我有这些范围：
7,10
11,13
11,15
14,20
23,39
我需要执行重叠范围的并集以给出不重叠的范围，所以在示例中：
7,20
23,39
我在Ruby中已经完成了这个操作，我已经在数组中推动了范围的开始和结束并对它们进行了排序，然后执行重叠范围的并集。有什么快速的方法在Python中执行此操作吗？
由于
答案 0 :(得分：10)
我们说(7, 10)和(11, 13)会导致(7, 13)：
a = [(7, 10), (11, 13), (11, 15), (14, 20), (23, 39)]
b = []
for begin,end in sorted(a):
if b and b[-1][1] >= begin - 1:
b[-1] = (b[-1][0], end)
else:
b.append((begin, end))
b现在
[(7, 20), (23, 39)]
修改：
正如@CentAu正确注意到，[(2,4), (1,6)]将返回(1,4)而不是(1,6)。以下是正确处理此案例的新版本：
a = [(7, 10), (11, 13), (11, 15), (14, 20), (23, 39)]
b = []
for begin,end in sorted(a):
if b and b[-1][1] >= begin - 1:
b[-1][1] = max(b[-1][1], end)
else:
b.append([begin, end])
答案 1 :(得分：6)
老问题。但是我想为将来的参考添加这个答案。 sympy可用于实现间隔的结合：
from sympy import Interval, Union
def union(data):
""" Union of a list of intervals e.g. [(1,2),(3,4)] """
intervals = [Interval(begin, end) for (begin, end) in data]
u = Union(*intervals)
return [list(u.args[:2])] if isinstance(u, Interval) \
else list(u.args)
如果Union的输出超过两个间隔是Union个对象，而当存在单个间隔时，输出是Interval个对象。这就是返回行中if statement的原因。
的示例：
In [26]: union([(10, 12), (14, 16), (15, 22)])
Out[26]: [[10, 12], [14, 22]]
In [27]: union([(10, 12), (9, 16)])
Out[27]: [[9, 16]]
答案 2 :(得分：1)
我尝试了存在（45,46）和（45,45）的特定情况 并且还测试在您的应用中不太可能发生的情况：存在（11,6），存在（-1，-5），存在（-9,5），存在（-3,10）。登记/> 无论如何，结果适合所有这些情况，这是重点。
算法：
def yi(li):
gen = (x for a,b in li for x in xrange(a,b+1))
start = p = gen.next()
for x in gen:
if x>p+2:
yield (start,p)
start = p = x
else:
p = x
yield (start,x)
如果以下代码中的aff设置为True，则会显示执行步骤。
def yi(li):
aff = 0
gen = (x for a,b in li for x in xrange(a,b+1))
start = p = gen.next()
for x in gen:
if aff:
print ('start %s p %d p+2 %d '
'x==%s' % (start,p,p+2,x))
if x>p+2:
if aff:
print 'yield range(%d,%d)' % (start,p+1)
yield (start,p)
start = p = x
else:
p = x
if aff:
print 'yield range(%d,%d)' % (start,x+1)
yield (start,x)
for li in ([(7,10),(23,39),(11,13),(11,15),(14,20),(45,46)],
[(7,10),(23,39),(11,13),(11,15),(14,20),(45,46),(45,45)],
[(7,10),(23,39),(11,13),(11,15),(14,20),(45,45)],
[(7,10),(23,39),(11,13),(11,6),(14,20),(45,46)],
#1 presence of (11, 6)
[(7,10),(23,39),(11,13),(-1,-5),(14,20),(45,45)],
#2 presence of (-1,-5)
[(7,10),(23,39),(11,13),(-9,-5),(14,20),(45,45)],
#3 presence of (-9, -5)
[(7,10),(23,39),(11,13),(-3,10),(14,20),(45,45)]):
#4 presence of (-3, 10)
li.sort()
print 'sorted li %s'%li
print '\n'.join(' (%d,%d) %r' % (a,b,range(a,b))
for a,b in li)
print 'list(yi(li)) %s\n' % list(yi(li))
结果
sorted li [(7, 10), (11, 13), (11, 15), (14, 20),
(23, 39), (45, 46)]
(7,10) [7, 8, 9]
(11,13) [11, 12]
(11,15) [11, 12, 13, 14]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,46) [45]
list(yi(li)) [(7, 20), (23, 39), (45, 46)]
sorted li [(7, 10), (11, 13), (11, 15), (14, 20),
