更好的解决方案

def get_abs_min(ranges):

return min([min(r) for r in ranges])

def get_abs_max(ranges):

return max([max(r) for r in ranges])

def count_appearances(i, ranges):

return sum([1 for r in ranges if i in r])

def create_histogram(ranges):

keys = [str(i) for i in range(len(ranges) + 1)]

histogram = dict.fromkeys(keys)

results = []

min = get_abs_min(range_list)

max = get_abs_max(range_list)

for i in range(min, max):

count = str(count_appearances(i, ranges))

if histogram[count] is None:

histogram[count] = dict(start=i, end=None)

elif histogram[count]['end'] is None:

histogram[count]['end'] = i

elif histogram[count]['end'] == i - 1:

histogram[count]['end'] = i

else:

start = histogram[count]['start']

end = histogram[count]['end']

results.append((range(start, end + 1), count))

histogram[count]['start'] = i

histogram[count]['end'] = None

for count, d in histogram.items():

if d is not None and d['start'] is not None and d['end'] is not None:

results.append((range(d['start'], d['end'] + 1), count))

return results

def main(ranges, top):

appearances = create_histogram(ranges)

return sorted(appearances, key=lambda t: t[1], reverse=True)[:top]

这里的想法很简单,就是迭代所有范围的叠加,并建立一个外观直方图(例如,当前原始范围的数量

i

出现在)

之后,根据选择的结果大小进行排序和切片。

只需使用所需的范围和最上面的数字调用main(或

None

下面的旧编辑

我(几乎)同意@Kasramvd的回答。

from collections import Counter

from itertools import combinations

def range_overlap(x, y):

common_part = list(set(x) & set(y))

if common_part:

return range(common_part[0], common_part[-1] +1)

else:

return False

def get_most_common(range_list, top_frequent):

overlaps = Counter(range_overlap(i, j) for i, j in

combinations(list_of_ranges, 2))

return [(r, i) for (r, i) in overlaps.most_common(top_frequent) if r]

你需要输入

range_list

top_frequent

你想要的。

编辑

此编辑将根据您的输入和结果进行测试,并给出正确答案:

from collections import Counter

from itertools import combinations

def range_overlap(*args):

sets = [set(r) for r in args]

common_part = list(set(args[0]).intersection(*sets))

if common_part:

return range(common_part[0], common_part[-1] +1)

else:

return False

def get_all_possible_combinations(range_list):

all_combos = []

for i in range(2, len(range_list)):

all_combos.append(combinations(range_list, i))

all_combos = [list(combo) for combo in all_combos]

return all_combos

def get_most_common_for_combo(combo):

return list(filter(None, [range_overlap(*option) for option in combo]))

def get_most_common(range_list, top_frequent):

all_overlaps = []

combos = get_all_possible_combinations(range_list)

for combo in combos:

all_overlaps.extend(get_most_common_for_combo(combo))

return [r for (r, i) in Counter(all_overlaps).most_common(top_frequent) if r]

为了得到结果

get_most_common(range_list, top_frequent)

在我的机器上测试(ubunut 16.04 with python 3.5.2)并输入

范围列表

top_frequent = 5

结果如下:

[range(3000, 4000), range(2500, 4000), range(1500, 4000), range(3000, 6000), range(1, 4000)]