Question

我有两个有序的时间间隔集（(表示“开始”，)表示“停止”）：

      1: (          )              (          )
         0---1------3------5---6---7------9---10----> time
      2: (   )(            )   (          )

在两个列表中，它看起来像：

intervals1 = [(0,3), (7,10)]
intervals2 = [(0,1), (1,5), (6,9)]

对时间序列的进一步评估将是两者的时间积分。为此，我想保留间隔字符，但要保留为通用间隔。在给定的示例中，时间序列和相应的列表如下所示：

common: (   )(     )(     )   (   )(     )(  )
        0---1------3------5---6---7------9---10----> time
intervals = [(0,1), (1,3), (3,5), (6,7), (7,9), (9,10)]

如何有效地将这两个时间序列结合起来？

Answer 1

我认为您可以使用基于堆栈的算法来有效解决此问题，因为您知道在任何给定位置最多可以重叠两个间隔：

def merge_intervals(a, b):
    stack = sorted(a+b, reverse=True)

    while len(stack) > 1:
        first = stack.pop()
        second = stack.pop()
        if first == second:         # identical intervals can be merged
            yield first
        elif first[1] <= second[0]: # no overlapping, yield first interval, put back second
            yield first
            stack.append(second)
        elif first[0] == second[0]: # overlap at start, yield shorter, put back rest of longer
            if first[1] > second[1]:
                first, second = second, first
            yield first
            stack.append((first[1], second[1]))
        elif first[1] < second[1]:  # partial overlap, yield first two parts, put back rest
            yield first[0], second[0]
            yield second[0], first[1]
            stack.append((first[1], second[1]))
        else: # first[1] >= second[1] # total envelopment
            yield first[0], second[0]
            yield second
            if first[1] != second[1]:
                stack.append((second[1], first[1]))

    yield from stack # there may or may not be one element left over

这是一个生成器，因此您将通过以下方式获得所需的输出：

intervals = list(merge_intervals(intervals1, intervals2))

将两组间隔合并为一组间隔

1 个答案: