Python 3二维列表理解

Question

我有两个列表，用于组装第三个列表。我将list_one与list_two进行了比较，如果list_one中的某个字段的值位于list_two中，则两个来自list_two的值都将被复制到list_final中。如果一个字段的值从list_two中丢失，那么我希望看到一个空值（无）放入list_final中。 list_final将有相同数量的项目，并在相同的顺序list_one：

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven'] 
list_two = [['seven','7'], ['five','5'], ['four','4'], ['three','3'], ['one','1']] 
list_final = [] 

list_final的值应为：

[['one','1'], [None,None], ['three','3'], ['four','4'], ['five','5'], [None,None], ['seven','7']] 

我已经得到最接近的是：

list_final = [x if [x,0] in list_two else [None,None] for x in list_one] 

但这只是填充list_final None。我已经看了一些教程，但我似乎无法围绕这个概念包围我的大脑。任何帮助，将不胜感激。

Answer 1

发生了什么事在你的代码：

list_final = [x if [x,0] in list_two else [None,None] for x in list_one] 

1. 以list_one与
取代所有的元素
2. 要么x（又名保持完好）是否存在于list_two[x,0]（但，见下文）
3. ELSE与[None, None]替换当前的元素。

而作为list_two不包含匹配[x,0]任何元素（无论是你定的例子x），所有的值被替换[None, None]。

工作液

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven'] 
list_two = [['seven','7'], ['five','5'], ['four','4'], ['three','3'], ['one','1']] 

# Turns list_two into a nice and convenient dict much easier to work with 
# (Could be inline, but best do it once and for all) 
list_two = dict(list_two) # {'one': '1', 'three': '3', etc} 

list_final = [[k, list_two[k]] if k in list_two else [None, None] for k in list_one] 

矿，在另一方面：

1. 获取你你想要什么，又名[k, dict(list_two)[k]]
2. 但只尝试这样做，如果k in list_two
3. ELSE用[None, None]替换此条目。

Answer 2

你可以试试这个：

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven'] 
list_two = [['seven','7'], ['five','5'], ['four','4'], ['three','3'], ['one','1']] 
final_list = [[None, None] if not any(i in b for b in list_two) else [c for c in list_two if i in c][0] for i in list_one] 
print(final_list) 

输出：

[['one', '1'], [None, None], ['three', '3'], ['four', '4'], ['five', '5'], [None, None], ['seven', '7']]