Question

我有一组向量（~30k），每个向量由fasttext生成的300个元素组成，每个向量代表一个实体的含义，我想计算所有实体之间的相似性，所以我迭代了嵌套物中的向量，O（N ^ 2）复杂度，这在时间上是不实际的。

您能否为我推荐另一种计算方法，或者如何将其并行化？

def calculate_similarity(v1, v2):
    """
    Calculate cosine distance between two vectors
    """
    n1 = np.linalg.norm(v1)
    n2 = np.linalg.norm(v2)
    return np.dot(v1, v2) / n1 / n2


similarities = {}
for ith_entity, ith_vector in vectors.items():
    for jth_entity, jth_vector in vectors.items():
        if ith_entity == jth_entity:
            continue
        if (ith_entity, jth_entity) in similarities.keys() or (jth_entity, ith_entity) in similarities.keys():
            continue
        similarities[(ith_entity, jth_entity)] = calculate_similarity(ith_vector, jth_vector)

Answer 1

使用scipy的距离模块可以摆脱嵌套循环，这很慢。

给定vectors = {'k1':v1, 'k2':v2, ..., 'km':vm} vi是长度为n的Python列表。

import numpy as np 
from scipy.spatial import distance

# transfrom vectors to m x n numpy array 
data = np.array(list(vectors.values())

# compute pairwise cosine distance 
pws = distance.pdist(data, metric='cosine')

pws是精简距离矩阵。它是一维的并按以下顺序保持距离：

pws = np.array([ (k1, k2), (k1, k3), (k1, k4), ..., (k1, km),
                           (k2, k3), (k2, k4), ..., (k2, km),
                                      ...,
                                                   (km-1, km) ])

另请注意distance.pdist计算余弦距离而不是余弦相似度。

Answer 2

我去了矢量化。

import numpy as np
from itertools import combinations

np.random.seed(1)

vector_data = np.random.randn(3, 3)

v1, v2, v3 = vector_data[0], vector_data[1], vector_data[2]

def similarities_vectorized(vector_data):
    norms = np.linalg.norm(vector_data, axis=1)
    combs = np.stack(combinations(range(vector_data.shape[0]),2))
    similarities = (vector_data[combs[:,0]]*vector_data[combs[:,1]]).sum(axis=1)/norms[combs][:,0]/norms[combs][:,1]
    return combs, similarities

combs, similarities = similarities_vectorized(vector_data)

for comb, similarity in zip(combs, similarities):
    print(comb, similarity)

输出：

[0 1] -0.217095007411
[0 2] 0.894174618451
[1 2] -0.630555641519

将结果与问题：

中的代码进行比较

def calculate_similarity(v1, v2):
    """
    Calculate cosine distance between two vectors
    """
    n1 = np.linalg.norm(v1)
    n2 = np.linalg.norm(v2)
    return np.dot(v1, v2) / n1 / n2

def calculate_simularities(vectors):
    similarities = {}
    for ith_entity, ith_vector in vectors.items():
        for jth_entity, jth_vector in vectors.items():
            if ith_entity == jth_entity:
                continue
            if (ith_entity, jth_entity) in similarities.keys() or (jth_entity, ith_entity) in similarities.keys():
                continue
            similarities[(ith_entity, jth_entity)] = calculate_similarity(ith_vector, jth_vector)
    return similarities

vectors = {'A': v1, 'B': v2, 'C': v3}

print(calculate_simularities(vectors))

输出：

{('A', 'B'): -0.21709500741113338, ('A', 'C'): 0.89417461845058566, ('B', 'C'): -0.63055564151883581}

当我在一组300个向量上运行时，矢量化版本的速度提高了大约3.3倍。

更新：

此版本比原版快约50倍：

def similarities_vectorized2(vector_data):
    norms = np.linalg.norm(vector_data, axis=1)
    combs = np.fromiter(combinations(range(vector_data.shape[0]),2), dtype='i,i')
    similarities = (vector_data[combs['f0']]*vector_data[combs['f1']]).sum(axis=1)/norms[combs['f0']]/norms[combs['f1']]
    return combs, similarities

combs, similarities = similarities_vectorized2(vector_data)

for comb, similarity in zip(combs, similarities):
    print(comb, similarity)

输出：

(0, 1) -0.217095007411
(0, 2) 0.894174618451
(1, 2) -0.630555641519

如何加速矢量集之间余弦相似度的计算

2 个答案: