如果不在另一個列表中返回項目

Question

如果該項目未出現在我的任務列表Task中，我想返回Project中的項目。我的代碼只返回所有內容並在Project。我究竟做錯了什麼？

Task = [['Task1','Project1',3],['Task2','Project4',6]] 
Project = [['Project1', 'Andrew'],['Project2','Bob'],['Project3','Bob']] 

not_in_list = [item for item in Project if item[0] not in Case] 

print not_in_list 

輸出：

[['Project1', 'Andrew'], ['Project2', 'Bob'], ['Project3', 'Bob']] 

預期結果：

[['Project2', 'Bob'],['Project3', 'Bob']] 

Answer 1

該做的伎倆：

Task = [['Task1','Project1',3],['Task2','Project4',6]] 
Project = [['Project1', 'Andrew'],['Project2','Bob'],['Project3','Bob']] 

no_tasks = [p for p in Project if all(p[0] not in t for t in Task)] 
print no_tasks 

，但將是大名單效率極其低下。有時間重新考慮你的數據結構！

Answer 2

只是稍微效率比以前的答案，如果你可以假設項目名稱是始終在任務的索引1：

>>> Task = [['Task1','Project1',3],['Task2','Project4',6]] 
>>> Project = [['Project1', 'Andrew'],['Project2','Bob'],['Project3','Bob']] 
>>> assigned = [t[1] for t in Task] 
>>> [p for p in Project if p[0] not in assigned] 
[['Project2', 'Bob'], ['Project3', 'Bob']] 

Answer 3

task = [['Task1','Project1',3],['Task2','Project4',6]] 
project = [['Project1', 'Andrew'],['Project2','Bob'],['Project3','Bob']] 
task_projects = set(pr for _, pr, _ in task) 

not_in_list = [item for item in project if item[0] not in task_projects] 

print not_in_list 

（請注意，我改變了變量名，使他們匹配建議）。

此代碼首先創建一組存在的項目名稱。檢查集合中項目的存在比列表中便宜得多。