我有一个元组列表,其中每个元组都是(start-time, end-time)
。我正在尝试合并所有重叠的时间范围并返回不同时间范围的列表。 例如合并具有重叠时间范围的时间范围元组列表
[(1, 5), (2, 4), (3, 6)] ---> [(1,6)]
[(1, 3), (2, 4), (5, 8)] ---> [(1, 4), (5,8)]
这是我如何实现它。
# Algorithm
# initialranges: [(a,b), (c,d), (e,f), ...]
# First we sort each tuple then whole list.
# This will ensure that a<b, c<d, e<f ... and a < c < e ...
# BUT the order of b, d, f ... is still random
# Now we have only 3 possibilities
#================================================
# b<c<d: a-------b Ans: [(a,b),(c,d)]
# c---d
# c<=b<d: a-------b Ans: [(a,d)]
# c---d
# c<d<b: a-------b Ans: [(a,b)]
# c---d
#================================================
def mergeoverlapping(initialranges):
i = sorted(set([tuple(sorted(x)) for x in initialranges]))
# initialize final ranges to [(a,b)]
f = [i[0]]
for c, d in i[1:]:
a, b = f[-1]
if c<=b<d:
f[-1] = a, d
elif b<c<d:
f.append((c,d))
else:
# else case included for clarity. Since
# we already sorted the tuples and the list
# only remaining possibility is c<d<b
# in which case we can silently pass
pass
return f
我想如果
- 弄清楚的是一个在某些Python模块的内置功能,可以更有效地做到这一点?或
- 是否有一种更为pythonic的方式来实现相同的目标?
您的帮助表示感谢。谢谢!然后
谢谢!同意我应该消除`set()`。循环会照顾它。就像根据需要产生元组而不是追加到列表一样。 – 2011-04-15 19:22:31
不幸的是,如果`len(times)== 0`,这会失败。 – phihag 2012-05-28 21:19:38