),
   datetime(2014, 10, 11),
  [datetime(2014, 10, 20), datetime(2014, 10, 22)]
]

from datetime import datetime, timedelta

def parse_date_ranges(dates):
    if(not dates or not len(dates) > 0):
        return False

    # make sure to order dates
    dates.sort()

    # init values
    result  = []
    tupl    = [dates[0], dates[0]]
    it      = iter(dates)
    date    = True

    def add_tuple_to_result(tuple):
        # if first part of tuple differs from last part -> add full tuple
        # else -> add first part of tuple only
        result.append(tupl if tupl[0] != tupl[1] else tupl[0])

    while date:
        # get next date or False if no next date
        date = next(it, False)

        # filter double dates
        if(date in tupl):
            continue

        elif(date):
            if(date - timedelta(days=1) == tupl[1]):
                # consecutive date, so add date to end of current tuple
                tupl[1] = date
            else:
                # gap larger than 1 day: add current tuple to result
                # and create new tuple
                add_tuple_to_result(tupl)
                tupl = [date, date]

        else:
            # date == false, so this is the last step.
            # add the current tuple to result
            add_tuple_to_result(tupl)

    return result

预期输出：

dates = [
  datetime(2014, 10, 11), 

  datetime(2014, 10, 1), 
  datetime(2014, 10, 2),
  datetime(2014, 10, 3),

  datetime(2014, 10, 5), 
  datetime(2014, 10, 5), 
  datetime(2014, 10, 6), 

  datetime(2014, 10, 22), 
  datetime(2014, 10, 20),
  datetime(2014, 10, 21),

  datetime(2014, 10, 9), 

  datetime(2014, 10, 7),
  datetime(2014, 10, 6)
]

expect = [
  [datetime(2014, 10, 1), datetime(2014, 10, 3)],
  [datetime(2014, 10, 5), datetime(2014, 10, 7)],
   datetime(2014, 10, 9),
   datetime(2014, 10, 11),
  [datetime(2014, 10, 20), datetime(2014, 10, 22)]
]

from datetime import datetime, timedelta

def parse_date_ranges(dates):
    if(not dates or not len(dates) > 0):
        return False

    # make sure to order dates
    dates.sort()

    # init values
    result  = []
    tupl    = [dates[0], dates[0]]
    it      = iter(dates)
    date    = True

    def add_tuple_to_result(tuple):
        # if first part of tuple differs from last part -> add full tuple
        # else -> add first part of tuple only
        result.append(tupl if tupl[0] != tupl[1] else tupl[0])

    while date:
        # get next date or False if no next date
        date = next(it, False)

        # filter double dates
        if(date in tupl):
            continue

        elif(date):
            if(date - timedelta(days=1) == tupl[1]):
                # consecutive date, so add date to end of current tuple
                tupl[1] = date
            else:
                # gap larger than 1 day: add current tuple to result
                # and create new tuple
                add_tuple_to_result(tupl)
                tupl = [date, date]

        else:
            # date == false, so this is the last step.
            # add the current tuple to result
            add_tuple_to_result(tupl)

    return result

用Python实现：

dates = [
  datetime(2014, 10, 11), 

  datetime(2014, 10, 1), 
  datetime(2014, 10, 2),
  datetime(2014, 10, 3),

  datetime(2014, 10, 5), 
  datetime(2014, 10, 5), 
  datetime(2014, 10, 6), 

  datetime(2014, 10, 22), 
  datetime(2014, 10, 20),
  datetime(2014, 10, 21),

  datetime(2014, 10, 9), 

  datetime(2014, 10, 7),
  datetime(2014, 10, 6)
]

expect = [
  [datetime(2014, 10, 1), datetime(2014, 10, 3)],
  [datetime(2014, 10, 5), datetime(2014, 10, 7)],
   datetime(2014, 10, 9),
   datetime(2014, 10, 11),
  [datetime(2014, 10, 20), datetime(2014, 10, 22)]
]

from datetime import datetime, timedelta

def parse_date_ranges(dates):
    if(not dates or not len(dates) > 0):
        return False

    # make sure to order dates
    dates.sort()

    # init values
    result  = []
    tupl    = [dates[0], dates[0]]
    it      = iter(dates)
    date    = True

    def add_tuple_to_result(tuple):
        # if first part of tuple differs from last part -> add full tuple
        # else -> add first part of tuple only
        result.append(tupl if tupl[0] != tupl[1] else tupl[0])

    while date:
        # get next date or False if no next date
        date = next(it, False)

