如何在 Python 中过滤字典？
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How to filter a dictionary in Python?
提问by user426795
d = {'foo': 'x',
'bar': 'y',
'zoo': 'None',
'foobar': 'None'}
I want to filter all the items whose value is 'None'and update the fooand baritems with a particular value. I tried:
我想过滤其值为 的所有项目并使用特定值'None'更新foo和bar项目。我试过：
for i in x.items():
....: if i[i] == 'None':
....: x.pop(i[0])
....: else:
....: x.update({i[0]:'updated'})
But it is not working.
但它不起作用。
采纳答案by khachik
It is not clear what is 'None'in the dictionary you posted. If it is a string, you can use the following:
不清楚'None'你贴的字典里有什么。如果是字符串，则可以使用以下内容：
dict((k, 'updated') for k, v in d.iteritems() if v != 'None')
If it is None, just replace the checking, for example:
如果是None，只需替换检查，例如：
dict((k, 'updated') for k, v in d.iteritems() if v is None)
回答by SilentGhost
it's not clear where you're getting your 'updated'value from, but in general it would look like this:
目前尚不清楚您'updated'从哪里获得价值，但总的来说，它看起来像这样：
{i: 'updated' for i, j in d.items() if j != 'None'}
in python2.7 or newer.
在 python2.7 或更新版本中。
回答by Rosh Oxymoron
new_d = dict((k, 'updated') for k, v in d.iteritems() if k != 'None')
回答by ismail
Something like this should work
这样的事情应该工作
>>> for x in [x for x in d.keys() if d[x] == 'None']:
d.pop(x)
回答by Woody1193
You could try writing a general filter function:
您可以尝试编写一个通用过滤器函数：
def filter(dict, by_key = lambda x: True, by_value = lambda x: True):
for k, v in dict.items():
if (by_key(k) and by_value(v)):
yield (k, v)
or
或者
def filter(dict, by_key = lambda x: True, by_value = lambda x: True):
return dict((k, v) for k, v in dict.items() if by_key(k) and by_value(v))
and then you can do this:
然后你可以这样做：
new_d = dict(filter(d, by_key = lambda k: k != 'None'))
Note that this was written to be compatible with Python 3. Mileage may vary.
请注意，这是为了与 Python 3 兼容而编写的。里程可能会有所不同。
The cool thing about this is that it allows you to pass arbitrary lambdas in to filter instead of hard-coding a particular condition. Plus, it's not any slower than any of the other solutions here. Furthermore, if you use the first solution I provided then you can iterate over the results.
很酷的一点是，它允许您将任意 lambda 传入过滤器，而不是对特定条件进行硬编码。另外，它并不比这里的任何其他解决方案慢。此外，如果您使用我提供的第一个解决方案，那么您可以迭代结果。
