我有一组10 万个向量,我需要根据余弦相似度检索前 25 个最接近的向量。
Scipy 和 Sklearn 有计算余弦距离/相似度 2 向量的实现,但我需要计算 100k X 100k 大小的余弦 Sim,然后取出前 25 个。python计算中是否有任何快速实现?
根据@Silmathoron 的建议,这就是我正在做的 -
#vectors is a list of vectors of size : 100K x 400 i.e. 100K vectors each of dimenions 400
vectors = numpy.array(vectors)
similarity = numpy.dot(vectors, vectors.T)
# squared magnitude of preference vectors (number of occurrences)
square_mag = numpy.diag(similarity)
# inverse squared magnitude
inv_square_mag = 1 / square_mag
# if it doesn't occur, set it's inverse magnitude to zero (instead of inf)
inv_square_mag[numpy.isinf(inv_square_mag)] = 0
# inverse of the magnitude
inv_mag = numpy.sqrt(inv_square_mag)
# cosine similarity (elementwise multiply by inverse magnitudes)
cosine = similarity * inv_mag
cosine = cosine.T * inv_mag
k = 26
box_plot_file = file("box_data.csv","w+")
for sim,query in itertools.izip(cosine,queries):
k_largest = heapq.nlargest(k, sim)
k_largest = map(str,k_largest)
result = query + "," + ",".join(k_largest) + "\n"
box_plot_file.write(result)
box_plot_file.close()