CODEWITHSUNDEEP
pythonarraysdictionaryfilter
user9410919
user9410919

Reputation:

Filter dictionary from array values

I have a dictionary which has values as follows:

dictionary = {(10,9): 1, (44,11): 2, (1,1): 99}

Basically my keys are pairs of integers and the values of each key are just integers.

I have an array which stores a set of keys:

array = [(1,1), (5,19), (58,7)]

I would like to filter my dictionary to contain only elements which keys are stored in the array. In my case, after filtering the dictionary I would obtain the following:

dictionary = {(1,1): 99}

since the only key of the dictionary which is stored in the array is (1,1)

What would be the most efficient way to do this?

Upvotes: 3

Views: 2229

Answers (4)

slider
slider

Reputation: 12990

What about just:

res = {i: dictionary[i] for i in array if i in dictionary}

This does it in O(n) where n is the number of items in array and doesn't require any additional data structures.

For your dictionary and array, this gives us:

{(1, 1): 99}

Upvotes: 0

sacuL
sacuL

Reputation: 51395

You could find the set intersection of the dictionary keys and the array tuples, then get your new values in a dict comprehension. This will reduce the complexity of searching for each key in your array:

dictionary = {(10,9): 1, (44,11): 2, (1,1): 99}
array = [(1,1), (5,19), (58,7)]

>>> {i:dictionary[i] for i in set(dictionary.keys()).intersection(array)}
{(1, 1): 99}

Upvotes: 1

timgeb
timgeb

Reputation: 78750

Here's a dict-comprehension:

>>> dictionary = {(10,9): 1, (44,11): 2, (1,1): 99}
>>> lst = [(1,1), (5,19), (58,7)]    
>>> d = {k:v for k,v in dictionary.items() if k in lst}
>>> {(1, 1): 99}

I renamed array to lst because it is a list. We should not confuse lists with numpy arrays, array.array or the bytearray type.

You could also write a traditional for loop if you are not yet comfortable with comprehensions:

>>> d = {}
>>> for key in dictionary:
...:    if key in lst:
...:        d[key] = dictionary[key]
...:        
>>> d
>>> {(1, 1): 99}

Upvotes: 0

Dani Mesejo
Dani Mesejo

Reputation: 61910

You could do something like this:

dictionary = {(10,9): 1, (44,11): 2, (1,1): 99}
array = [(1,1), (5,19), (58,7)]

result = { k:v for k, v in dictionary.items() if k in array}

Output

{(1, 1): 99}

Or even faster, transforming the list into a set:

s = set(array)
result = {k: v for k, v in dictionary.items() if k in s}

Upvotes: 6

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