如何將一組重疊范圍划分為非重疊范圍？

Question

假設您有一組范圍：

• 0 - 100：“一”
• 0 - 75：'b'
• 95 - 150：'c'
• 120 - 130：'d'

顯然，這些范圍在某些點重疊。 您將如何剖析這些范圍以生成不重疊范圍的列表，同時保留與其原始范圍相關的信息（在這種情況下，范圍后面的字母）？

例如，上面運行算法后的結果將是：

• 0 - 75：'a'，'b'
• 76 - 94：'一個'
• 95 - 100：'a'，'c'
• 101 - 119：'c'
• 120 - 130：'c'，'d'
• 131 - 150：'c'

Answer 1

在編寫混合（部分重疊）音頻樣本的程序時，我遇到了同樣的問題。

我所做的是將“開始事件”和“停止事件”（針對每個項目）添加到列表中，按時間點對列表進行排序，然后按順序處理。 您可以這樣做，除了使用 integer 點而不是時間，而不是混合聲音，您將向與范圍相對應的集合中添加符號。 無論您是生成空范圍還是忽略它們都是可選的。

Edit也許一些代碼......

# input = list of (start, stop, symbol) tuples
points = [] # list of (offset, plus/minus, symbol) tuples
for start,stop,symbol in input:
    points.append((start,'+',symbol))
    points.append((stop,'-',symbol))
points.sort()

ranges = [] # output list of (start, stop, symbol_set) tuples
current_set = set()
last_start = None
for offset,pm,symbol in points:
    if pm == '+':
         if last_start is not None:
             #TODO avoid outputting empty or trivial ranges
             ranges.append((last_start,offset-1,current_set))
         current_set.add(symbol)
         last_start = offset
    elif pm == '-':
         # Getting a minus without a last_start is unpossible here, so not handled
         ranges.append((last_start,offset-1,current_set))
         current_set.remove(symbol)
         last_start = offset

# Finish off
if last_start is not None:
    ranges.append((last_start,offset-1,current_set))

顯然，完全未經測試。

Answer 2

我會說創建一個端點列表並對其進行排序，還可以通過起點和終點索引范圍列表。 然后遍歷排序的端點列表，並為每個端點檢查范圍以查看哪些端點在該點開始/停止。

這可能在代碼中更好地表示......如果您的范圍由元組表示：

ranges = [(0,100,'a'),(0,75,'b'),(95,150,'c'),(120,130,'d')]
endpoints = sorted(list(set([r[0] for r in ranges] + [r[1] for r in ranges])))
start = {}
end = {}
for e in endpoints:
    start[e] = set()
    end[e] = set()
for r in ranges:
    start[r[0]].add(r[2])
    end[r[1]].add(r[2])
current_ranges = set()
for e1, e2 in zip(endpoints[:-1], endpoints[1:]):
    current_ranges.difference_update(end[e1])
    current_ranges.update(start[e1])
    print '%d - %d: %s' % (e1, e2, ','.join(current_ranges))

雖然回想起來，如果沒有更有效（或至少看起來更干凈）的方法來做到這一點，我會感到驚訝。

Answer 3

你描述的是集合論的一個例子。 有關計算集合的並集、交集和差集的通用算法，請參見：

www.gvu.gatech.edu/~jarek/graphics/papers/04PolygonBooleansMargalit.pdf

雖然本文針對的是圖形，但它也適用於一般集合論。 不完全是輕閱讀材料。

Answer 4

對 Edmunds 的類似回答，經過測試，包括對 (1,1) 等間隔的支持：

class MultiSet(object):
    def __init__(self, intervals):
        self.intervals = intervals
        self.events = None

    def split_ranges(self):
        self.events = []
        for start, stop, symbol in self.intervals:
            self.events.append((start, True, stop, symbol))
            self.events.append((stop, False, start, symbol))

        def event_key(event):
            key_endpoint, key_is_start, key_other, _ = event
            key_order = 0 if key_is_start else 1
            return key_endpoint, key_order, key_other

        self.events.sort(key=event_key)

        current_set = set()
        ranges = []
        current_start = -1

        for endpoint, is_start, other, symbol in self.events:
            if is_start:
                if current_start != -1 and endpoint != current_start and \
                       endpoint - 1 >= current_start and current_set:
                    ranges.append((current_start, endpoint - 1, current_set.copy()))
                current_start = endpoint
                current_set.add(symbol)
            else:
                if current_start != -1 and endpoint >= current_start and current_set:
                    ranges.append((current_start, endpoint, current_set.copy()))
                current_set.remove(symbol)
                current_start = endpoint + 1

        return ranges


if __name__ == '__main__':
    intervals = [
        (0, 100, 'a'), (0, 75, 'b'), (75, 80, 'd'), (95, 150, 'c'), 
        (120, 130, 'd'), (160, 175, 'e'), (165, 180, 'a')
    ]
    multiset = MultiSet(intervals)
    pprint.pprint(multiset.split_ranges())


[(0, 74, {'b', 'a'}),
 (75, 75, {'d', 'b', 'a'}),
 (76, 80, {'d', 'a'}),
 (81, 94, {'a'}),
 (95, 100, {'c', 'a'}),
 (101, 119, {'c'}),
 (120, 130, {'d', 'c'}),
 (131, 150, {'c'}),
 (160, 164, {'e'}),
 (165, 175, {'e', 'a'}),
 (176, 180, {'a'})]

Answer 5

偽代碼：

unusedRanges = [ (each of your ranges) ]
rangesInUse = []
usedRanges = []
beginningBoundary = nil

boundaries = [ list of all your ranges' start and end values, sorted ]
resultRanges = []

for (boundary in boundaries) {
    rangesStarting = []
    rangesEnding = []

    // determine which ranges begin at this boundary
    for (range in unusedRanges) {
        if (range.begin == boundary) {
            rangesStarting.add(range)
        }
    }

    // if there are any new ones, start a new range
    if (rangesStarting isn't empty) {
        if (beginningBoundary isn't nil) {
            // add the range we just passed
            resultRanges.add(beginningBoundary, boundary - 1, [collected values from rangesInUse])
        }

        // note that we are starting a new range
        beginningBoundary = boundary

        for (range in rangesStarting) {
            rangesInUse.add(range)
            unusedRanges.remove(range)
        }
    }

    // determine which ranges end at this boundary
    for (range in rangesInUse) {
        if (range.end == boundary) {
            rangesEnding.add(range)
        }
    }

    // if any boundaries are ending, stop the range
    if (rangesEnding isn't empty) {
        // add the range up to this boundary
        resultRanges.add(beginningBoundary, boundary, [collected values from rangesInUse]

        for (range in rangesEnding) {
            usedRanges.add(range)
            rangesInUse.remove(range)
        }

        if (rangesInUse isn't empty) {
            // some ranges didn't end; note that we are starting a new range
            beginningBoundary = boundary + 1
        }
        else {
            beginningBoundary = nil
        }
    }
}

單元測試：

最后，resultRanges 應該有你正在尋找的結果，unusedRanges 和 rangeInUse 應該是空的，beginningBoundary 應該是 nil，並且 usedRanges 應該包含未使用的Ranges 曾經包含的內容（但按 range.end 排序）。