用字典值替換鍵列表的有效方法？

Question

說我有一本字典屬於鍵列表：

dict = {"a":[1,2], "b":[3,4], "c":[5,6]} 
keys = ["a","b","c"] 

什麼是替換在字典中的值的列表中所有鍵的最有效方法是什麼？

即

keys = ["a","b","c"] 

成爲

keys = [[1,2],[3,4],[5,6]] 

Answer 1

map很容易：

map(d.get, keys) 

（或者，在Python 3.x中，list(map(d.get, keys))）

Answer 2

使用list comprehension像這樣：

>>> # Please don't name a dictionary dict -- it overrides the built-in 
>>> dct = {"a":[1,2], "b":[3,4], "c":[5,6]} 
>>> keys = ["a","b","c"] 
>>> id(keys) 
28590032 
>>> keys[:] = [dct[k] for k in keys] 
>>> keys 
[[1, 2], [3, 4], [5, 6]] 
>>> id(keys) 
28590032 
>>> 

[:]如果您想要列表對象保持不變時，才需要。否則，你可以將其刪除：

>>> dct = {"a":[1,2], "b":[3,4], "c":[5,6]} 
>>> keys = ["a","b","c"] 
>>> id(keys) 
28561280 
>>> keys = [dct[k] for k in keys] 
>>> keys 
[[1, 2], [3, 4], [5, 6]] 
>>> id(keys) 
28590032 
>>> 

Answer 3

如果你想把所有的值的列表從字典中，使用

>>> d={1: 'a', 2: 'b', 3: 'c', 4: 'd'} 
>>> d.values() 
['a', 'b', 'c', 'd'] 

它返回所有包含值的列表。

如果你想鍵的子集，你可以使用：

>>> d={1: 'a', 2: 'b', 3: 'c', 4: 'd'} 
>>> l=[1, 3] 
>>> [d[x] for x in l] 
['a', 'c'] 

請讓我知道，你所追求的......