如何在 Python 中正确合并重叠的日期时间范围

Question

我正在尝试合并重叠的日期时间范围。我将日期时间范围列表作为列表中的元组：

data = [(datetime.datetime(2016, 1, 10, 13, 0), datetime.datetime(2016, 1, 10, 16, 0)), (datetime.datetime(2016, 1, 10, 14, 0), datetime.datetime(2016, 1, 10, 14, 0)), (datetime.datetime(2016, 1, 10, 22, 0), datetime.datetime(2016, 1, 10, 22, 0)), (datetime.datetime(2016, 1, 10, 23, 0), datetime.datetime(2016, 1, 11, 0, 30)), (datetime.datetime(2016, 1, 11, 2, 30), datetime.datetime(2016, 1, 11, 3, 30)), (datetime.datetime(2016, 1, 11, 13, 0), datetime.datetime(2016, 1, 11, 16, 0)), (datetime.datetime(2016, 1, 11, 14, 0), datetime.datetime(2016, 1, 11, 14, 0)), (datetime.datetime(2016, 1, 11, 20, 30), datetime.datetime(2016, 1, 11, 21, 30)), (datetime.datetime(2016, 1, 11, 22, 0), datetime.datetime(2016, 1, 11, 22, 0)), (datetime.datetime(2016, 1, 12, 2, 30), datetime.datetime(2016, 1, 12, 3, 30)), (datetime.datetime(2016, 1, 12, 13, 0), datetime.datetime(2016, 1, 12, 16, 0)), (datetime.datetime(2016, 1, 12, 14, 0), datetime.datetime(2016, 1, 12, 14, 0)), (datetime.datetime(2016, 1, 12, 19, 30), datetime.datetime(2016, 1, 12, 20, 30)), (datetime.datetime(2016, 1, 12, 22, 0), datetime.datetime(2016, 1, 12, 22, 0)), (datetime.datetime(2016, 1, 13, 2, 30), datetime.datetime(2016, 1, 13, 3, 30)), (datetime.datetime(2016, 1, 13, 13, 0), datetime.datetime(2016, 1, 13, 16, 0)), (datetime.datetime(2016, 1, 13, 14, 0), datetime.datetime(2016, 1, 13, 14, 0)), (datetime.datetime(2016, 1, 13, 20, 0), datetime.datetime(2016, 1, 13, 21, 0)), (datetime.datetime(2016, 1, 13, 21, 30), datetime.datetime(2016, 1, 13, 22, 0)), (datetime.datetime(2016, 1, 13, 22, 0), datetime.datetime(2016, 1, 13, 22, 0)), (datetime.datetime(2016, 1, 14, 2, 30), datetime.datetime(2016, 1, 14, 3, 30)), (datetime.datetime(2016, 1, 14, 13, 0), datetime.datetime(2016, 1, 14, 16, 0)), (datetime.datetime(2016, 1, 14, 14, 0), datetime.datetime(2016, 1, 14, 14, 0)), (datetime.datetime(2016, 1, 14, 22, 0), datetime.datetime(2016, 1, 14, 22, 0)), (datetime.datetime(2016, 1, 14, 22, 0), datetime.datetime(2016, 1, 14, 23, 0)), (datetime.datetime(2016, 1, 15, 2, 30), datetime.datetime(2016, 1, 15, 3, 30)), (datetime.datetime(2016, 1, 15, 13, 0), datetime.datetime(2016, 1, 15, 16, 0)), (datetime.datetime(2016, 1, 15, 14, 0), datetime.datetime(2016, 1, 15, 14, 0)), (datetime.datetime(2016, 1, 15, 20, 30), datetime.datetime(2016, 1, 15, 22, 0)), (datetime.datetime(2016, 1, 15, 22, 0), datetime.datetime(2016, 1, 15, 22, 0)), (datetime.datetime(2016, 1, 16, 2, 30), datetime.datetime(2016, 1, 16, 3, 30)), (datetime.datetime(2016, 1, 16, 13, 0), datetime.datetime(2016, 1, 16, 16, 0)), (datetime.datetime(2016, 1, 17, 2, 30), datetime.datetime(2016, 1, 17, 3, 30))]

这是我当前的代码：

import datetime

def merge_date_ranges(data):

    result = []

    for t1, t2 in ((data[i], data[i+1]) for i in range(len(data)-1)):   
        if t1[1] >= t2[0]:
            result.append((min(t1[0], t2[0]), max(t1[1], t2[1])))
        else:
            result.append(t1)

