## 问题：如何根据任意条件函数过滤字典？

``>>> points={'a':(3,4), 'b':(1,2), 'c':(5,5), 'd':(3,3)}``

``````>>> points.items()
[('a', (3, 4)), ('c', (5, 5)), ('b', (1, 2)), ('d', (3, 3))]``````

``````>>> for item in [i for i in points.items() if i[1][0]<5 and i[1][1]<5]:
...     points_small[item[0]]=item[1]
...
>>> points_small
{'a': (3, 4), 'b': (1, 2), 'd': (3, 3)}``````

I have a dictionary of points, say:

``````>>> points={'a':(3,4), 'b':(1,2), 'c':(5,5), 'd':(3,3)}
``````

I want to create a new dictionary with all the points whose x and y value is smaller than 5, i.e. points ‘a’, ‘b’ and ‘d’.

According to the the book, each dictionary has the `items()` function, which returns a list of `(key, pair)` tuple:

``````>>> points.items()
[('a', (3, 4)), ('c', (5, 5)), ('b', (1, 2)), ('d', (3, 3))]
``````

So I have written this:

``````>>> for item in [i for i in points.items() if i[1][0]<5 and i[1][1]<5]:
...     points_small[item[0]]=item[1]
...
>>> points_small
{'a': (3, 4), 'b': (1, 2), 'd': (3, 3)}
``````

Is there a more elegant way? I was expecting Python to have some super-awesome `dictionary.filter(f)` function…

## 回答 0

``{k: v for k, v in points.iteritems() if v[0] < 5 and v[1] < 5}``

``{k: v for k, v in points.items() if v[0] < 5 and v[1] < 5}``

Nowadays, in Python 2.7 and up, you can use a dict comprehension:

``````{k: v for k, v in points.iteritems() if v[0] < 5 and v[1] < 5}
``````

And in Python 3:

``````{k: v for k, v in points.items() if v[0] < 5 and v[1] < 5}
``````

## 回答 1

``dict((k, v) for k, v in points.items() if all(x < 5 for x in v))``

`all(x < 5 for x in v)`如果您确定每个点始终都是二维的，则可能会过大（在这种情况下，您可能会用表示相同的约束`and`），但效果很好；-)。

``````dict((k, v) for k, v in points.items() if all(x < 5 for x in v))
``````

You could choose to call `.iteritems()` instead of `.items()` if you’re in Python 2 and `points` may have a lot of entries.

`all(x < 5 for x in v)` may be overkill if you know for sure each point will always be 2D only (in that case you might express the same constraint with an `and`) but it will work fine;-).

## 回答 2

``points_small = dict(filter(lambda (a,(b,c)): b<5 and c < 5, points.items()))``
``````points_small = dict(filter(lambda (a,(b,c)): b<5 and c < 5, points.items()))
``````

## 回答 3

``````>>> points = {'a': (3, 4), 'c': (5, 5), 'b': (1, 2), 'd': (3, 3)}
>>> dict(filter(lambda x: (x[1][0], x[1][1]) < (5, 5), points.items()))

{'a': (3, 4), 'b': (1, 2), 'd': (3, 3)}``````
``````>>> points = {'a': (3, 4), 'c': (5, 5), 'b': (1, 2), 'd': (3, 3)}
>>> dict(filter(lambda x: (x[1][0], x[1][1]) < (5, 5), points.items()))

{'a': (3, 4), 'b': (1, 2), 'd': (3, 3)}
``````

## 回答 4

``dict((k, v) for (k, v) in points.iteritems() if v[0] < 5 and v[1] < 5)``
``````dict((k, v) for (k, v) in points.iteritems() if v[0] < 5 and v[1] < 5)
``````

## 回答 5

``````class FilterDict(dict):
def __init__(self, input_dict):
for key, value in input_dict.iteritems():
self[key] = value
def filter(self, criteria):
for key, value in self.items():
if (criteria(value)):
self.pop(key)

my_dict = FilterDict( {'a':(3,4), 'b':(1,2), 'c':(5,5), 'd':(3,3)} )
my_dict.filter(lambda x: x[0] < 5 and x[1] < 5)``````

I think that Alex Martelli’s answer is definitely the most elegant way to do this, but just wanted to add a way to satisfy your want for a super awesome `dictionary.filter(f)` method in a Pythonic sort of way:

``````class FilterDict(dict):
def __init__(self, input_dict):
for key, value in input_dict.iteritems():
self[key] = value
def filter(self, criteria):
for key, value in self.items():
if (criteria(value)):
self.pop(key)

my_dict = FilterDict( {'a':(3,4), 'b':(1,2), 'c':(5,5), 'd':(3,3)} )
my_dict.filter(lambda x: x[0] < 5 and x[1] < 5)
``````

Basically we create a class that inherits from `dict`, but adds the filter method. We do need to use `.items()` for the the filtering, since using `.iteritems()` while destructively iterating will raise exception.

## 回答 6

``dict((k, v) for (k, v) in points.iteritems() if v[0] < 5 and v[1] < 5)``
``````dict((k, v) for (k, v) in points.iteritems() if v[0] < 5 and v[1] < 5)
``````