Запуск уравнения для всех пар в кадре данных, выходная матрица

Вот снимок с использованием некоторого изменения формы из tidyr и операций группирования:

library(dplyr)
library(tidyr)
longdf <- df %>%
  pivot_longer(
    cols = min_01:max_12,
    names_to = c(".value", "set"),
    names_pattern = "(.*)_(.*)"
  )
longdf
# # A tibble: 60 x 5
#    sample   start set     min   max
#    <fct>    <dbl> <chr> <dbl> <dbl>
#  1 sample01   233 01     2.94  13.1
#  2 sample01   233 02     4.22  15.8
#  3 sample01   233 03     5.7   18.5
#  4 sample01   233 04     7.15  22.9
#  5 sample01   233 05    10.5   27.5
#  6 sample01   233 06    13.8   31.9
#  7 sample01   233 07    16.2   35.2
#  8 sample01   233 08    15.3   34.7
#  9 sample01   233 09    14.3   32.4
# 10 sample01   233 10     9.83  26.7
# # ... with 50 more rows

Отсюда давайте снова присоединимся к нему по отдельности, удалив сравнения с самим собой и убедившись, что наборы совпадают (min_01 с min_01 из другого образца):

crossing(
  rename_all(longdf, ~ paste0(., "1")),
  rename_all(longdf, ~ paste0(., "2"))
) %>% 
  filter(sample1 != sample2, set1 == set2)
# # A tibble: 240 x 10
#    sample1  start1 set1   min1  max1 sample2  start2 set2   min2  max2
#    <fct>     <dbl> <chr> <dbl> <dbl> <fct>     <dbl> <chr> <dbl> <dbl>
#  1 sample01    233 01     2.94  13.1 sample02     99 01     3.26  14.1
#  2 sample01    233 01     2.94  13.1 sample03    288 01     3.15  14.9
#  3 sample01    233 01     2.94  13.1 sample04    313 01     2.55  14.5
#  4 sample01    233 01     2.94  13.1 sample05    346 01     2.59  13.3
#  5 sample01    233 02     4.22  15.8 sample02     99 02     4.97  16.9
#  6 sample01    233 02     4.22  15.8 sample03    288 02     3.51  16.9
#  7 sample01    233 02     4.22  15.8 sample04    313 02     4.14  16.4
#  8 sample01    233 02     4.22  15.8 sample05    346 02     4.12  15.7
#  9 sample01    233 03     5.7   18.5 sample02     99 03     6.61  19.8
# 10 sample01    233 03     5.7   18.5 sample03    288 03     4.86  19.2
# # ... with 230 more rows

Отсюда, это "просто" математика:

crossing(
  rename_all(longdf, ~ paste0(., "1")),
  rename_all(longdf, ~ paste0(., "2"))
) %>% 
  filter(sample1 != sample2, set1 == set2) %>%
  mutate(
    out = (max1 - min2) / sqrt((max1 - min1) * (max2 - min2))
  )
# # A tibble: 240 x 11
#    sample1  start1 set1   min1  max1 sample2  start2 set2   min2  max2   out
#    <fct>     <dbl> <chr> <dbl> <dbl> <fct>     <dbl> <chr> <dbl> <dbl> <dbl>
#  1 sample01    233 01     2.94  13.1 sample02     99 01     3.26  14.1 0.939
#  2 sample01    233 01     2.94  13.1 sample03    288 01     3.15  14.9 0.911
#  3 sample01    233 01     2.94  13.1 sample04    313 01     2.55  14.5 0.960
#  4 sample01    233 01     2.94  13.1 sample05    346 01     2.59  13.3 1.01 
#  5 sample01    233 02     4.22  15.8 sample02     99 02     4.97  16.9 0.922
#  6 sample01    233 02     4.22  15.8 sample03    288 02     3.51  16.9 0.990
#  7 sample01    233 02     4.22  15.8 sample04    313 02     4.14  16.4 0.982
#  8 sample01    233 02     4.22  15.8 sample05    346 02     4.12  15.7 1.01 
#  9 sample01    233 03     5.7   18.5 sample02     99 03     6.61  19.8 0.916
# 10 sample01    233 03     5.7   18.5 sample03    288 03     4.86  19.2 1.01 
# # ... with 230 more rows

