如何在排序日期列表中查找缺少的日期?

使用集合

>>> from datetime import date, timedelta
>>> d = [date(2010, 2, 23), date(2010, 2, 24), date(2010, 2, 25),
         date(2010, 2, 26), date(2010, 3, 1), date(2010, 3, 2)]
>>> date_set = set(d[0] + timedelta(x) for x in range((d[-1] - d[0]).days))
>>> missing = sorted(date_set - set(d))
>>> missing
[datetime.date(2010, 2, 27), datetime.date(2010, 2, 28)]
>>> 

对日期列表排序并迭代,记住上一个条目。如果上一个条目和当前条目之间的差异超过一天,则表示缺少天数。

有一种方法可以实现:

from datetime import date, timedelta
from itertools import tee, izip

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    b.next()
    return izip(a, b)

def missing_dates(dates):
    for prev, curr in pairwise(sorted(dates)):
        i = prev
        while i + timedelta(1) < curr:
            i += timedelta(1)
            yield i

dates = [ date(2010, 1, 8),
          date(2010, 1, 2),
          date(2010, 1, 5),
          date(2010, 1, 1),
          date(2010, 1, 7) ]

for missing in missing_dates(dates):
    print missing

输出:

2010-01-03
2010-01-04
2010-01-06

性能是O(n*log(n)),其中n是输入未排序时跨度中的天数。由于您的列表已经排序,它将以O(n)运行。

>>> from datetime import datetime, timedelta
>>> date_list = [datetime(2010, 2, 23),datetime(2010, 2, 24),datetime(2010, 2, 25),datetime(2010, 2, 26),datetime(2010, 3, 1),datetime(2010, 3, 2)]
>>> 
>>> date_set=set(date_list)         # for faster membership tests than list
>>> one_day = timedelta(days=1)
>>> 
>>> test_date = date_list[0]
>>> missing_dates=[]
>>> while test_date < date_list[-1]:
...     if test_date not in date_set:
...         missing_dates.append(test_date)
...     test_date += one_day
... 
>>> print missing_dates
[datetime.datetime(2010, 2, 27, 0, 0), datetime.datetime(2010, 2, 28, 0, 0)]

这也适用于 datetime.date datetime.datetime 物体

把日期放在一个盒子里 set 然后使用 datetime.timedelta()

您需要的导入:

import datetime
from datetime import date, timedelta

假设您有一个排序列表,名为 dates 有几个日期不见了。

start_date = dates[0]
end_date = dates[len(dates)-1]

numdays = (end_date - start_date).days

开始日期和结束日期之间的日期:

all_dates = []
for x in range (0, (numdays+1)):
all_dates.append(start_date + datetime.timedelta(days = x))

而不是检查日期 all_dates 通过使用范围为的for循环并将这些日期添加到缺少的日期:

dates_missing = []

for i in range (0, len(all_dates)):
   if (all_dates[i] not in dates):
       dates_missing.append(all_dates[i])
   else:
       pass

现在你会有一个叫做 dates_missing

import datetime

DAY = datetime.timedelta(days=1)
# missing dates: a list of [start_date, end)
missing = [(d1+DAY, d2) for d1, d2 in zip(dates, dates[1:]) if (d2 - d1) > DAY]

def date_range(start_date, end, step=DAY):
    d = start_date
    while d < end:
        yield d
        d += step

missing_dates = [d for d1, d2 in missing for d in date_range(d1, d2)]

使用列表理解

>>> from datetime import date, timedelta
>>> d = [date(2010, 2, 23),date(2010, 2, 24),date(2010, 2, 25),date(2010, 2, 26),date(2010, 3, 1),date(2010, 3, 2)]
>>> date_set=set(d)
>>> missing = [x for x in (d[0]+timedelta(x) for x in range((d[-1]-d[0]).days)) if x not in date_set]

>>> missing
[datetime.date(2010, 2, 27), datetime.date(2010, 2, 28)]

创建一个函数,为您提供从 start_date 到 end_date . 使用它。

import datetime

def get_dates(start_date, end_date):
    span_between_dates = range(end_date - start_date).days
    for index in span_between_dates + 1:
        # +1 is to make start and end dates inclusive.
        yield start_date + datetime.timedelta(index)

my_date_list = ['2017-03-05', '2017-03_07', ...]
# Edit my_date_list as per your requirement.
start_date = min(my_date_list)
end_date = max(my_date_list)
for current_date in get_dates(start_date, end_date)
    if date not in my_date_list:
        print date

get_dates

my_other_date_list = []  # your other date range
start_date = min(my_date_list)
end_date = max(my_date_list)
for current_date in get_dates(start_date, end_date)
    if (date in my_date_range) and (date in my_other_date_list):
        print ('overlapping dates between 2 lists:')
        print date
    elif (date in my_date_range) and (date not in my_other_date_list):
        print ('missing dates:')
        print date