CODEWITHSUNDEEP
pythonpython-3.xdictionarytuples
artemis
artemis

Reputation: 7281

Compute dot product of two dictionaries with tuple values in Python

I have two dictionaries that look like so:

dict_of_items = tf_idf_by_doc {1: [('dog', 3), ('bird', 0)], 2: [('egret', 2), ('cat', 3), ('bird', 0), ('aardvark', 1)], 3: [('fish', 6), ('bird', 0), ('dog', 1), ('aardvark', 5)], 4: [('fish', 6), ('bird', 0), ('dog', 1), ('aardvark', 2)], 5: [('egret', 4), ('bird', 0)], 6: [('bird', 0)], 7: [('dog', 5), ('bird', 0)], 8: [('bird', 0), ('aardvark', 1)]}

dict_of_search = {1: [('bird', 0), ('dog', 1), ('cat', 3)]}

I need to compute the dot product between the dict_of_search and each of the keys in the dict_of_items, and then store the resulting dot product values and keep track by key. What I mean is...

In dict_of_items, 1 and the item in dict_of_search have a vectors of:

|      | dict_of_items_1 | dict_of_search |
|:----:|:---------------:|:--------------:|
| bird |        0        |        0       |
|  dog |        3        |        1       |
|  cat |        0        |        3       |

And so my dot product would be: 3

Desired results would be a dictionary of keys in dict_of_items and their respective dot products as compared to dict_of_search (this will only ever be one item), sorted in descending order by dot product.

However, I am not sure how to translate the shape of my dictionaries into two arrays to perform a vector calculation, especially when to handle when one of the terms does not appear (for example, in the example above cat did not appear in key 1 in dict_of_items_1.

I have tried something like this using numpy...

import numpy as numpy

def main():
    test_arr_1 = [1,2,3]
    test_arr_2 = [3,2,6]

    first_dot_product = numpy.dot(test_arr_1, test_arr_2)

    print("First Example: ", first_dot_product)

    test_arr_3 = [3,0,1]
    test_arr_4 = [2,10]

    second_dot_product = numpy.dot(test_arr_3, test_arr_4)

    print("Second Example Missing Value: ", second_dot_product)

main()

But that fails since the vectors are not of the same size and shape.

ValueError: shapes (3,) and (2,) not aligned: 3 (dim 0) != 2 (dim 0)

I have also tried to reshape the dictionary values into lists:

def main():
    dict_of_items = {'1': [('bird', 0), ('dog', 3), ('egret', 2), ('bird', 0), ('aardvark', 1), ('cat', 3), ('dog', 1), ('bird', 0), ('fish', 6), ('aardvark', 5), ('dog', 1), ('bird', 0), ('fish', 6), ('aardvark', 2), ('egret', 4), ('bird', 0), ('bird', 0), ('bird', 0), ('dog', 5), ('bird', 0), ('aardvark', 1)]}

    test_list_of_lists = []
    for k, v in dict_of_items.items():
        curr_list = []
        for aTuple in v:
            curr_list.append(aTuple[1])
        test_list_of_lists.append(curr_list)

    print(test_list_of_lists)   

main()

But that simply merges everything into one list incorrectly: [[0, 3, 2, 0, 1, 3, 1, 0, 6, 5, 1, 0, 6, 2, 4, 0, 0, 0, 5, 0, 1]]

I also took a look at this post, but that dictionary is in a much more simple format.

Upvotes: 2

Views: 715

Answers (2)

Dev Khadka
Dev Khadka

Reputation: 5461

it will be easier if you convert your tuples to dictionary like below. then we can use list comprehension like that

dict_of_items = {key:dict(value) for key, value in dict_of_items.items()}
dict_of_search = {key:dict(value) for key, value in dict_of_search.items()}

{item_key: sum([search[key]*item.get(key,0)  for key in search.keys()]) 
     for item_key, item in dict_of_items.items() 
     for search in dict_of_search.values()}

Upvotes: 0

Dani Mesejo
Dani Mesejo

Reputation: 61910

To compute the doc product of the values on dict_of_search vs dict_of_items, you could do:

def prod(source, target):
    return sum(source.get(key, 0) * target.get(key, 0) for key in source.keys() | target.keys())


dict_of_items = {1: [('dog', 3), ('bird', 0)], 2: [('egret', 2), ('cat', 3), ('bird', 0), ('aardvark', 1)],
                 3: [('fish', 6), ('bird', 0), ('dog', 1), ('aardvark', 5)],
                 4: [('fish', 6), ('bird', 0), ('dog', 1), ('aardvark', 2)], 5: [('egret', 4), ('bird', 0)],
                 6: [('bird', 0)], 7: [('dog', 5), ('bird', 0)], 8: [('bird', 0), ('aardvark', 1)]}

dict_of_search = {1: [('bird', 0), ('dog', 1), ('cat', 3)]}

for k, v in dict_of_items.items():
    for se in dict_of_search.values():
        print(k, prod(dict(v), dict(se)))

Output

1 3
2 9
3 1
4 1
5 0
6 0
7 5
8 0

If you want to store the results in a dictionary, do:

result = {}
for k, v in dict_of_items.items():
    for se in dict_of_search.values():
        result[k] = prod(dict(v), dict(se))

print(result)

Output

{1: 3, 2: 9, 3: 1, 4: 1, 5: 0, 6: 0, 7: 5, 8: 0}

Upvotes: 1

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