... и суммирование по группам:

crossing(
  rename_all(longdf, ~ paste0(., "1")),
  rename_all(longdf, ~ paste0(., "2"))
) %>% 
  filter(sample1 != sample2, set1 == set2) %>%
  mutate(
    out = (max1 - min2) / sqrt((max1 - min1) * (max2 - min2))
  ) %>%
  group_by(sample1, sample2) %>%
  summarize(out = sum(out)) %>%
  ungroup()
# # A tibble: 20 x 3
#    sample1  sample2    out
#    <fct>    <fct>    <dbl>
#  1 sample01 sample02  11.1
#  2 sample01 sample03  11.9
#  3 sample01 sample04  11.8
#  4 sample01 sample05  11.8
#  5 sample02 sample01  12.9
#  6 sample02 sample03  12.8
#  7 sample02 sample04  12.7
#  8 sample02 sample05  12.6
#  9 sample03 sample01  12.1
# 10 sample03 sample02  11.3
# 11 sample03 sample04  12.0
# 12 sample03 sample05  11.9
# 13 sample04 sample01  12.2
# 14 sample04 sample02  11.3
# 15 sample04 sample03  12.1
# 16 sample04 sample05  11.9
# 17 sample05 sample01  12.3
# 18 sample05 sample02  11.4
# 19 sample05 sample03  12.1
# 20 sample05 sample04  12.1

А если они нужны в матричном макете, то

crossing(
  rename_all(longdf, ~ paste0(., "1")),
  rename_all(longdf, ~ paste0(., "2"))
) %>% 
  filter(sample1 != sample2, set1 == set2) %>%
  mutate(
    out = (max1 - min2) / sqrt((max1 - min1) * (max2 - min2))
  ) %>%
  group_by(sample1, sample2) %>%
  summarize(out = sum(out)) %>%
  ungroup() %>%
  pivot_wider(sample1, names_from = "sample2", values_from = "out") %>%
  select(c("sample1", setdiff(sort(colnames(.)), "sample1")))
# # A tibble: 5 x 6
#   sample1  sample01 sample02 sample03 sample04 sample05
#   <fct>       <dbl>    <dbl>    <dbl>    <dbl>    <dbl>
# 1 sample01     NA       11.1     11.9     11.8     11.8
# 2 sample02     12.9     NA       12.8     12.7     12.6
# 3 sample03     12.1     11.3     NA       12.0     11.9
# 4 sample04     12.2     11.3     12.1     NA       11.9
# 5 sample05     12.3     11.4     12.1     12.1     NA  

Вот попытка без дополнительных зависимостей. Он использует combn для генерации всех пар индивидов, затем вычисляет желаемые значения, используя apply с функцией пользовательского значения.

# Helpers
run_eqn <- function(sample_1, sample_2, eqn_ind) {
  # eqn_ind is a character string, e.g. "01", "02", ..., "12"
  max_id <- paste0("max_", eqn_ind)
  min_id <- paste0("min_", eqn_ind)
  (sample_1[[max_id]] - sample_2[[min_id]]) / sqrt((sample_1[[max_id]] - sample_1[[min_id]]) * (sample_2[[max_id]] - sample_2[[min_id]]))
}

sum_eqn <- function(sample_1, sample_2) {
  eqn_ind <- c("01", "02", "03", "04", "05", "06", "07", "08", "09", "10", "11", "12")
  sum(sapply(eqn_ind, run_eqn, sample_1 = sample_1, sample_2 = sample_2))
}

# Main
pairs <- t(combn(1:5, 2))
pairs_value <- apply(pairs, 1, function(pair) {
  sum_eqn(df[pair[1], ], df[pair[2], ])
})
res <- cbind(pairs, pairs_value)
colnames(res) <- c("sample_1_id", "sample_2_id", "value")

res
# sample_1_id sample_2_id    value
# [1,]           1           2 11.11671
# [2,]           1           3 11.89706
# [3,]           1           4 11.84859
# [4,]           1           5 11.75426
# [5,]           2           3 12.75344
# [6,]           2           4 12.73372
# [7,]           2           5 12.61932
# [8,]           3           4 11.96217
# [9,]           3           5 11.87361
# [10,]           4           5 11.91347