通过R中的余弦相似度检索矩阵中每一行的前k个相似行答案

【问题标题】：Retrieving top k similar rows in a matrix for each row via cosine similarity in R通过R中的余弦相似度检索矩阵中每一行的前k个相似行
【发布时间】：2015-12-23 21:33:33
【问题描述】：

How to efficiently retrieve top K-similar vectors by cosine similarity using R? 询问如何计算一个矩阵的每个向量相对于另一个矩阵的顶部相似向量。它是satisfactorily answered，我想调整它以在单个矩阵上运行。

也就是说，我希望矩阵中每一行的前 k 个相似的 other 行。我怀疑解决方案非常相似，但可以优化。

【问题讨论】：

标签： r similarity cosine-similarity

【解决方案1】：

此功能基于链接的答案：

CosineSimilarities <- function(m, top.k) {
  # Computes cosine similarity between each row and all other rows in a matrix.
  #
  # Args:
  #   m: Matrix of values.
  #   top.k: Number of top rows to show for each row.
  #
  # Returns:
  #   Data frame with columns for pair of rows, and cosine similarity, for top
  #   `top.k` rows per row.
  #   
  # Similarity computation
  cp <- tcrossprod(m)
  mm <- rowSums(m ^ 2)
  result <- cp / sqrt(outer(mm, mm))
  # Top similar rows from train (per row)
  # Use `top.k + 1` to remove the self-reference (similarity = 1)
  top <- apply(result, 2, order, decreasing=TRUE)[seq(top.k + 1), ]
  result.df <- data.frame(row.id1=c(col(top)), row.id2=c(top))
  result.df$cosine.similarity <- result[as.matrix(result.df[, 2:1])]
  # Remove same-row records and return
  return(result.df[result.df$row.id1 != result.df$row.id2, ])
}

例如：

(m <- matrix(1:9, nrow=3))
#      [,1] [,2] [,3]
# [1,]    1    4    7
# [2,]    2    5    8
# [3,]    3    6    9
CosineSimilarities(m, 1)
#   row.id1 row.id2 cosine.similarity
# 2       1       2            0.9956
# 4       2       3            0.9977
# 6       3       2            0.9977

【讨论】：