Question

我有一个包含许多词典的大型列表。在每个字典中，我想迭代3个特定键，然后转储到新列表中。每个字典的键都相同。

例如，我想从下面列表中的所有词组中获取键c，d，e，输出到List2。

List = [{'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6...},
        {'a':10, 'b':20, 'c':30, 'd':40, 'e':50, 'f':60...},
        {'a':100, 'b':200, 'c':300, 'd':400, 'e':500, 'f':600...},]

List2 = [{'c':3, 'd':4, 'e':5},
         {'c':30, 'd':40, 'e':50},
         {'c':300, 'd':400, 'e':500}]

Answer 1

您可以使用嵌套的字典理解：

keys = ('c', 'd', 'e')
[{k: d[k] for k in keys} for d in List]

如果缺少这些键，您可以使用dictionary view object（Python 2中的dict.viewkeys()，Python 3中的dict.keys()）来查找交叉点仅包括实际存在的密钥：

keys = {'c', 'd', 'e'}
[{k: d[k] for k in d.viewkeys() & keys} for d in List]

演示：

>>> List = [{'a':1, 'b':2, 'c':3, 'd':4, 'e':5, 'f':6},
...         {'a':10, 'b':20, 'c':30, 'd':40, 'e':50, 'f':60},
...         {'a':100, 'b':200, 'c':300, 'd':400, 'e':500, 'f':600}]
>>> keys = ('c', 'd', 'e')
>>> [{k: d[k] for k in keys} for d in List]
[{'c': 3, 'e': 5, 'd': 4}, {'c': 30, 'e': 50, 'd': 40}, {'c': 300, 'e': 500, 'd': 400}]
>>> List = [{'a':1, 'b':2, 'd':4, 'e':5, 'f':6},
...         {'a':10, 'b':20, 'c':30, 'd':40, 'f':60},
...         {'a':100, 'b':200, 'e':500, 'f':600}]
>>> keys = {'c', 'd', 'e'}
>>> [{k: d[k] for k in d.viewkeys() & keys} for d in List]
[{'e': 5, 'd': 4}, {'c': 30, 'd': 40}, {'e': 500}]

Answer 2

试试这段代码：

keys = ['c', 'd']  
for dictionary in List1:
        d = {}
        for key in dictionary:
            if key in keys:
                d[key] = dictionary[key]
        List2.append(d)

如何迭代dict中的特定键来获取值

2 个答案: