  # Word analogies¶

کدها با تغییرات برگرفته از کورس Sequence Models پروفسور Andrew NG است.

https://www.coursera.org/learn/nlp-sequence-models

In :
import numpy as np
import os

In :
glove_dir = 'D:/data/'

embeddings_index = {}
f = open(os.path.join(glove_dir, 'glove.6B.100d.txt'), encoding="utf8")
for line in f:
values = line.split()
word = values
coefs = np.asarray(values[1:], dtype='float32')
embeddings_index[word] = coefs
f.close()


# 1 - Cosine similarity¶

To measure how similar two words are, we need a way to measure the degree of similarity between two embedding vectors for the two words. Given two vectors $u$ and $v$, cosine similarity is defined as follows:

$$\text{CosineSimilarity(u, v)} = \frac {u . v} {||u||_2 ||v||_2} = cos(\theta) \tag{1}$$

where $u.v$ is the dot product (or inner product) of two vectors, $||u||_2$ is the norm (or length) of the vector $u$, and $\theta$ is the angle between $u$ and $v$. This similarity depends on the angle between $u$ and $v$. If $u$ and $v$ are very similar, their cosine similarity will be close to 1; if they are dissimilar, the cosine similarity will take a smaller value. **Figure 1**: The cosine of the angle between two vectors is a measure of how similar they are

Exercise: Implement the function cosine_similarity() to evaluate similarity between word vectors.

Reminder: The norm of $u$ is defined as $||u||_2 = \sqrt{\sum_{i=1}^{n} u_i^2}$

In :
from sklearn.metrics.pairwise import cosine_similarity

def similarity(u, v):
return np.squeeze(cosine_similarity(u.reshape(1, -1), v.reshape(1, -1)))

In :
father = embeddings_index["father"]
mother = embeddings_index["mother"]
ball = embeddings_index["ball"]
crocodile = embeddings_index["crocodile"]
france = embeddings_index["france"]
italy = embeddings_index["italy"]
paris = embeddings_index["paris"]
rome = embeddings_index["rome"]
print("cosine_similarity(father, mother) = ", similarity(father, mother))
print("cosine_similarity(ball, crocodile) = ",similarity(ball, crocodile))
print("cosine_similarity(france - paris, iran - tehran) = ",similarity(france - paris, rome - italy))

cosine_similarity(father, mother) =  0.86566603
cosine_similarity(ball, crocodile) =  0.15206575
cosine_similarity(france - paris, iran - tehran) =  -0.70562387


## 2 - Word analogy task¶

In the word analogy task, we complete the sentence "*a* is to *b* as *c* is to **____**". An example is '*man* is to *woman* as *king* is to *queen*' . In detail, we are trying to find a word d, such that the associated word vectors $e_a, e_b, e_c, e_d$ are related in the following manner: $e_b - e_a \approx e_d - e_c$. We will measure the similarity between $e_b - e_a$ and $e_d - e_c$ using cosine similarity.

Exercise: Complete the code below to be able to perform word analogies!

In :
embeddings_index["father"]

Out:
array([ 0.64706 , -0.068067,  0.15468 , -0.17408 , -0.29134 ,  0.76999 ,
-0.3192  , -0.25663 , -0.25082 , -0.036737, -0.25509 ,  0.29636 ,
0.5776  ,  0.49641 ,  0.19167 , -0.83888 ,  0.58482 , -0.38717 ,
-0.71591 ,  0.9519  , -0.37966 , -0.1131  ,  0.47154 ,  0.20921 ,
0.38197 ,  0.067582, -0.92879 , -1.1237  ,  0.84831 ,  0.68744 ,
-0.15472 ,  0.92714 ,  0.53371 , -0.037392, -0.856   ,  0.19056 ,
-0.014594,  0.15186 ,  0.53514 , -0.20306 , -0.35164 ,  0.33152 ,
1.1306  , -0.72787 , -0.19724 ,  0.031659, -0.24041 , -0.057617,
0.60473 , -0.49233 , -0.24405 , -0.3184  ,  0.96156 ,  1.0895  ,
0.21534 , -2.0542  , -1.0615  ,  0.052439,  0.57958 ,  0.2748  ,
0.91587 ,  0.85195 ,  0.36113 , -0.31901 ,  0.7784  , -0.36865 ,
0.64387 ,  0.33104 , -0.27181 ,  0.58524 , -0.15143 ,  0.11121 ,
0.2126  , -0.60345 ,  0.16148 ,  0.32952 , -0.1354  , -0.30629 ,
-0.89143 ,  0.091912,  0.49753 ,  0.55932 ,  0.19329 ,  0.044859,
-1.0416  , -0.41566 , -0.54174 , -0.7244  , -0.57492 , -1.1188  ,
0.087097, -0.2992  ,  0.87227 ,  0.86996 , -0.89641 , -0.28259 ,
-0.47295 , -0.74062 , -0.39    , -0.78099 ], dtype=float32)
In :
def complete_analogy(word_a, word_b, word_c, embeddings_index):
"""
Performs the word analogy task as explained above: a is to b as c is to ____.

Arguments:
word_a -- a word, string
word_b -- a word, string
word_c -- a word, string
word_to_vec_map -- dictionary that maps words to their corresponding vectors.

Returns:
best_word --  the word such that v_b - v_a is close to v_best_word - v_c, as measured by cosine similarity
"""

# convert words to lower case
word_a, word_b, word_c = word_a.lower(), word_b.lower(), word_c.lower()

# Get the word embeddings v_a, v_b and v_c
e_a, e_b, e_c = embeddings_index[word_a], embeddings_index[word_b], embeddings_index[word_c]

words = embeddings_index.keys()
max_cosine_sim = -100              # Initialize max_cosine_sim to a large negative number
best_word = None                   # Initialize best_word with None, it will help keep track of the word to output

# loop over the whole word vector set
for w in words:
# to avoid best_word being one of the input words, pass on them.
if w in [word_a, word_b, word_c] :
continue

# Compute cosine similarity between the vector (e_b - e_a) and the vector ((w's vector representation) - e_c)  (≈1 line)
cosine_sim = similarity(e_b - e_a, embeddings_index[w] - e_c)

# If the cosine_sim is more than the max_cosine_sim seen so far,
# then: set the new max_cosine_sim to the current cosine_sim and the best_word to the current word (≈3 lines)
if cosine_sim > max_cosine_sim:
max_cosine_sim = cosine_sim
best_word = w

return best_word


Run the cell below to test your code, this may take 1-2 minutes.

In :
complete_analogy('china', 'chinese', 'iran', embeddings_index)

Out:
'iranian'
In :
complete_analogy('india', 'delhi', 'iran', embeddings_index)

Out:
'tehran'
In :
complete_analogy('man', 'woman', 'boy', embeddings_index)

Out:
'girl'
In :
complete_analogy('small', 'smaller', 'big', embeddings_index)

Out:
'bigger'
دانشگاه تربیت دبیر شهید رجایی
مباحث ویژه 2 - یادگیری عمیق پیشرفته
علیرضا اخوان پور
97-98
SRTTU.edu - Class.Vision - AkhavanPour.ir