        # filter double dates
        if(date in tupl):
            continue

        elif(date):
            if(date - timedelta(days=1) == tupl[1]):
                # consecutive date, so add date to end of current tuple
                tupl[1] = date
            else:
                # gap larger than 1 day: add current tuple to result
                # and create new tuple
                add_tuple_to_result(tupl)
                tupl = [date, date]

        else:
            # date == false, so this is the last step.
            # add the current tuple to result
            add_tuple_to_result(tupl)

    return result

有关更多测试，请参阅

问题

---

实现是可行的，但我对Python还是新手。所以我想知道是否有更好的方法来解决这个问题？还是很好？

我为您编写了另一个解决方案，并写了一些注释来解释代码

from datetime import datetime, timedelta

dates = [
    datetime(2014, 10, 11),
    datetime(2014, 10, 1),
    datetime(2014, 10, 2),
    datetime(2014, 10, 3),
    datetime(2014, 10, 5),
    datetime(2014, 10, 5),
    datetime(2014, 10, 6),
    datetime(2014, 10, 22),
    datetime(2014, 10, 20),
    datetime(2014, 10, 21),
    datetime(2014, 10, 9),
    datetime(2014, 10, 7),
    datetime(2014, 10, 6)
]
# Remove duplicates, and sort the dates ascending
sorted_dates = sorted(set(dates))
# Set initial first and last element as the current element
first, last = sorted_dates[0], sorted_dates[0]
date_ranges = []

# Loop over the sorted list from the second value
for d in sorted_dates[1:]:
    # Check if the current date is exactly one day later then the current
    # "last" date
    if d - last != timedelta(days=1):
        date_ranges.append(tuple(sorted({first, last})))
        first, last = d, d
    else:
        last = d

# Handle last element
if first == last:
    date_ranges.append((first,))
else:
    date_ranges.append((first, last))

for dt_pair in date_ranges:
    print dt_pair

输出：

(datetime.datetime(2014, 10, 1, 0, 0), datetime.datetime(2014, 10, 3, 0, 0))
(datetime.datetime(2014, 10, 5, 0, 0), datetime.datetime(2014, 10, 7, 0, 0))
(datetime.datetime(2014, 10, 9, 0, 0),)
(datetime.datetime(2014, 10, 11, 0, 0),)
(datetime.datetime(2014, 10, 20, 0, 0), datetime.datetime(2014, 10, 22, 0, 0))

我喜欢：

结果:

(datetime.datetime(2014, 10, 1, 0, 0), datetime.datetime(2014, 10, 3, 0, 0))
(datetime.datetime(2014, 10, 5, 0, 0), datetime.datetime(2014, 10, 7, 0, 0))
datetime.datetime(2014, 10, 9, 0, 0)
datetime.datetime(2014, 10, 11, 0, 0)
(datetime.datetime(2014, 10, 20, 0, 0), datetime.datetime(2014, 10, 22, 0, 0))
(1, 3)
(5, 7)
9
11
(20, 22)

---

通过将其与标题类似的
使用

datetime

对象的问题：

def parse_date_ranges(dates):

    def group_consecutive(dates):
        dates_iter = iter(sorted(set(dates)))  # de-dup and sort

        run = [next(dates_iter)]
        for d in dates_iter:
            if (d.toordinal() - run[-1].toordinal()) == 1:  # consecutive?
                run.append(d)
            else:  # [start, end] of range else singleton
                yield [run[0], run[-1]] if len(run) > 1 else run[0]
                run = [d]

        yield [run[0], run[-1]] if len(run) > 1 else run[0]

    return list(group_consecutive(dates)) if dates else False

适应使用日期时间对象。这包括非唯一和非排序输入，并且与python3兼容：

import itertools
from datetime import datetime, timedelta

def datetimes_to_ranges(iterable):
    iterable = sorted(set(iterable))
    keyfunc = lambda t: t[1] - timedelta(days=t[0])
    for key, group in itertools.groupby(enumerate(iterable), keyfunc):
        group = list(group)
        if len(group) == 1:
            yield group[0][1]
        else:
            yield group[0][1], group[-1][1]

例如：

>>> for i in datetimes_to_ranges(dates): i
... 
(datetime.datetime(2014, 10, 1, 0, 0), datetime.datetime(2014, 10, 3, 0, 0))
(datetime.datetime(2014, 10, 5, 0, 0), datetime.datetime(2014, 10, 7, 0, 0))
datetime.datetime(2014, 10, 9, 0, 0)
datetime.datetime(2014, 10, 11, 0, 0)
(datetime.datetime(2014, 10, 20, 0, 0), datetime.datetime(2014, 10, 22, 0, 0))

---

为什么9和11不在一起？这真的不是一个好问题。我们可以帮助您修复不起作用的代码，但是要求人们编写比您的工作代码更好的代码，或者承诺不存在这样的东西，在这里是不合适的。可能更好，但我也不知道这是否合适。不过，你可能想看看，这是我在帮助有类似问题的人之后写的。（他试图使用itertools，结果卡住了，所以我展示了各种方法来摆脱困境。首先，这到底是不是最好的解决方案，是一个更主观、更难回答的问题……）9。。(10) .. 11，所以他们没有互相跟踪。不知道代码审查。谢谢你的提示。