如何将列表中的子列表成对组合? 例如:
list1 = [[1,2,3],[4,5],[6],[7,8],[9,10]]
结果将会是:
[[1,2,3,4,5],[6,7,8],[9,10]]
如何将列表中的子列表成对组合? 例如:
list1 = [[1,2,3],[4,5],[6],[7,8],[9,10]]
[[1,2,3,4,5],[6,7,8],[9,10]]
您可以使用zip_longest
和填充值(以防列表包含奇数个子列表)来压缩对list1
的迭代器。在zip生成器对象上运行列表推导式允许您连接连续的列表对:
>>> from itertools import zip_longest # izip_longest in Python 2.x
>>> x = iter(list1)
>>> [a+b for a, b in zip_longest(x, x, fillvalue=[])]
[[1, 2, 3, 4, 5], [6, 7, 8], [9, 10]]
list1 = [[1, 2, 3], [4, 5], [6], [7, 8], [9, 10]]
n = len(list1)
[list1[i] + (list1[i+1] if i+1 < n else []) for i in xrange(0, n, 2)]
=> [[1, 2, 3, 4, 5], [6, 7, 8], [9, 10]]
list1=[[1,2,3],[4,5],[6],[7,8],[9,10]]
length = len(list1)
new_list = [ list1[i]+list1[i+1] if i+1 < length
else [list1[i]] for i in range(0,length,2) ]
print(new_list)
for i in range(len(l)/2):#here we go only to last even item
l[i]+=l[i+1]#adding odd sublist to even sublist
l.pop(i+1)#removing even sublist
>>> list1=[[1,2,3],[4,5],[6],[7,8],[9,10]]
>>> list1
[[1, 2, 3], [4, 5], [6], [7, 8], [9, 10]]
现在我们可以做到:
>>> test = [list1[0]+list1[1]]+[list1[2]+list1[3]]+list1[4]
>>> test
[[1, 2, 3, 4, 5], [6, 7, 8], 9, 10]
>>>
我相信有更好的方法,但这是我能想到的方法!
list1 = [[1, 2, 3], [4, 5], [6], [7, 8], [9, 10]]
from itertools import islice, chain
print([list(chain.from_iterable(islice(list1, i, i + 2)))
for i in range(0, len(list1), 2)])
[[1, 2, 3, 4, 5], [6, 7, 8], [9, 10]]
或者不使用islice:
print([list(chain.from_iterable(list1[i:i+2]))
for i in range(0, len(list1), 2)])
[[1, 2, 3, 4, 5], [6, 7, 8], [9, 10]]
list1=[[1,2,3],[4,5],[6],[7,8],[9,10]]
newlist = []
for i in range(0, len(list1), 2):
newlist.append(list1[i] + list1[i+1])
if len(list1) % 2 > 0:
newlist.append(list1[-1])
print newlist
这是(我希望)一个正确的解决方案:
def pair_up(ls):
new_list = []
every_other1 = ls[::2]
every_other2 = ls[1::2]
for i in range(len(every_other2)):
new_list.append(every_other1[i]+every_other2[i])
if len(ls) % 2 == 1:
new_list.append(ls[-1])
return new_list