更新列表

Question

考虑到与列表字典作为值和一个单独的列表：

myDict = {0: [0, 1, 2], 1: [], 2: [20, 25, 26, 28, 31, 34], 3: [], 4: [0, 1, 2, 3, 4, 7, 10], 5: [], 6: [10, 20, 24]} 

myList = [1, 34, 10] 

如何删除从列表中值myDict，如果他们匹配myList中值是多少？

所以对于例如字典我期待在年底这样一个字典：

myDict = {0: [0, 2], 1: [], 2: [20, 25, 26, 28, 31], 3: [], 4: [0, 2, 3, 4, 7], 5: [], 6: [20, 24]} 

Answer 1

这工作：

>>> myDict = {0: [0, 1, 2], 1: [], 2: [20, 25, 26, 28, 31, 34], 3: [], 4: [0, 1, 2, 3, 4, 7, 10], 5: [], 6: [10, 20, 24]} 
>>> myList = [1, 34, 10] 
>>> {x:[z for z in y if z not in myList] for x,y in myDict.items()} 
{0: [0, 2], 1: [], 2: [20, 25, 26, 28, 31], 3: [], 4: [0, 2, 3, 4, 7], 5: [], 6: [20, 24]} 
>>> 

Answer 2

让myList一组，并使用字典和列表理解组合：

mySet = set(myList) 

myDict = {k: [i for i in v if i not in mySet] for k, v in myDict.items()} 

使用一组使得这更高效，因为in成员测试的集合比列表要快得多。

Answer 3

这个答案也行：

updatedDict = {k:list(set(v) - set(myList)) for k, v in myDict.items()}