(23, 39), (45, 45), (45, 46)]
(7,10) [7, 8, 9]
(11,13) [11, 12]
(11,15) [11, 12, 13, 14]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,45) []
(45,46) [45]
list(yi(li)) [(7, 20), (23, 39), (45, 46)]
sorted li [(7, 10), (11, 13), (11, 15), (14, 20),
(23, 39), (45, 45)]
(7,10) [7, 8, 9]
(11,13) [11, 12]
(11,15) [11, 12, 13, 14]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,45) []
list(yi(li)) [(7, 20), (23, 39), (45, 45)]
sorted li [(7, 10), (11, 6), (11, 13), (14, 20),
(23, 39), (45, 46)]
(7,10) [7, 8, 9]
(11,6) []
(11,13) [11, 12]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,46) [45]
list(yi(li)) [(7, 20), (23, 39), (45, 46)]
sorted li [(-1, -5), (7, 10), (11, 13), (14, 20),
(23, 39), (45, 45)]
(-1,-5) []
(7,10) [7, 8, 9]
(11,13) [11, 12]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,45) []
list(yi(li)) [(7, 20), (23, 39), (45, 45)]
sorted li [(-9, -5), (7, 10), (11, 13), (14, 20),
(23, 39), (45, 45)]
(-9,-5) [-9, -8, -7, -6]
(7,10) [7, 8, 9]
(11,13) [11, 12]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,45) []
list(yi(li)) [(-9, -5), (7, 20), (23, 39), (45, 45)]
sorted li [(-3, 10), (7, 10), (11, 13), (14, 20),
(23, 39), (45, 45)]
(-3,10) [-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
(7,10) [7, 8, 9]
(11,13) [11, 12]
(14,20) [14, 15, 16, 17, 18, 19]
(23,39) [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38]
(45,45) []
list(yi(li)) [(-3, 20), (23, 39), (45, 45)]
答案 3 :(得分：0)
def range_overlap_adjust(list_ranges):
overlap_corrected = []
for start, stop in sorted(list_ranges):
if overlap_corrected and start-1 <= overlap_corrected [-1][1] and stop >= overlap_corrected [-1][1]:
overlap_corrected [-1] = min(overlap_corrected [-1][0], start), stop
elif overlap_corrected and start <= overlap_corrected [-1][1] and stop <= overlap_corrected [-1][1]:
break
else:
overlap_corrected.append((start,stop))
return overlap_corrected
list_ranges = [(7, 10), (11, 13), (11, 15), (14, 20), (23, 39)]
print range_overlap_adjust(list_ranges)
答案 4 :(得分：0)
这里是使用functools.reduce的单线（假设（x，10）和（11，y）重叠）：
reduce(
lambda acc, el: acc[:-1:] + [(min(*acc[-1], *el), max(*acc[-1], *el))]
if acc[-1][1] >= el[0] - 1
else acc + [el],
ranges[1::],
ranges[0:1]
)
这从第一个范围开始，并使用reduce遍历其余范围。它将最后一个元素（acc[-1]）与下一个范围（el）比较。如果它们重叠，它将用两个范围的最小值和最大值（acc[:-1:] + [min, max]）替换最后一个元素。如果它们不重叠，则仅将这个新范围放在列表的末尾（acc + [el]）。
示例：
from functools import reduce
example_ranges = [(7, 10), (11, 13), (11, 15), (14, 20), (23, 39)]
def combine_overlaps(ranges):
return reduce(
lambda acc, el: acc[:-1:] + [(min(*acc[-1], *el), max(*acc[-1], *el))]
if acc[-1][1] >= el[0] - 1
else acc + [el],
ranges[1::],
ranges[0:1],
)
print(combine_overlaps(example_ranges))
输出：
[(7, 20), (23, 39)]