從一組區間 Python 中獲取不重疊的不同區間

Question

給定一組區間，我想從一組區間中找到不重疊的不同區間。

例如：

輸入： [[1,10], [5,20], [6,21], [17,25], [22,23], [24,50],[30,55], [60,70]]

Output: [[1,5], [21,22], [23,24], [25,30], [50,55], [60,70]]

我怎樣才能做到這一點？

我目前嘗試過的：

gene_bounds_list = [[1,10],[5,20], [6,21],[17,25],[22,23], [24,50],[30,55],[60,70]]
overlap_list = []
nonoverlap_list = []
nonoverlap_list.append(gene_bounds_list[0])

for i in range(1, len(gene_bounds_list)):
    curr_gene_bounds = gene_bounds_list[i]
    prev_gene_bounds = nonoverlap_list[-1]

    if curr_gene_bounds[0]<prev_gene_bounds[0]:
        if curr_gene_bounds[1]<prev_gene_bounds[0]: #case1
            continue
        if curr_gene_bounds[1] < prev_gene_bounds[1]:  #case2
            nonoverlap_list[-1][0] = curr_gene_bounds[1]
        if curr_gene_bounds[1]>prev_gene_bounds[1]:
            # previous gene was completely overlapping within current gene,
            # so replace previous gene by current (bigger) gene and put previous gene into overlap list
            overlap_list.append(nonoverlap_list[-1])
            new_bound = [gene_bounds_list[i][0], gene_bounds_list[i][1]]
            nonoverlap_list.pop()
            nonoverlap_list.append([new_bound[0], new_bound[1]])

    elif curr_gene_bounds[0] > prev_gene_bounds[0] and curr_gene_bounds[1] < prev_gene_bounds[1]:
        # completely within another gene
        overlap_list.append([curr_gene_bounds[0], curr_gene_bounds[1]])

    elif curr_gene_bounds[0] < prev_gene_bounds[1]:
        # partially overlapping with another gene
        new_bound = [nonoverlap_list[-1][1], curr_gene_bounds[1]]
        nonoverlap_list[-1][1] = curr_gene_bounds[0]
        nonoverlap_list.append([new_bound[0], new_bound[1]])

    else:
        # not overlapping with another gene
        nonoverlap_list.append([gene_bounds_list[i][0], gene_bounds_list[i][1]])

unique_data = [list(x) for x in set(tuple(x) for x in gene_bounds_list)]
within_overlapping_intervals = []

for small in overlap_list:
    for master in unique_data:
        if (small[0]==master[0] and small[1]==master[1]):
            continue
        if (small[0]>master[0] and small[1]<master[1]):
            if(small not in within_overlapping_intervals):
                within_overlapping_intervals.append([small[0], small[1]])

for o in within_overlapping_intervals:
    nonoverlap_list.append(o)  # append the overlapping intervals

nonoverlap_list.sort(key=lambda tup: tup[0])
flat_data = sorted([x for sublist in nonoverlap_list for x in sublist])
new_gene_intervals = [flat_data[i:i + 2] for i in range(0, len(flat_data), 2)]
print(new_gene_intervals)

但是，這給了我一個 output： [[1, 5], [6, 10], [17, 20], [21, 22], [23, 24], [25, 30], [50, 55], [60, 70]]

關於如何刪除不需要的間隔的任何想法？

Answer 1

這是一種方法。 這個想法是在任何時候跟蹤間隔的層數。 為此，我們在進入區間時添加一層，退出時移除一層。

我們首先構建開始和結束的排序列表。 為了識別一個值是開始還是結束，我們創建元組(start, 1)或(end, -1) 。

然后，我們合並這兩個列表，按值排序，並遍歷結果列表（使用heapq.merge使這很容易）。 每次層數變為 1 時，我們就有一個非重疊區間的開始。 當它再次改變時，它就結束了。

from heapq import merge


def non_overlapping(data):
    out = []
    starts = sorted([(i[0], 1) for i in data])  # start of interval adds a layer of overlap
    ends = sorted([(i[1], -1) for i in data])   # end removes one
    layers = 0
    current = []
    for value, event in merge(starts, ends):    # sorted by value, then ends (-1) before starts (1)
        layers += event
        if layers ==1:  # start of a new non-overlapping interval
            current.append(value)
        elif current:  # we either got out of an interval, or started an overlap
            current.append(value)
            out.append(current)
            current = []
    return out


data = [[1,10], [5,20], [6,21], [17,25], [22,23], [24,50] ,[30,55], [60,70]]

non_overlapping(data)
# [[1, 5], [21, 22], [23, 24], [25, 30], [50, 55], [60, 70]]

請注意，您在問題中輸入的預期答案是錯誤的（例如，它包含不屬於輸入數據的 45）

Answer 2

Plot 時間線上的間隔。 由於范圍僅為10**5 ，我們可以使用 memory。 繪圖和掃描可以在線性時間內完成。

intervals = [[1,10], [5,20], [6,21], [17,25], [22,23], [24,50] ,[30,55], [60,70]]

max_ = max(intervals, key=lambda x: x[1])[1]
timeline = [0] * (max_ + 1)

# mark plots
for start, end in intervals:
    timeline[start] += 1
    timeline[end] -= 1


# make the timeline
for i in range(1, len(timeline)):
    timeline[i] += timeline[i - 1]


# scan
result = []
for i, item in enumerate(timeline):
    if i == 0:
        continue
    prev = timeline[i - 1]
    if item == 1 and prev != 1:
        result.append([i, i + 1])
    elif item == 1 and prev == 1:
        result[-1][1] = i + 1
        end = i


print(result)

編輯：隨着范圍更新為 ~ 10^8 ，這將不起作用。