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Question

考慮到與列表字典作爲值和一個單獨的列表：

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] 

如何刪除從列表中值myDict，如果他們匹配myList中值是多少？

所以對於例如字典我期待在年底這樣一個字典：

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

Answer 1

這工作：

>>> 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

讓myList一組，並使用字典和列表理解組合：

mySet = set(myList) 

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

使用一組使得這更高效，因爲in成員測試的集合比列表要快得多。

Answer 3

這個答案也行：

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