Python - 交错两个不同长度的列表

给定两个不同长度的列表，任务是编写一个Python程序来交替获取它们的元素，并重复较小列表的列表元素，直到用尽较大的列表元素。

例子：

Input : test_list1 = [‘a’, ‘b’, ‘c’], test_list2 = [5, 7, 3, 0, 1, 8, 4]

Output : [‘a’, 5, ‘b’, 7, ‘c’, 3, ‘a’, 0, ‘b’, 1, ‘c’, 8, ‘a’, 4]

Explanation : Alternate elements from 1st list are printed in cyclic manner once it gets exhausted. Then after exhaustion, again 1st list starts from ‘a’, with elements left in 2nd list.

Input : test_list1 = [3, 8, 7], test_list2 = [5, 7, 3, 0, 1, 8]

Output : [3, 5, 8, 7, 7, 3, 3, 0, 8, 1, 7, 8]

Explanation : Alternate elements from 1st list are printed in cyclic manner once it gets exhausted. 3, 5, 8, 7, 7, 3.. Then after exhaustion, again 1st list starts from 3, with elements left in 2nd list

方法 #1：使用zip() + cycle() +列表理解

在这种情况下，较小列表中元素的重复使用 cycle() 处理，连接使用 zip() 完成。列表理解同时执行交错任务。

Python3

# Python3 code to demonstrate working of
# Repetitive Interleave 2 lists
# Using zip() + cycle() + list comprehension 
from itertools import cycle
  
# initializing lists
test_list1 = list('abc')
test_list2 = [5, 7, 3, 0, 1, 8, 4]
  
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
  
# zip() combining list, after Repetitiveness using cycle()
res = [ele for comb in zip(cycle(test_list1), test_list2) for ele in comb]
  
# printing result
print("The interleaved list : " + str(res))

输出：

The original list 1 is : [‘a’, ‘b’, ‘c’]

The original list 2 is : [5, 7, 3, 0, 1, 8, 4]

The interleaved list : [‘a’, 5, ‘b’, 7, ‘c’, 3, ‘a’, 0, ‘b’, 1, ‘c’, 8, ‘a’, 4]

方法#2：使用chain() + zip() + cycle()

大多数操作与上述方法相同，唯一的区别是使用 chain() 执行交错任务。

蟒蛇3

# Python3 code to demonstrate working of
# Repetitive Interleave 2 lists
# Using chain() + zip() + cycle()
from itertools import cycle, chain
  
# initializing lists
test_list1 = list('abc')
test_list2 = [5, 7, 3, 0, 1, 8, 4]
  
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
  
# zip() combining list, after Repetitiveness using cycle()
# chain() gets interleaved done
res = list(chain(*zip(cycle(test_list1), test_list2)))
  
# printing result
print("The interleaved list : " + str(res))

输出：

The original list 1 is : [‘a’, ‘b’, ‘c’]

The original list 2 is : [5, 7, 3, 0, 1, 8, 4]

The interleaved list : [‘a’, 5, ‘b’, 7, ‘c’, 3, ‘a’, 0, ‘b’, 1, ‘c’, 8, ‘a’, 4]