将日期列表拆分为连续日期的子集

Question

我有一个日期数组，其中可以包含多个日期范围。

dates = [
 '2020-01-01',
 '2020-01-02',
 '2020-01-03',
 '2020-01-06',
 '2020-01-07',
 '2020-01-08'
]

在此示例中，列表包含 2 个单独的连续日期范围（2020-01-01 到 2020-01-03 和 2020-01-06 到 2020-01-08）

我试图弄清楚如何循环遍历这个列表并找到所有连续的日期范围。

我正在查看的一篇文章 (How to detect if dates are consecutive in Python?) 似乎有一个很好的方法，但是，我正在努力在我的用例中实现这个逻辑。

Answer 1

更多 itertools 有一个名为 consecutive_groups 的函数可以为您执行此操作：

或者您可以查看源代码并复制它的方法：

from datetime import datetime
from itertools import groupby
from operator import itemgetter

def consecutive_groups(iterable, ordering=lambda x: x):
    for k, g in groupby(enumerate(iterable), key=lambda x: x[0] - ordering(x[1])):
        yield map(itemgetter(1), g)

---

for g in consecutive_groups(dates, lambda x: datetime.strptime(x, '%Y-%m-%d').toordinal()):
    print(list(g))

---

['2020-01-01', '2020-01-02', '2020-01-03']
['2020-01-06', '2020-01-07', '2020-01-08']

Answer 2

这假设单日期“范围”仍由 2 个日期表示：

def makedate(s):
    return datetime.strptime( s, "%Y-%m-%d" )
def splitIntoRanges( dates ):
    ranges = []
    start_s = last_s = dates[0]
    last = makedate(start_s)
    for curr_s in dates[1:]:
        curr = makedate(curr_s)
        if (curr - last).days > 1:
            ranges.append((start_s,last_s))
            start_s = curr_s
        last_s = curr_s
        last = curr
    return ranges + [(start_s,last_s)]

Answer 3

我采用了类似的方法，但绝对不如@Scott 优雅：

ranges = []

dates = [datetime.strptime(date, '%Y-%m-%d') for date in dates]
start = dates[0]

for i in range(1, len(dates)):
    if (dates[i] - dates[i-1]).days == 1 and i==len(dates)-1:
        end = dates[i]
        ranges.append(f'{start} to {end}')
        start = dates[i]
    elif (dates[i] - dates[i - 1]).days > 1:
        end = dates[i - 1]
        ranges.append(f'{start} to {end}')
        start = dates[i]
    else:
        continue

Answer 4

我在第二篇文章中找到了解决方案的关键并将其拼凑在一起。

我的问题有两个部分：

1. 如何有效地表示日期列表

回答：https://stackoverflow.com/a/9589929/2150673

pto = [
    '2020-01-03',
    '2020-01-08',
    '2020-01-02',
    '2020-01-07',
    '2020-01-01',
    '2020-01-06'
]

ordinal_dates = [datetime.datetime.strptime(i, '%Y-%m-%d').toordinal() for i in pto]

1. 获得整数表示的日期列表后，您可以简单地查找连续整数并获取每个范围的上限和下限，然后转换回 yyyy-mm-dd 格式。

回答：https://stackoverflow.com/a/48106843

def ranges(nums):
    nums = sorted(set(nums))
    gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s+1 < e]
    edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])
    return list(zip(edges, edges))

我的完整功能：

def get_date_ranges(pto_list: list) -> list:
    pto_dates = [datetime.datetime.strptime(i, '%Y-%m-%d').toordinal() for i in pto_list]
    nums = sorted(set(pto_dates))
    gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s + 1 < e]
    edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])
    ordinal_ranges = list(zip(edges, edges))
    date_bounds = []
    for start, end in ordinal_ranges:
        date_bounds.append((
            datetime.datetime.fromordinal(start).strftime('%Y-%m-%d'),
            datetime.datetime.fromordinal(end).strftime('%Y-%m-%d')
        ))
    return date_bounds

Answer 5

您可以找到所有连续的日期范围并将它们附加到列表列表中并根据索引访问您的范围，但我更喜欢使用字典中的键以提高可读性。

方法如下：（注意：请阅读 cmets）

dates = [datetime.strptime(d, "%Y-%m-%d") for d in dates] # new datetime parsed from a string
date_ints = [d.toordinal() for d in dates]  # toordinal() returns the day count from the date 01/01/01 in integers
ranges = {}; arange = []; prev=0; index=0; j=1
for i in date_ints: # iterate through date integers
    if i+1 == date_ints[index] + 1 and i - 1 == prev: # check and compare if integers are in sequence
        arange.append(dates[index].strftime("%Y-%m-%d"))
    elif prev == 0: # append first date to 'arange' list since 'prev' has not been updated
        arange.append(dates[index].strftime("%Y-%m-%d"))
    else:
        ranges.update({f'Range{j}': tuple(arange)}) # integer are no longer in sequence, update dictionary with new range  
        arange = []; j += 1                                   # clear 'arange' and start appending to new range  
        arange.append(dates[index].strftime("%Y-%m-%d"))
    index += 1; prev = i
ranges.update({f'Range{j}': tuple(arange)})
print(ranges)  
print(ranges['Range1'])  # access a range based on the associated key
print(ranges['Range2']) 

输出：

{'Range1': ('2020-01-01', '2020-01-02', '2020-01-03'), 'Range2': ('2020-01-06', '2020-01-07', '2020-01-08')}
('2020-01-01', '2020-01-02', '2020-01-03')
('2020-01-06', '2020-01-07', '2020-01-08')