Assignment1词向量

提前导包：

import sys
assert sys.version_info[0]==3
assert sys.version_info[1] >= 5

from gensim.models import KeyedVectors
from gensim.test.utils import datapath
import pprint
import matplotlib.pyplot as plt
plt.rcParams['figure.figsize'] = [10, 5]
import nltk
from nltk.corpus import reuters
import numpy as np
import random
import scipy as sp
from sklearn.decomposition import TruncatedSVD
from sklearn.decomposition import PCA

START_TOKEN = '<START>'
END_TOKEN = '<END>'

np.random.seed(0)
random.seed(0)

Tip：from nltk.corpus import reuters这一句需要提前下载reuters，解压后放置C:UsersAdministrator ltk_datacorpora目录下，这样不用运行时再下载。

1.基于共现矩阵得到的词向量

1.1加载语料，给每句话都加上start_token和end_token

def read_corpus(category="crude"):
    """ Read files from the specified Reuter's category.
        Params:
            category (string): category name
        Return:
            list of lists, with words from each of the processed files
    """
    files = reuters.fileids(category)
    return [[START_TOKEN] + [w.lower() for w in list(reuters.words(f))] + [END_TOKEN] for f in files]

测试一下：

reuters_corpus = read_corpus()
pprint.pprint(reuters_corpus[:1], compact=True, width=100)   #pprint美观打印

[['<START>', 'japan', 'to', 'revise', 'long', '-', 'term', 'energy', 'demand', 'downwards', 'the',
'ministry', 'of', 'international', 'trade', 'and', 'industry', '(', 'miti', ')', 'will', 'revise',
'its', 'long', '-', 'term', 'energy', 'supply', '/', 'demand', 'outlook', 'by', 'august', 'to',
'meet', 'a', 'forecast', 'downtrend', 'in', 'japanese', 'energy', 'demand', ',', 'ministry',
'officials', 'said', '.', 'miti', 'is', 'expected', 'to', 'lower', 'the', 'projection', 'for',
'primary', 'energy', 'supplies', 'in', 'the', 'year', '2000', 'to', '550', 'mln', 'kilolitres',
'(', 'kl', ')', 'from', '600', 'mln', ',', 'they', 'said', '.', 'the', 'decision', 'follows',
'the', 'emergence', 'of', 'structural', 'changes', 'in', 'japanese', 'industry', 'following',
'the', 'rise', 'in', 'the', 'value', 'of', 'the', 'yen', 'and', 'a', 'decline', 'in', 'domestic',
'electric', 'power', 'demand', '.', 'miti', 'is', 'planning', 'to', 'work', 'out', 'a', 'revised',
'energy', 'supply', '/', 'demand', 'outlook', 'through', 'deliberations', 'of', 'committee',
'meetings', 'of', 'the', 'agency', 'of', 'natural', 'resources', 'and', 'energy', ',', 'the',
'officials', 'said', '.', 'they', 'said', 'miti', 'will', 'also', 'review', 'the', 'breakdown',
'of', 'energy', 'supply', 'sources', ',', 'including', 'oil', ',', 'nuclear', ',', 'coal', 'and',
'natural', 'gas', '.', 'nuclear', 'energy', 'provided', 'the', 'bulk', 'of', 'japan', "'", 's',
'electric', 'power', 'in', 'the', 'fiscal', 'year', 'ended', 'march', '31', ',', 'supplying',
'an', 'estimated', '27', 'pct', 'on', 'a', 'kilowatt', '/', 'hour', 'basis', ',', 'followed',
'by', 'oil', '(', '23', 'pct', ')', 'and', 'liquefied', 'natural', 'gas', '(', '21', 'pct', '),',
'they', 'noted', '.', '<END>']]

1.2得到词典（去掉重复词）

def distinct_words(corpus):
    """ Determine a list of distinct words for the corpus.
        Params:
            corpus (list of list of strings): corpus of documents
        Return:
            corpus_words (list of strings): list of distinct words across the corpus, sorted (using python 'sorted' function)
            num_corpus_words (integer): number of distinct words across the corpus
    """
    corpus_words = []
    num_corpus_words = -1
    
