我想我曾经写过这个。只需遍历列表即可完成此操作。
alist = [[2, 3], [4, 5], [7, 8], [8, 9], [11, 12], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [20, 21]]
l = [alist[0][:]]
for e in alist[1:]:
if l[-1][-1] == e[0]:
l[-1].append(e[1])
else:
l.append(e[:])
代码读取为从第一对开始。循环其余部分。检查最后一个列表的最后一个元素是否与该对的第一个元素相同。如果是,则附加第二个元素,否则将该对附加到列表中。
这导致l
being:
[[2, 3], [4, 5], [7, 8, 9], [11, 12], [14, 15, 16, 17, 18, 19], [20, 21]]
如果您只想要最大的子列表,我建议:
>>> l = [[2, 3], [4, 5], [7, 8, 9], [11, 12], [14, 15, 16, 17, 18, 19], [20, 21]]
>>> max(l, key=len)
[14, 15, 16, 17, 18, 19]
并评价:
>>> alist = [[2, 3], [4, 5], [7, 8], [8, 9], [11, 12], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [20, 21]]
>>>
>>> l = [alist[0][:]]
>>> for e in alist[1:]:
... if l[-1][-1] == e[0]:
... l[-1].append(e[1])
... else:
... l.append(e[:])
...
>>> l
[[2, 3], [4, 5], [7, 8, 9], [11, 12], [14, 15, 16, 17, 18, 19], [20, 21]]
>>> alist
[[2, 3], [4, 5], [7, 8], [8, 9], [11, 12], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [20, 21]]
并进行了比较。减少解决方案需要 6.4 usecs:
$ python -mtimeit "list = [[2, 3], [4, 5], [7, 8], [8, 9], [11, 12], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [20, 21]]" "reduce(lambda x,y: x[:-1] + [x[-1] + y[1:]] if x[-1][-1] == y[0] else x + [y], list[1:], [list[0]])"
100000 loops, best of 3: 6.4 usec per loop
for 循环需要 3.62 usecs:
$ python -mtimeit "alist = [[2, 3], [4, 5], [7, 8], [8, 9], [11, 12], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [20, 21]]" "l = [alist[0][:]]" "for e in alist[1:]:" " if l[-1][-1] == e[0]:" " l[-1].append(e[1])" " else:" " l.append(e[:])"
100000 loops, best of 3: 3.62 usec per loop
在 Python 2.7.3 上。 for 循环快了 56%。输入越大,差异可能会更加明显,因为列表串联的成本取决于两个列表的长度之和。而附加到列表则稍微便宜一些。