首页 > 编程语言 > 详细

python 交错列表合并

时间:2019-01-11 21:51:02      阅读:233      评论:0      收藏:0      [点我收藏+]
l1=[1,2,3]
l2=[a,"b","c"]
list1=list(chain.from_iterable(zip(l1,l2)))
list2=[]
for i in list1:
    list2.append(i)
print(list2)

技术分享图片

def xmerge(a, b):
    alen, blen = len(a), len(b)
    mlen = min(alen, blen)
    for i in range(mlen):
        yield a[i]
        yield b[i]

    if alen > blen:
        for i in range(mlen, alen):
            yield a[i]
    else:
        for i in range(mlen, blen):
            yield b[i]

a = [1, 2, 3]
b = [5, 6, 7, 8, 9, 10]

c = [i for i in xmerge(a, b)]
print (c)

c = [i for i in xmerge(b, a)]
print (c)

技术分享图片

a = [1, 2, 3]
b = [5, 6, 7]
result = [list(zip(a, b))[i][j] for i in range(len(a)) for j in range(len(list(zip(a, b))[0]))]
print(result)

技术分享图片

三种方法足够了,如果你有其他方法也可以留言

python 交错列表合并

原文:https://www.cnblogs.com/liangliangzz/p/10257356.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!