    # ------------------
    # Write your implementation here.
    corpus_words = sorted(list(set([word for wordlst in corpus for word in wordlst])))
    num_corpus_words = len(corpus_words)
    # ------------------

    return corpus_words, num_corpus_words

 测试一下：

# Define toy corpus
test_corpus = ["{} All that glitters isn't gold {}".format(START_TOKEN, END_TOKEN).split(" "), "{} All's well that ends well {}".format(START_TOKEN, END_TOKEN).split(" ")]
test_corpus_words, num_corpus_words = distinct_words(test_corpus)

# Correct answers
ans_test_corpus_words = sorted([START_TOKEN, "All", "ends", "that", "gold", "All's", "glitters", "isn't", "well", END_TOKEN])
ans_num_corpus_words = len(ans_test_corpus_words)

# Test correct number of words
assert(num_corpus_words == ans_num_corpus_words), "Incorrect number of distinct words. Correct: {}. Yours: {}".format(ans_num_corpus_words, num_corpus_words)

# Test correct words
assert (test_corpus_words == ans_test_corpus_words), "Incorrect corpus_words.
Correct: {}
Yours:   {}".format(str(ans_test_corpus_words), str(test_corpus_words))

# Print Success
print ("-" * 80)
print("Passed All Tests!")
print ("-" * 80)

--------------------------------------------------------------------------------
Passed All Tests!
--------------------------------------------------------------------------------

1.3计算共现矩阵

举个栗子： Co-Occurrence with Fixed Window of n=1

Document 1: "all that glitters is not gold"

Document 2: "all is well that ends well"

def compute_co_occurrence_matrix(corpus, window_size=4):
    """ Compute co-occurrence matrix for the given corpus and window_size (default of 4).
    
        Note: Each word in a document should be at the center of a window. Words near edges will have a smaller
              number of co-occurring words.
              
              For example, if we take the document "<START> All that glitters is not gold <END>" with window size of 4,
              "All" will co-occur with "<START>", "that", "glitters", "is", and "not".
    
        Params:
            corpus (list of list of strings): corpus of documents
            window_size (int): size of context window
        Return:
            M (a symmetric numpy matrix of shape (number of unique words in the corpus , number of unique words in the corpus)): 
                Co-occurence matrix of word counts. 
                The ordering of the words in the rows/columns should be the same as the ordering of the words given by the distinct_words function.
            word2Ind (dict): dictionary that maps word to index (i.e. row/column number) for matrix M.
    """
    words, num_words = distinct_words(corpus)
    M = None
    word2Ind = {}
    
    # ------------------
    # Write your implementation here.
    M = np.zeros((num_words, num_words))
    word2Ind = dict(zip(words, range(num_words)))
    
    for sen in corpus:
        for i in range(len(sen)):                   #sen[i]为中心词
            center_idx = word2Ind[sen[i]]           #center_idx为中心词对应的索引
            for w in sen[i-window_size : i] + sen[i+1 : i+window_size+1]:    #遍历中心词的上下文    
                context_idx = word2Ind[w]                                    #context_idx为上下文对应的索引  
                M[center_idx, context_idx] += 1      
    # ------------------

    return M, word2Ind

 测试一下：

# Define toy corpus and get student's co-occurrence matrix
test_corpus = ["{} All that glitters isn't gold {}".format(START_TOKEN, END_TOKEN).split(" "), "{} All's well that ends well {}".format(START_TOKEN, END_TOKEN).split(" ")]
M_test, word2Ind_test = compute_co_occurrence_matrix(test_corpus, window_size=1)