如果 T1（第一个日期时间范围）和 T2（第二个日期时间范围）不重叠，那么我只需将 T1 添加到新列表中（结果）。如果 T1 和 T2 DO 重叠，那么我将合并的元组添加到新列表中（结果）。

我的问题是合并后会发生什么。例如：

T1 = (datetime.datetime(2016, 1, 10, 13, 0), datetime.datetime(2016, 1, 10, 16, 0))
T2 = (datetime.datetime(2016, 1, 10, 14, 0), datetime.datetime(2016, 1, 10, 14, 0))

T1 和 T2 合并，以下添加到我的新列表中：

(datetime.datetime(2016, 1, 10, 13, 0), datetime.datetime(2016, 1, 10, 16, 0))

所以现在我希望我的代码（在 for 循环的下一次迭代中）将合并的元组（新 T1）与我列表中的下一个日期时间范围进行比较：

T1 = (datetime.datetime(2016, 1, 10, 13, 0), datetime.datetime(2016, 1, 10, 16, 0))
T2 = (datetime.datetime(2016, 1, 10, 22, 0), datetime.datetime(2016, 1, 10, 22, 0))

但是，T1 和 T2 是这样的：

T1 = (datetime.datetime(2016, 1, 10, 14, 0), datetime.datetime(2016, 1, 10, 14, 0))
T2 = (datetime.datetime(2016, 1, 10, 22, 0), datetime.datetime(2016, 1, 10, 22, 0))

并且 T1 被添加到我的新列表中（我不想要），因为它之前已经合并了。

但我就是想不通如何做到这一点。如果我能够通过用合并的元组替换 T2 并删除 T1 来更新我的原始列表会更容易。但据我所知，这是不可能的，甚至是一种好的做法。

在拔掉头发一周后，我在这里发布我的第一个问题，希望有人可以帮助我恢复理智。 :)

更新 基本上我想得到一个没有日期时间范围重叠的新列表。

Answer 1

我想这就是你想要的。 给它一个测试和评论：

def merge_date_ranges(data):
    result = []
    t_old = data[0]
    for t in data[1:]:
        if t_old[1] >= t[0]:  #I assume that the data is sorted already
            t_old = ((min(t_old[0], t[0]), max(t_old[1], t[1])))
        else:
            result.append(t_old)
            t_old = t
    else:
        result.append(t_old)
    return result

我假设日期已经排序。

顺便说一句，我发现唯一奇怪的日期是单日，也许你应该改正你的输入数据。

salida = merge_date_ranges(data)
for item in [t for t in data if t not in salida]:
    print item

(datetime.datetime(2016, 1, 10, 14, 0), datetime.datetime(2016, 1, 10, 14, 0))
(datetime.datetime(2016, 1, 11, 14, 0), datetime.datetime(2016, 1, 11, 14, 0))
(datetime.datetime(2016, 1, 12, 14, 0), datetime.datetime(2016, 1, 12, 14, 0))
(datetime.datetime(2016, 1, 13, 14, 0), datetime.datetime(2016, 1, 13, 14, 0))
(datetime.datetime(2016, 1, 13, 22, 0), datetime.datetime(2016, 1, 13, 22, 0))
(datetime.datetime(2016, 1, 14, 14, 0), datetime.datetime(2016, 1, 14, 14, 0))
(datetime.datetime(2016, 1, 14, 22, 0), datetime.datetime(2016, 1, 14, 22, 0))
(datetime.datetime(2016, 1, 15, 14, 0), datetime.datetime(2016, 1, 15, 14, 0))
(datetime.datetime(2016, 1, 15, 22, 0), datetime.datetime(2016, 1, 15, 22, 0))

Answer 2

编辑：鉴于新信息，您仍应执行循环，但基于布尔值并检查重叠。

import datetime

def merge_date_ranges(data):

    input = data
    result = []
    overlap = True # we assume there's overlap to begin with

    while(overlap):
        overlap = False # will remain false unless overlap is found

        for t1, t2 in ((input[i], input[i+1]) for i in range(len(input)-1)):   
            if t1[1] >= t2[0]:
                overlap = True # an overlap was found, so loop will continue
                result.append((min(t1[0], t2[0]), max(t1[1], t2[1])))
            else:
                result.append(t1)

        if(overlap):
            input = result # preparing the next round

    return result