计算R列发生的相似性

Question

我有以下数据

df <- data.frame(
  group = c('r1','r2','r3','r4'),
  X1 = c('A','B','C','K'),
  X2 = c('A','C','M','K'),
  X3 = c('D','A','C','K')
)

> df
  group X1 X2 X3
1    r1  A  A  D
2    r2  B  C  A
3    r3  C  M  C
4    r4  K  K  K

我想根据列X1，X2

> df
  group X1 X2 X3 similarity_score
1    r1  A  A  D .67
2    r2  B  C  A .33
3    r3  C  M  C .67
4    r4  K  K  K 1

我怎样才能做到这一点？

Answer 1

tidyverse解决方案：

library(tidyverse)

df %>% 
  rowwise() %>% 
  mutate(
    similarity_score = max(colMeans(outer(c_across(-group), c_across(-group), `==`)))
  ) 

或者，你可以做一个nest解决方案，而不是cross：

df %>% 
  group_by(group) %>% 
  nest(data = -group) %>% 
  rowwise() %>% 
  mutate(
    similarity_score = max(colMeans(outer(unlist(data), unlist(data), `==`)))
  ) %>% 
  unnest(data)

  group X1    X2    X3    similarity_score
  <chr> <chr> <chr> <chr>            <dbl>
1 r1    A     A     D                0.667
2 r2    B     C     A                0.333
3 r3    C     M     C                0.667
4 r4    K     K     K                1   

Answer 2

你可以做的

df$similarity <- round(apply(df[-1], 1, function(x) max(table(x))/length(x)), 2)

df
#>   group X1 X2 X3 similarity
#> 1    r1  A  A  D       0.67
#> 2    r2  B  C  A       0.33
#> 3    r3  C  M  C       0.67
#> 4    r4  K  K  K       1.00

2018年4月2日由reprex软件包（v2.0.1）创建

Answer 3

另一种可能的解决方案：

library(dplyr)

df %>% 
  rowwise %>% 
  mutate(score = max(prop.table(table(c_across(X1:X3))))) %>% 
  ungroup

#> # A tibble: 4 × 5
#>   group X1    X2    X3    score
#>   <chr> <chr> <chr> <chr> <dbl>
#> 1 r1    A     A     D     0.667
#> 2 r2    B     C     A     0.333
#> 3 r3    C     M     C     0.667
#> 4 r4    K     K     K     1

甚至更短：

library(tidyverse)
df %>% mutate(score = pmap_dbl(across(X1:X3), ~ max(prop.table(table(c(...))))))