# Correct M and word2Ind
M_test_ans = np.array( 
    [[0., 0., 0., 0., 0., 0., 1., 0., 0., 1.,],
     [0., 0., 1., 1., 0., 0., 0., 0., 0., 0.,],
     [0., 1., 0., 0., 0., 0., 0., 0., 1., 0.,],
     [0., 1., 0., 0., 0., 0., 0., 0., 0., 1.,],
     [0., 0., 0., 0., 0., 0., 0., 0., 1., 1.,],
     [0., 0., 0., 0., 0., 0., 0., 1., 1., 0.,],
     [1., 0., 0., 0., 0., 0., 0., 1., 0., 0.,],
     [0., 0., 0., 0., 0., 1., 1., 0., 0., 0.,],
     [0., 0., 1., 0., 1., 1., 0., 0., 0., 1.,],
     [1., 0., 0., 1., 1., 0., 0., 0., 1., 0.,]]
)
ans_test_corpus_words = sorted([START_TOKEN, "All", "ends", "that", "gold", "All's", "glitters", "isn't", "well", END_TOKEN])
word2Ind_ans = dict(zip(ans_test_corpus_words, range(len(ans_test_corpus_words))))

# Test correct word2Ind
assert (word2Ind_ans == word2Ind_test), "Your word2Ind is incorrect:
Correct: {}
Yours: {}".format(word2Ind_ans, word2Ind_test)

# Test correct M shape
assert (M_test.shape == M_test_ans.shape), "M matrix has incorrect shape.
Correct: {}
Yours: {}".format(M_test.shape, M_test_ans.shape)

# Test correct M values
for w1 in word2Ind_ans.keys():
    idx1 = word2Ind_ans[w1]
    for w2 in word2Ind_ans.keys():
        idx2 = word2Ind_ans[w2]
        student = M_test[idx1, idx2]
        correct = M_test_ans[idx1, idx2]
        if student != correct:
            print("Correct M:")
            print(M_test_ans)
            print("Your M: ")
            print(M_test)
            raise AssertionError("Incorrect count at index ({}, {})=({}, {}) in matrix M. Yours has {} but should have {}.".format(idx1, idx2, w1, w2, student, correct))

# Print Success
print ("-" * 80)
print("Passed All Tests!")
print ("-" * 80)

--------------------------------------------------------------------------------
Passed All Tests!
--------------------------------------------------------------------------------

1.4使用奇异值分解对共现矩阵降维（降成二维是为了在坐标轴中展示）

def reduce_to_k_dim(M, k=2):
    """ Reduce a co-occurence count matrix of dimensionality (num_corpus_words, num_corpus_words)
        to a matrix of dimensionality (num_corpus_words, k) using the following SVD function from Scikit-Learn:
            - http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.TruncatedSVD.html
    
        Params:
            M (numpy matrix of shape (number of unique words in the corpus , number of unique words in the corpus)): co-occurence matrix of word counts
            k (int): embedding size of each word after dimension reduction
        Return:
            M_reduced (numpy matrix of shape (number of corpus words, k)): matrix of k-dimensioal word embeddings.
                    In terms of the SVD from math class, this actually returns U * S
    """    
    n_iters = 10     # Use this parameter in your call to `TruncatedSVD`
    M_reduced = None
    print("Running Truncated SVD over %i words..." % (M.shape[0]))
    
    # ------------------
    # Write your implementation here.
    svd = TruncatedSVD(n_components=k, n_iter=n_iters)
    M_reduced = svd.fit_transform(M)
    # ------------------
    
    print("Done.")
    return M_reduced

 测试一下：

# Define toy corpus and run student code
test_corpus = ["{} All that glitters isn't gold {}".format(START_TOKEN, END_TOKEN).split(" "), "{} All's well that ends well {}".format(START_TOKEN, END_TOKEN).split(" ")]
M_test, word2Ind_test = compute_co_occurrence_matrix(test_corpus, window_size=1)
M_test_reduced = reduce_to_k_dim(M_test, k=2)

# Test proper dimensions
assert (M_test_reduced.shape[0] == 10), "M_reduced has {} rows; should have {}".format(M_test_reduced.shape[0], 10)
assert (M_test_reduced.shape[1] == 2), "M_reduced has {} columns; should have {}".format(M_test_reduced.shape[1], 2)

# Print Success
print ("-" * 80)
print("Passed All Tests!")
print ("-" * 80)

Running Truncated SVD over 10 words...
Done.
--------------------------------------------------------------------------------
Passed All Tests!
--------------------------------------------------------------------------------

1.5使用matplotlib展示

def plot_embeddings(M_reduced, word2Ind, words):
    """ Plot in a scatterplot the embeddings of the words specified in the list "words".
        NOTE: do not plot all the words listed in M_reduced / word2Ind.
        Include a label next to each point.
        
