使用一個列表來排序另一個

Question

我想用Python 3來使用list_one對list_two進行排序並輸出到list_sorted。如果字段的值丟失，那麼我希望看到一個空值。輸出將具有相同數量的項目作爲list_one

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven'] 
list_two = ['seven', 'five', 'four', 'three', 'one'] 
list_sorted = [] 

for i in list_one: 
    for field in list_two: 
     if i == field: 
      list_sorted.append(field) 
     else: 
      list_sorted.append(None) 

print ('final output:', list_sorted) 

所需的輸出是：

['one', None, 'three', 'four', 'five', None, 'seven'] 

但它的實際輸出：

[None, None, None, None, 'one', None, None, None, None, None, None, None, None, 'three', None, None, None, 'four', None, None, None, 'five', None, None, None, None, None, None, None, None, 'seven', None, None, None, None] 

我現在的想法是答案涉及enumerate但我不確定。任何幫助將大大讚賞。

Answer 1

將list_two轉換成一個集合，然後根據該元素是否在list_two建立一個列表理解。

set_two = set(list_two) 
list_sorted = [x if x in set_two else None for x in list_one] 

print(list_sorted) 
['one', None, 'three', 'four', 'five', None, 'seven'] 

Answer 2

你不是真的排序任何東西，但它看起來像你可以達到你想要一部測試i是否存在於list_two。刪除內部for循環。

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven'] 
list_two = ['seven', 'five', 'four', 'three', 'one'] 
list_sorted = [] 

for i in list_one: 
    if i in list_two: 
     list_sorted.append(i) 
    else: 
     list_sorted.append(None) 

print ('final output:', list_sorted)