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R中余弦相似矩阵的前N值

cosine-similarity matrix r

3

nak5120 · 技术社区 · 4 年前

如何获得余弦相似矩阵的顶部对,如下所示:

southpark_matrix <- structure(c(0, 0.165272735625452, 0.386480286121192, 0.170696960480773, 
0.0869562860988618, 0.165272735625452, 0, 0.251690602341816, 
0.472701602991984, 0.137486001150133, 0.386480286121192, 0.251690602341816, 
0, 0.255849200006255, 0.0972813221214626, 0.170696960480773, 
0.472701602991984, 0.255849200006255, 0, 0.156449701347234, 0.0869562860988618, 
0.137486001150133, 0.0972813221214626, 0.156449701347234, 0), .Dim = c(5L, 
5L), .Dimnames = list(Docs = c("Mr. Garrison_2", "Cartman_3", 
"Mr. Garrison_3", "Cartman_4", "Jimbo_5"), Docs = c("Mr. Garrison_2", 
"Cartman_3", "Mr. Garrison_3", "Cartman_4", "Jimbo_5")))

南方公园矩阵

                Docs
Docs             Mr. Garrison_2 Cartman_3 Mr. Garrison_3 Cartman_4    Jimbo_5
  Mr. Garrison_2     0.00000000 0.1652727     0.38648029 0.1706970 0.08695629
  Cartman_3          0.16527274 0.0000000     0.25169060 0.4727016 0.13748600
  Mr. Garrison_3     0.38648029 0.2516906     0.00000000 0.2558492 0.09728132
  Cartman_4          0.17069696 0.4727016     0.25584920 0.0000000 0.15644970
  Jimbo_5            0.08695629 0.1374860     0.09728132 0.1564497 0.00000000

我怎样才能得到前两对呢?

Cartman_3 Cartman_4             0.4727016
Mr. Garrison_2 Mr. Garrison_3   0.38648029

2 回复 | 直到 4 年前

1

4

Joel Kandiah 4 年前

我要做的就是把矩阵转换成tible。我们可以按照以下步骤将矩阵的上三角部分转换为2列的数据帧(请参见此处: Convert upper triangular part of a matrix to 3-column long format

在此之后,我们可以简单地使用由我们的值指定的top(2,val)函数。此步骤的另一种方法是使用arrange(desc(val))按降序排列值,然后使用head(2)函数获取前2个值。

library(tidyverse)
#> Warning: package 'ggplot2' was built under R version 4.0.4
#> Warning: package 'tibble' was built under R version 4.0.4
#> Warning: package 'tidyr' was built under R version 4.0.4
#> Warning: package 'readr' was built under R version 4.0.3
#> Warning: package 'dplyr' was built under R version 4.0.4
#> Warning: package 'forcats' was built under R version 4.0.4


southpark_matrix <- structure(c(0, 0.165272735625452, 0.386480286121192, 0.170696960480773, 
                                0.0869562860988618, 0.165272735625452, 0, 0.251690602341816, 
                                0.472701602991984, 0.137486001150133, 0.386480286121192, 0.251690602341816, 
                                0, 0.255849200006255, 0.0972813221214626, 0.170696960480773, 
                                0.472701602991984, 0.255849200006255, 0, 0.156449701347234, 0.0869562860988618, 
                                0.137486001150133, 0.0972813221214626, 0.156449701347234, 0), .Dim = c(5L, 
                                                                                                       5L), .Dimnames = list(Docs = c("Mr. Garrison_2", "Cartman_3", 
                                                                                                                                      "Mr. Garrison_3", "Cartman_4", "Jimbo_5"), Docs = c("Mr. Garrison_2", 
                                                                                                                                                                                          "Cartman_3", "Mr. Garrison_3", "Cartman_4", "Jimbo_5")))

# Convert the matrix to an upper diagonal form
ind <- which(upper.tri(southpark_matrix, diag = TRUE), arr.ind = TRUE)
dimnam <- dimnames(southpark_matrix)
df <- data.frame(row = dimnam[[1]][ind[, 1]],
           col = dimnam[[2]][ind[, 2]],
           val = southpark_matrix[ind])
#top n method
df %>%
  tibble() %>% 
  top_n(2, val)
#> # A tibble: 2 x 3
#>   row            col              val
#>   <chr>          <chr>          <dbl>
#> 1 Mr. Garrison_2 Mr. Garrison_3 0.386
#> 2 Cartman_3      Cartman_4      0.473

#arrange and head method
df %>% 
  arrange(desc(val)) %>% 
  head(2)
#> # A tibble: 2 x 3
#>   row            col              val
#>   <chr>          <chr>          <dbl>
#> 1 Cartman_3      Cartman_4      0.473
#> 2 Mr. Garrison_2 Mr. Garrison_3 0.386

^{创建于2021-04-04

reprex package}

2

1

Waldi 4 年前

与 lapply :

best <- head(unique(sort(southpark_matrix,decreasing=T)),2)
lapply(best,function(x) {list(score=x, names = rownames(which(southpark_matrix == x,arr.ind=T)))})

[[1]]
[[1]]$score
[1] 0.4727016

[[1]]$names
[1] "Cartman_4" "Cartman_3"


[[2]]
[[2]]$score
[1] 0.3864803

[[2]]$names
[1] "Mr. Garrison_3" "Mr. Garrison_2"