向量空间模型：余弦相似度与欧几里得距离

Question

我有分类文本的语料库。从这些我创建向量。每个向量对应一个文档。向量分量是本文档中的单词权重，计算为 TFIDF 值。接下来，我构建了一个模型，其中每个类都由一个向量表示。模型具有与语料库中的类一样多的向量。模型向量的分量被计算为取自此类向量的所有分量值的平均值。对于未分类的向量，我通过计算这些向量之间的余弦来确定与模型向量的相似性。

问题：

1) 我可以使用未分类向量和模型向量之间的欧几里得距离来计算它们的相似度吗？

2) 为什么欧几里得距离不能作为相似度度量，而不是两个向量之间夹角的余弦值，反之亦然？

谢谢！

Answer 1

考虑这一点的一种非正式但相当直观的方式是考虑向量的两个分量：方向和幅度。

方向是向量的“偏好”/“风格”/“情感”/“潜在变量”，而幅度是它对那个方向的强度方向。

在对文档进行分类时，我们希望根据它们的整体情绪对其进行分类，因此我们使用角距离。

欧几里得距离容易受到文档被它们的 L2 范数（大小，在二维情况下）而不是方向聚类的影响。 IE。具有完全不同方向的向量将被聚类，因为它们与原点的距离相似。

Answer 2

我将按相反的顺序回答问题。对于第二个问题，余弦相似度和欧几里得距离是衡量向量相似度的两种不同方法。前者测量向量相对于原点的相似性，而后者测量沿向量的特定兴趣点之间的距离。您可以单独使用，将它们组合并同时使用，或者查看许多其他方法中的一种来确定相似性。有关更多信息，请参阅 Michael Collins 讲座中的 these 幻灯片。

您的第一个问题不是很清楚，但是无论您是在比较文档还是您的“模型”（传统上将其描述为集群），您都应该能够使用任一度量来找到两个向量之间的距离，其中模型是所有聚类的总和）。

Answer 3

计算时间明智（python）：

import time
import numpy as np

for i in range(10):
    start = time.time() 
    for i in range(10000):
        a, b = np.random.rand(100), np.random.rand(100) 
        np.dot(a, b) / ( np.linalg.norm(a) * np.linalg.norm(b))
    print 'Cosine similarity took', time.time() - start

    start = time.time() 
    for i in range(10000):
        a, b = np.random.rand(100), np.random.rand(100) 
        2 * (1 - np.dot(a, b) / ( np.linalg.norm(a) * np.linalg.norm(b)))
    print 'Euclidean from 2*(1 - cosine_similarity) took', time.time() - start


    start = time.time() 
    for i in range(10000):
        a, b = np.random.rand(100), np.random.rand(100) 
        np.linalg.norm(a-b)
    print 'Euclidean Distance using np.linalg.norm() took', time.time() - start


    start = time.time() 
    for i in range(10000):
        a, b = np.random.rand(100), np.random.rand(100) 
        np.sqrt(np.sum((a-b)**2))
    print 'Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took', time.time() - start
    print '--------------------------------------------------------'

[出]：

Cosine similarity took 0.15826010704
Euclidean from 2*(1 - cosine_similarity) took 0.179041862488
Euclidean Distance using np.linalg.norm() took 0.10684299469
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.113723039627
--------------------------------------------------------
Cosine similarity took 0.161732912064
Euclidean from 2*(1 - cosine_similarity) took 0.178358793259
Euclidean Distance using np.linalg.norm() took 0.107393980026
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.111194849014
--------------------------------------------------------
Cosine similarity took 0.16274189949
Euclidean from 2*(1 - cosine_similarity) took 0.178978919983
Euclidean Distance using np.linalg.norm() took 0.106336116791
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.111373186111
--------------------------------------------------------
Cosine similarity took 0.161939144135
Euclidean from 2*(1 - cosine_similarity) took 0.177414178848
Euclidean Distance using np.linalg.norm() took 0.106301784515
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.11181807518
--------------------------------------------------------
Cosine similarity took 0.162333965302
Euclidean from 2*(1 - cosine_similarity) took 0.177582979202
Euclidean Distance using np.linalg.norm() took 0.105742931366
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.111120939255
--------------------------------------------------------
Cosine similarity took 0.16153883934
Euclidean from 2*(1 - cosine_similarity) took 0.176836967468
Euclidean Distance using np.linalg.norm() took 0.106392860413
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.110891103745
--------------------------------------------------------
Cosine similarity took 0.16018986702
Euclidean from 2*(1 - cosine_similarity) took 0.177738189697
Euclidean Distance using np.linalg.norm() took 0.105060100555
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.110497951508
--------------------------------------------------------
Cosine similarity took 0.159607887268
Euclidean from 2*(1 - cosine_similarity) took 0.178565979004
Euclidean Distance using np.linalg.norm() took 0.106383085251
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.11084485054
--------------------------------------------------------
Cosine similarity took 0.161075115204
Euclidean from 2*(1 - cosine_similarity) took 0.177822828293
Euclidean Distance using np.linalg.norm() took 0.106630086899
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.110257148743
--------------------------------------------------------
Cosine similarity took 0.161051988602
Euclidean from 2*(1 - cosine_similarity) took 0.181928873062
Euclidean Distance using np.linalg.norm() took 0.106360197067
Euclidean Distance using np.sqrt(np.sum((a-b)**2)) took 0.111301898956
--------------------------------------------------------

Answer 4

我建议确定在给定应用程序中哪种距离测量更好的唯一可靠方法是同时尝试两种方法，看看哪一种能提供更令人满意的结果。我猜在大多数情况下，效果的差异不会很大，但在您的特定应用中可能并非如此。