连接 list1和的最简单方法list2:
merged = list1 + list2
zip返回一个元组列表,其中第i个元组包含每个参数序列或可迭代对象中的第i个元素:
alist = ['a1', 'a2', 'a3']
blist = ['b1', 'b2', 'b3']
for a, b in zip(alist, blist):
print(a, b)
# 输出:
# a1 b1
# a2 b2
# a3 b3
如果列表的长度不同,则结果将仅包含与最短的元素一样多的元素:
alist = ['a1', 'a2', 'a3']
blist = ['b1', 'b2', 'b3', 'b4']
for a, b in zip(alist, blist):
print(a, b)
# 输出:
# a1 b1
# a2 b2
# a3 b3
alist = []
len(list(zip(alist, blist)))
# 输出:
# 0
对于长度不等长的填充列表,请使用Nones itertools.zip_longest(itertools.izip_longest在Python 2中)
alist = ['a1', 'a2', 'a3']
blist = ['b1']
clist = ['c1', 'c2', 'c3', 'c4']
for a,b,c in itertools.zip_longest(alist, blist, clist):
print(a, b, c)
# 输出:
# a1 b1 c1
# a2无c2
# a3无c3
# 无无c4
插入到特定索引值:
alist = [123, 'xyz', 'zara', 'abc']
alist.insert(3, [2009])
print("最终名单:", alist)
输出:
最终名单: [123, 'xyz', 'zara', 2009, 'abc']