比较两个列表并使用循环打印下一个元素

Question

我有两个列表：

list1=['lo0','lo1','te123','te234']
list2=['lo0','first','lo1','second','lo2','third','te123','fourth']

我想要写一段python代码来打印list2的下一个元素，其中list1的项出现在list2中，否则写成"no-match"，即，我希望输出为：

first
second
no-match
fourth

我想出了以下代码：

for i1 in range(len(list2)):
        for i2 in range(len(list1)):
            if list1[i2]==rlist2[i1]:
                 desc.write(list2[i1+1])
                 desc.write('\n')

但它给出的输出为：

first
second
fourth

而且我不知道如何在list2中不存在元素的情况下导致“不匹配”。请指点一下！提前谢谢。

Answer 1

list1=['lo0','lo1','te123','te234']
list2=['lo0','first','l01','second','lo2','third','te123','fourth']

for i in list1:
    if i not in list2:
        print('no-match')
    else:
        print(list2[list2.index(i)+1])

或者，您可以包含一个try，但如果该项是list2中的最后一个值，则可以包含一个例程。

list1=['lo0','lo1','te123','te234','fourth']
list2=['lo0','first','l01','second','lo2','third','te123','fourth']

for i in list1:
    if i not in list2:
        print('no-match')
    else:
        try:            
            print(list2[list2.index(i)+1])
        except IndexError:
            print(str(i)+" is last item in the list2")

Answer 2

您可以使用带有集合的enumerate来测试成员资格，如果找到集合中的元素，则使用当前元素索引+1打印list2中的下一个元素：

list1=['lo0','lo1','te123','te234',"tel23"]
list2=['lo0','first','l01','second','lo2','third','te123','fourth']
st = set(list1)

# set start to one to always be one index ahead
for ind, ele in enumerate(list2, start=1):
    # if we get a match and it is not the last element from list2
    # print the following element.
    if ele in st and ind < len(list2):
        print(list2[ind])
    else:
        print("No match")

正确答案也是：

first
No match
second
No match
No match
No match
fourth
No match

'l01'不等于'lo1'，你也不能像重复单词一样使用索引，你总是会得到第一个匹配。

要将您自己的逻辑与double for循环相匹配，并进行O(n^*2)比较：

for ind, ele in enumerate(list2, start=1):
    for ele2 in list1:
        if ele == ele2 and ind < len(list2):
            print(list2[ind])
        else:
            print("No match")

Answer 3

list1=['lo0','lo1','te123','te234']
list2=['lo0','first','lo1','second','lo2','third','te123','fourth']
res=[]
for elm in list1:
    if elm in list2:
        print list2[list2.index(elm)+1]
    else : 
        print 'No match'

输出：

first
second
fourth
No match