比較兩個列表，並使用循環

Question

我有兩個列表打印下一個元素：

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

我想寫一個Python代碼打印列表2的下一個元素，其中列表1的產品出現在列表2，其他的寫「不匹配」，即我想要的輸出：

first 
second 
no-match 
fourth 

我想出了下面的代碼：

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') 

但給出的輸出爲：

first 
second 
fourth 

我不知道如何誘導「不匹配」，其中的元素不在列表2中。請指導！提前致謝。

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]) 

或者你可能包括一試，除了包括常規，如果該項目是在list2中的最後一個值。

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

您可以使用枚舉有設置來測試會員，如果發現是集合的元素，然後從列表2使用當前的元素索引+ 1打印下一個元素：

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") 

正確的答案也是：

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

'l01'不等於'lo1'，你也不能像使用重複單詞那樣使用索引，你總是會得到第一個匹配。

要與雙for循環和做Ø符合自己的邏輯（N^* 2）比較：

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' 

輸出地說：

first 
second 
fourth 
No match