时间戳

首页 > 解决方案 > 从一组区间Python中获取不重叠的不同区间

python - 从一组区间Python中获取不重叠的不同区间

问题描述

给定一组区间，我想从一组区间中找到不重叠的不同区间。

例如：

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

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

我怎样才能做到这一点？

我目前尝试过的：

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，这给了我一个输出：[[1, 5], [6, 10], [17, 20], [21, 22], [23, 24], [25, 30], [50, 55], [60, 70]]

关于如何删除不需要的间隔的任何想法？

标签： pythonintervalsoverlapping

解决方案

这是一种方法。这个想法是在任何时候跟踪间隔的层数。为此，我们在进入区间时添加一层，退出时移除一层。

我们首先构建开始和结束的排序列表。为了识别一个值是开始还是结束，我们创建元组(start, 1)或(end, -1)。

然后，我们合并这两个列表，按值排序，并遍历结果列表（使用heapq.merge使这很容易）。每次层数变为 1 时，我们就有一个非重叠区间的开始。当它再次改变时，它就结束了。

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]]

请注意，您在问题中输入的预期答案是错误的（例如，它包含不属于输入数据的 45）

推荐阅读