获取 Python 字典的子集
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Get a sub-set of a Python dictionary
提问by Oli
I have a dictionary:
我有一本字典：
{'key1':1, 'key2':2, 'key3':3}
I need to pass a sub-set of that dictionary to third-party code. It only wants a dictionary containing keys ['key1', 'key2', 'key99']and if it gets another key (eg 'key3'), it explodes in a nasty mess. The code in question is out of my control so I'm left in a position where I have to clean my dictionary.
我需要将该字典的一个子集传递给第三方代码。它只需要一个包含键的字典['key1', 'key2', 'key99']，如果它得到另一个键（例如'key3'），它就会爆炸成一团糟。有问题的代码超出了我的控制范围，所以我不得不清理我的字典。
What's the best, way to limit a dictionary to a set of keys?
将字典限制为一组键的最佳方法是什么？
Given the example dictionary and allowed keys above, I want:
鉴于上面的示例字典和允许的键，我想要：
{'key1':1, 'key2':2}
回答by unutbu
In [38]: adict={'key1':1, 'key2':2, 'key3':3}
In [41]: dict((k,adict[k]) for k in ('key1','key2','key99') if k in adict)
Out[41]: {'key1': 1, 'key2': 2}
In Python3 (or Python2.7 or later) you can do it with a dict-comprehensiontoo:
在 Python3（或 Python2.7 或更高版本）中，您也可以使用dict-comprehension 来实现：
>>> {k:adict[k] for k in ('key1','key2','key99') if k in adict}
{'key2': 2, 'key1': 1}
回答by Bj?rn Pollex
dict(filter(lambda i:i[0] in validkeys, d.iteritems()))
回答by watsonic
In modern Python (2.7+,3.0+), use a dictionary comprehension:
在现代 Python (2.7+,3.0+) 中，使用字典理解：
d = {'key1':1, 'key2':2, 'key3':3}
included_keys = ['key1', 'key2', 'key99']
{k:v for k,v in d.items() if k in included_keys}
回答by Lei Zhao
My way to do this is.
我这样做的方法是。
from operator import itemgetter
def subdict(d, ks):
return dict(zip(ks, itemgetter(*ks)(d)))
my_dict = {'key1':1, 'key2':2, 'key3':3}
subdict(my_dict, ['key1', 'key3'])
Update
更新
I have to admit though, the above implementation doesn't handle the case when the length of ksis 0 or 1. The following code handles the situation and it is no longer an one-liner.
我不得不承认，上面的实现没有处理长度ks为 0 或 1 的情况。下面的代码处理这种情况，它不再是单行的。
def subdict(d, ks):
vals = []
if len(ks) >= 1:
vals = itemgetter(*ks)(d)
if len(ks) == 1:
vals = [vals]
return dict(zip(ks, vals))
回答by Nazime Lakehal
An other solution without if in dict comprehension.
没有 if in dict 理解的另一种解决方案。
>>> a = {'key1':1, 'key2':2, 'key3':3}
>>> b = {'key1':1, 'key2':2}
>>> { k:a[k] for k in b.keys()}
{'key2': 2, 'key1': 1}
回答by theo olsthoorn
With a complex class Myclassbeing a subclass of collections.UserDict. To select a subset of it, i.e keeping all its container properties, it's convenient to define a method, e.g. named sublike so:
复杂类Myclass是collections.UserDict. 要选择它的一个子集，即保留其所有容器属性，定义一个方法很方便，例如命名sub如下：
def sub(self, keys):
subset = Myclass() # no arguments; works if defined with only keyword arguments
for key in keys:
subset[key] = self[key]
return subset
It is then used as Myclass.sub([key1, key2 ...])
然后用作Myclass.sub([key1, key2 ...])
