list1 = ["experience","as","a","java","developer"]
list2 = ["B","O","O","B","I"]
list3 = ["java","developer"]
number = 0
print(list2)
for i in list1:
for j in list3:
if(i!=j):
for l in range(number,len(list2)):
list2[number] = "O"
number += 1
print(list2)
这里"B"
表示“experience”,"O"
表示“as”等等。Python:我需要以下程序的一些建议
预期输出:
["O","O","O","B","I"]
我的输出:
["O","O","O","O","O"]
你想设置o根据该单词是否在list3中找到,但是您多次设置多个元素,它们是list2中的ne元素。 – stark
我要求(i!= j)意味着(经验!= java)等等,如果这是真的,那么list2索引应该替换为“O”,而不是i [0]到j [0] – Shireesh