        Params:
            M_reduced (numpy matrix of shape (number of unique words in the corpus , 2)): matrix of 2-dimensioal word embeddings
            word2Ind (dict): dictionary that maps word to indices for matrix M
            words (list of strings): words whose embeddings we want to visualize
    """
    # ------------------
    # Write your implementation here.
    for word in words:
        idx = word2Ind[word]
        x = M_reduced[idx, 0]
        y = M_reduced[idx, 1]
        
        plt.scatter(x, y, marker='x', color='red')
        plt.text(x, y, word)    
    # ------------------

测试一下：

print ("-" * 80)
print ("Outputted Plot:")

M_reduced_plot_test = np.array([[1, 1], [-1, -1], [1, -1], [-1, 1], [0, 0]])
word2Ind_plot_test = {'test1': 0, 'test2': 1, 'test3': 2, 'test4': 3, 'test5': 4}
words = ['test1', 'test2', 'test3', 'test4', 'test5']
plot_embeddings(M_reduced_plot_test, word2Ind_plot_test, words)

print ("-" * 80)

--------------------------------------------------------------------------------
Outputted Plot:
--------------------------------------------------------------------------------

1.6调用以上方法

reuters_corpus = read_corpus()
M_co_occurrence, word2Ind_co_occurrence = compute_co_occurrence_matrix(reuters_corpus)
M_reduced_co_occurrence = reduce_to_k_dim(M_co_occurrence, k=2)

# Rescale (normalize) the rows to make them each of unit-length
M_lengths = np.linalg.norm(M_reduced_co_occurrence, axis=1)
M_normalized = M_reduced_co_occurrence / M_lengths[:, np.newaxis] # broadcasting

words = ['barrels', 'bpd', 'ecuador', 'energy', 'industry', 'kuwait', 'oil', 'output', 'petroleum', 'venezuela']

plot_embeddings(M_normalized, word2Ind_co_occurrence, words)

Running Truncated SVD over 8185 words...
Done.

2.基于Glove得到的词向量

 2.1加载向量模型

def load_embedding_model():
    """ Load GloVe Vectors
        Return:
            wv_from_bin: All 400000 embeddings, each lengh 200
    """
    import gensim.downloader as api
    wv_from_bin = api.load("glove-wiki-gigaword-200")
    print("Loaded vocab size %i" % len(wv_from_bin.vocab.keys()))
    return wv_from_bin

wv_from_bin = load_embedding_model()

[==================================================] 100.0% 252.1/252.1MB downloaded
Loaded vocab size 400000

2.2计算嵌入矩阵

def get_matrix_of_vectors(wv_from_bin, required_words=['barrels', 'bpd', 'ecuador', 'energy', 'industry', 'kuwait', 'oil', 'output', 'petroleum', 'venezuela']):
    """ Put the GloVe vectors into a matrix M.
        Param:
            wv_from_bin: KeyedVectors object; the 400000 GloVe vectors loaded from file
        Return:
            M: numpy matrix shape (num words, 200) containing the vectors
            word2Ind: dictionary mapping each word to its row number in M
    """
    import random
    words = list(wv_from_bin.vocab.keys())
    print("Shuffling words ...")
    random.seed(224)
    random.shuffle(words)
    words = words[:10000]
    print("Putting %i words into word2Ind and matrix M..." % len(words))
    word2Ind = {}
    M = []
    curInd = 0
    for w in words:
        try:
            M.append(wv_from_bin.word_vec(w))
            word2Ind[w] = curInd
            curInd += 1
        except KeyError:
            continue
    for w in required_words:
        if w in words:
            continue
        try:
            M.append(wv_from_bin.word_vec(w))
            word2Ind[w] = curInd
            curInd += 1
        except KeyError:
            continue
    M = np.stack(M)
    print("Done.")
    return M, word2Ind


