Python中确定两个范围中哪些值重叠的最佳方法是什么？
例如:
x = range(1,10)
y = range(8,20)
(The answer I am looking for would be the integers 8 and 9.)
给定一个范围,x,迭代另一个范围的最佳方法是什么,y并输出两个范围共享的所有值？在此先感谢您的帮助.
编辑:
作为后续行动,我意识到我还需要知道x是否与y重叠.我正在寻找一种迭代范围列表的方法,并做一些重叠范围的额外事情.是否有一个简单的True/False语句来实现这一目标？
And*_*ark 69
如果步骤始终为+1(这是范围的默认值),则以下应该比将每个列表转换为集合或迭代任一列表更有效:
range(max(x[0], y[0]), min(x[-1], y[-1])+1)
joa*_*uin 48
尝试使用set intersection:
>>> x = range(1,10)
>>> y = range(8,20)
>>> xs = set(x)
>>> xs.intersection(y)
set([8, 9])
请注意,intersection接受任何iterable作为参数(y不需要将其转换为操作的集合).有一个等效于该intersection方法的运算符:&但是,在这种情况下,它需要两个参数都是集合.
pla*_*aes 14
您可以使用set s,但要注意set(list)从以下位置删除所有重复的条目list:
>>> x = range(1,10)
>>> y = range(8,20)
>>> list(set(x) & set(y))
[8, 9]
Tim*_*man 10
一种选择是使用列表理解,如:
x = range(1,10)
y = range(8,20)
z = [i for i in x if i in y]
print z
小智 10
如果您寻找两个实值有界区间之间的重叠，那么这非常好：
def overlap(start1, end1, start2, end2):
"""how much does the range (start1, end1) overlap with (start2, end2)"""
return max(max((end2-start1), 0) - max((end2-end1), 0) - max((start2-start1), 0), 0)
我在网上找不到这个，所以我想出了这个，我在这里发帖。
上面的答案似乎大多过于复杂。这个 liner 在 Python3 中完美运行，将范围作为输入和输出。它还处理非法范围。如果不是 None ，要获取值只需迭代结果。
# return overlap range for two range objects or None if no ovelap
# does not handle step!=1
def range_intersect(r1, r2):
return range(max(r1.start,r2.start), min(r1.stop,r2.stop)) or None
这是 step=1 情况下的简单范围的答案（99% 的时间），当使用集合比较长范围时，它可以比基准快 2500 倍（当您只想知道是否有重叠时） ：
x = range(1,10)
y = range(8,20)
def range_overlapping(x, y):
if x.start == x.stop or y.start == y.stop:
return False
return x.start <= y.stop and y.start <= x.stop
>>> range_overlapping(x, y)
True
要查找重叠值：
def overlap(x, y):
if not range_overlapping(x, y):
return set()
return set(range(max(x.start, y.start), min(x.stop, y.stop)+1))
视觉帮助：
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基准：
x = range(1,10)
y = range(8,20)
In [151]: %timeit set(x).intersection(y)
2.74 µs ± 11.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [152]: %timeit range_overlapping(x, y)
1.4 µs ± 2.91 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
结论：即使对于小范围，它的速度也快两倍。
x = range(1,10000)
y = range(50000, 500000)
In [155]: %timeit set(x).intersection(y)
43.1 ms ± 158 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [156]: %timeit range_overlapping(x, y)
1.75 µs ± 88.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
结论：你想在这种情况下使用 range_overlapping 函数，因为它快了 2500 倍（我在加速方面的个人记录）
对于“如果x与y重叠或不重叠”：
for a,b,c,d in ((1,10,10,14),
(1,10,9,14),
(1,10,4,14),
(1,10,4,10),
(1,10,4,9),
(1,10,4,7),
(1,10,1,7),
(1,10,-3,7),
(1,10,-3,2),
(1,10,-3,1),
(1,10,-11,-5)):
x = range(a,b)
y = range(c,d)
print 'x==',x
print 'y==',y
b = not ((x[-1]<y[0]) or (y[-1]<x[0]))
print ' x %s y' % ("does not overlap"," OVERLAPS ")[b]
print
结果
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [10, 11, 12, 13]
x does not overlap y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [9, 10, 11, 12, 13]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [4, 5, 6, 7, 8, 9]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [4, 5, 6, 7, 8]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [4, 5, 6]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [1, 2, 3, 4, 5, 6]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [-3, -2, -1, 0, 1, 2, 3, 4, 5, 6]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [-3, -2, -1, 0, 1]
x OVERLAPS y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [-3, -2, -1, 0]
x does not overlap y
x== [1, 2, 3, 4, 5, 6, 7, 8, 9]
y== [-11, -10, -9, -8, -7, -6]
x does not overlap y
速度比较：
from time import clock
x = range(-12,15)
y = range(-5,3)
te = clock()
for i in xrange(100000):
w = set(x).intersection(y)
print ' set(x).intersection(y)',clock()-te
te = clock()
for i in xrange(100000):
w = range(max(x[0], y[0]), min(x[-1], y[-1])+1)
print 'range(max(x[0], y[0]), min(x[-1], y[-1])+1)',clock()-te
结果
set(x).intersection(y) 0.951059981087
range(max(x[0], y[0]), min(x[-1], y[-1])+1) 0.377761978129
这些执行时间的比率为2.5
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