How to filter a dictionary according to an arbitrary condition function?


Question

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...

1
184
11/20/2015 10:23:15 PM

Accepted Answer

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}
378
4/27/2016 12:13:42 PM

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;-).


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