CODEWITHSUNDEEP
pythondictionarytext-processing
DannyBoy
DannyBoy

Reputation: 77

How to deal with keys not found in Python Dictionaries

I have the following code:

from math import sqrt
from collections import Counter

def forSearch():
    words = {'bit':{1:3,2:4,3:19,4:0},'shoe':{1:0,2:0,3:0,4:0},'dog':{1:3,2:0,3:4,4:5}, 'red':{1:0,2:0,3:15,4:0}}
    search = {'bit':1,'dog':3,'shoe':5}
num_files = 4

    file_relevancy = Counter()
    c = sqrt(sum([x**2 for x in search.values()]))
    for i in range(1, num_files+1):
        words_ith_val = [words[x][i] for x in search.keys() ]
        a = sum([search[key] * words[key][i] for key in search.keys()])
        b = sqrt(sum([x**2 for x in words_ith_val]))
        file_relevancy[i] = (a / (b * c))

    return [x[0] for x in file_relevancy.most_common(num_files)]

print forSearch()

However, this has a problem with words which are contained in search but not in words:

I want to say something like this here:

for i in range(1, num_files+1):
    if corresponding key in words cannot be found
        insert it and make its value = 0
    words_ith_val = [words[x][i] for x in search.keys() ]

Then it should work?

Unless anyone else has any better suggestions?

Upvotes: 0

Views: 134

Answers (3)

Salvatore Avanzo
Salvatore Avanzo

Reputation: 2786

What about this code:

if key not in words:
    words[key] = {k+1: 0 for k in range(num_files)}

In your code you can try to do

for key in search.keys():
    if key not in words:
        words[key] = {k+1: 0 for k in range(num_files)}
    words_ith_val = [words[key][k + 1] for k in range(num_files)]

Upvotes: 0

Klaus D.
Klaus D.

Reputation: 14404

You can use the defaultdict:

from collections import defaultdict
d = defaultdict(int)

This will initialize a dictionary where the keys are created on access and the default value is 0. You can use other types as well:

defaultdict(dict)
defaultdict(list)

They will be initialized with an empty dictionary/list. You can also overwrite the default value with a factory method. See https://docs.python.org/2/library/collections.html#collections.defaultdict for details.

Upvotes: 2

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799210

collections.defaultdict

import collections

D = collections.defaultdict(int)
D['foo'] = 42
print D['foo'], D['bar']

Upvotes: 2

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