M, word2Ind = get_matrix_of_vectors(wv_from_bin)
M_reduced = reduce_to_k_dim(M, k=2)                         #M_reduced.shape= (10010, 2)

# Rescale (normalize) the rows to make them each of unit-length
M_lengths = np.linalg.norm(M_reduced, axis=1)               #求L2范数，M_lengths.shape= (10010,)
M_reduced_normalized = M_reduced / M_lengths[:, np.newaxis] # broadcasting，np.newaxis插入新维度

words = ['barrels', 'bpd', 'ecuador', 'energy', 'industry', 'kuwait', 'oil', 'output', 'petroleum', 'venezuela']
plot_embeddings(M_reduced_normalized, word2Ind, words)

Shuffling words ...
Putting 10000 words into word2Ind and matrix M...
Done.
Running Truncated SVD over 10010 words...
Done.

和共现矩阵得到的结果类似。

2.3前十个最相似的单词，根据给定单词的余弦相似度对表中的其他单词进行排序

# ------------------
# Write your implementation here.
similarwords = wv_from_bin.most_similar('leaves')
print(similarwords)
# ------------------

[('ends', 0.6128067970275879), ('leaf', 0.6027014255523682), ('stems', 0.5998532772064209), ('takes', 0.5902855396270752), ('leaving', 0.5761634111404419), ('grows', 0.5663397312164307), ('flowers', 0.5600922107696533), ('turns', 0.5536050796508789), ('leave', 0.5496848225593567), ('goes', 0.5434924960136414)]

2.4计算两个单词间的余弦距离

# ------------------
# Write your implementation here.
w1, w2, w3 = 'happy', 'cheerful', 'sad'
w1_w2_dis = wv_from_bin.distance(w1, w2)
w1_w3_dis = wv_from_bin.distance(w1, w3)

print("Synonyms {}, {} have cosine distance: {}".format(w1, w2, w1_w2_dis))
print("Antonyms {}, {} have cosine distance: {}".format(w1, w3, w1_w3_dis))
# ------------------

Synonyms happy, cheerful have cosine distance: 0.5172466933727264
Antonyms happy, sad have cosine distance: 0.40401363372802734

pprint.pprint(wv_from_bin.most_similar(positive=['woman', 'king'], negative=['man']))

[('queen', 0.6978678703308105),
('princess', 0.6081745028495789),
('monarch', 0.5889754891395569),
('throne', 0.5775108933448792),
('prince', 0.5750998258590698),
('elizabeth', 0.5463595986366272),
('daughter', 0.5399125814437866),
('kingdom', 0.5318052172660828),
('mother', 0.5168544054031372),
('crown', 0.5164473056793213)]

2.5嵌入向量存在偏见

pprint.pprint(wv_from_bin.most_similar(positive=['woman', 'worker'], negative=['man']))
print()
pprint.pprint(wv_from_bin.most_similar(positive=['man', 'worker'], negative=['woman']))

[('employee', 0.6375863552093506),
('workers', 0.6068919897079468),
('nurse', 0.5837947130203247),
('pregnant', 0.5363885760307312),
('mother', 0.5321309566497803),
('employer', 0.5127025842666626),
('teacher', 0.5099577307701111),
('child', 0.5096741914749146),
('homemaker', 0.5019455552101135),
('nurses', 0.4970571994781494)]

[('workers', 0.611325740814209),
('employee', 0.5983108878135681),
('working', 0.5615329742431641),
('laborer', 0.5442320108413696),
('unemployed', 0.5368517637252808),
('job', 0.5278826951980591),
('work', 0.5223963260650635),
('mechanic', 0.5088937282562256),
('worked', 0.5054520964622498),
('factory', 0.4940453767776489)]