CODEWITHSUNDEEP
pythondictionarydataframeencodingtext-processing
Jai Kotia
Jai Kotia

Reputation: 176

How to encode words in a list in Python

I have a dictionary where I have each word as a key and a corresponding integer value, such as : 

 {'me': 41, 'are': 21, 'the': 0}

I have a data frame with a column of lists of words that are already tokenized, such as :

['I', 'liked', 'the', 'color', 'of', 'this', 'top']
['Just', 'grabbed', 'this', 'today', 'great', 'find']

How can I encode each of these words into their corresponding values from the dictionary. For example:

[56, 78, 5, 1197, 556, 991, 40]

Upvotes: 1

Views: 1678

Answers (4)

Neil
Neil

Reputation: 14313

Using a dictionary and list

The following uses a dictionary (final_dictionary) to determine the id's of the words. This is great if you have a preset dictionary of id's.

def encode_tokens(tokens):
    encoded_tokens = tokens[:]
    for i, token in enumerate(tokens):
        if token in final_dictionary:
            encoded_tokens[i] = final_dictionary[token]
    return encoded_tokens

print(encode_tokens(tokens))

Adding and maintaining id's

If you are dynamically assigning id's I would implement a class to do so (see bellow). If you, however, have a dictionary of id's that you have already defined ahead of time, you can pass in the keyword argument di:

token_words_1 = ['I', 'liked', 'the', 'color', 'of', 'this', 'top']
token_words_2 = ['I', 'liked', 'to', 'test', 'repeat', 'words']

class AutoId:
    def __init__(self, **kwargs):
        self.di = kwargs.get("di", {})
        self.loc = 0
    def get(self, value):
        if value not in self.di:
            self.di[value] = self.loc
            self.loc += 1
        return self.di[value]
    def get_list(self, li):
        return [*map(self.get, li)]

encoding = AutoId()
print(encoding.get_list(token_words_1))
print(encoding.get_list(token_words_2))

Upvotes: 1

mab
mab

Reputation: 2774

from itertools import chain
import numpy as np

# d = {'me': 41, 'are': 21, 'the': 0}
l1 = ['I', 'liked', 'the', 'color', 'of', 'this', 'top']
l2 = ['Just', 'grabbed', 'this', 'today', 'great', 'find']

# This is just for data generation for the sake of a complete example.
# Use your already given d here instead.
d = {k: np.random.randint(10) for k in chain(l1, l2)}
print(d)

l1_d = [d.get(k, 0) for k in l1]  # <- this is the actual command you need
print(l1_d)

l2_d = [d.get(k, 0) for k in l2]
print(l2_d)

Outcome:

{'I': 3, 'liked': 3, 'the': 8, 'color': 7, 'of': 3, 'this': 5,
 'top': 3, 'Just': 6, 'grabbed': 0, 'today': 0, 'great': 7, 'find': 0}
[3, 3, 8, 7, 3, 5, 3]
[6, 0, 5, 0, 7, 0]

Upvotes: 1

BiBi
BiBi

Reputation: 7908

What about doing

word2key = {'me': 41, 'are': 21, 'the': 0}
words = ['Just', 'grabbed', 'this', 'today', 'great', 'find']
default = 'unknown'
output = [word2key.get(x, default) for x in words]

You might want to use x.lower() if you want 'Just' and 'just' to be mapped to the same value.

Upvotes: 5

politinsa
politinsa

Reputation: 3720

Assume your dict is in a variable named d and your list is named l:

d = {'me': 41, 'are': 21, 'the': 0}
l = ['I', 'liked', 'the', 'color', 'of', 'this', 'top']

print(l)
c = 0
while c < len(l):
    try:
        l[c] = d[l[c]]
    except:
        l[c] = None
    c += 1

print(l)

Upvotes: 1

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