Count similarity of occurrences across columns R

user_10984004 0
Viewed 925

I have the following data:

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

I want to estimate a 'similarity score' based on columns X1, X2 & X3. For example, within group r1 (or row 1), 2 out of 3 elements are similar, so score is 2/3 (~67%). And the group r4 (or row 4), the score would be 3/3 (100%). The desired outcome is below:

> 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

How can I achieve this?

dplyr r
6 Answers

Another possible solution:

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

Or even shorter:

library(tidyverse)
df %>% mutate(score = pmap_dbl(across(X1:X3), ~ max(prop.table(table(c(...))))))
user_115644870

You could do

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

Created on 2022-04-18 by the reprex package (v2.0.1)

user_125003150

A tidyverse solution:

library(tidyverse)

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

Or instead of c_across, you could do a nest solution:

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
user_64365450

As an other alternative, we could store all occurences in a single pass (instead of operating by row):

tab = table(rep(df[, 1], ncol(df) - 1), as.matrix(df[, -1]))

And, then, retrieve the proportion of the most element for each row:

tab = tab / rowSums(tab)
tab[cbind(1:nrow(df), max.col(tab))]
#[1] 0.6666667 0.3333333 0.6666667 1.0000000
user_24149480

Another possible option with first pivoting longer to summarize then join back to dataframe.

library(tidyverse)

df %>%
  left_join(pivot_longer(., -group) %>%
              group_by(group) %>%
              summarise(score = round(max(table(value))/length(value), 2)))

Output

  group X1 X2 X3 score
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
user_152931910

Here's another way in base R:

df$score <- round(sapply(apply(df[,c(2:4)], 1, table), first) / 3, 2) 

#   group X1 X2 X3 similarity_score
# 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
user_57837450
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