计算R中的重复阵容

Question

我运行模拟并创建10,000个阵容。我希望列出创建的阵容数量。例如，这里有5个阵容......

col_1 <- c("Mary", "Jane", "Latoya", "Sandra", "Ebony", "Jada")
col_2 <- c("Jack", "Malik", "Brett", "Demetrius", "Jalen","David")
col_3 <- c("Mary", "Jane", "Latoya", "Sandra", "Ebony", "Jada")
col_4 <- c("Katie", "Emily", "Tara", "Imani", "Molly", "Claire")
col_5 <- c("Mary", "Jane", "Latoya", "Sandra", "Ebony", "Jada")
df <- data.frame(col_1, col_2, col_3,col_4,col_5)

我想要的输出是大约...

阵容A = col_1，col_3，col5 = 3

阵容B = col_2 = 1

阵容C = col_5 = 1

我把头撞到了调查dplyr包装解决方案的墙上。任何帮助，将不胜感激。谢谢。

Answer 1

这将是我的解决方案：

df_t <- df %>% 
  # Transpose the dataset, make sure people are sorted alphabetically
  gather(lineup_number, person_name) %>% # Lineup/Person Level
  arrange(lineup_number, person_name) %>% # Arrange alphabetically
  group_by(lineup_number) %>% 
  mutate(person_order = paste0("person", row_number())) %>%  
  ungroup() %>% 
  spread(person_order, person_name) # Row: Lineup. Column: Person

df_t %>% 
  select(starts_with("person")) %>% 
  group_by_all() %>% 
  summarise(num_lineups = n())

Answer 2

这里是一个tidyverse唯一的解决方案，我们安排所有cols，折叠，获取唯一值，转置和分组以获得计数。这种方法也为团队成员提供了帮助。

library(tidyverse)

df2 <- df %>%
  arrange_all() %>%
  mutate_all(funs(paste0(., collapse = ","))) %>% 
  distinct() %>% 
  t() %>%
  as.data.frame %>%
  mutate(col       = colnames(df)) %>% 
  group_by(team    = V1) %>% 
  summarise(count  = n(), 
            lineup = paste(col, collapse = ","))


print(df2)
# A tibble: 3 x 3
  team                                   count lineup           
  <fct>                                  <int> <chr>            
1 Ebony,Jada,Jane,Latoya,Mary,Sandra         3 col_1,col_3,col_5
2 Jalen,David,Malik,Brett,Jack,Demetrius     1 col_2            
3 Molly,Claire,Emily,Tara,Katie,Imani        1 col_4    

Answer 3

首先，我们确保数据框的所有列中的级别匹配并剥离它们以获得数字。

(d2 <- sapply(d, function(x) as.numeric(factor(x, levels=sort(unique(unlist(d)))))))
#      col_1 col_2 col_3 col_4 col_5
# [1,]     5    10     5    16     5
# [2,]     3    12     3    14     3
# [3,]     4     7     4    18     4
# [4,]     6     9     6    15     6
# [5,]     1    11     1    17     1
# [6,]     2     8     2    13     2

然后我们可以在列上应用toString，对它们进行分解并将它们分解为因子级别;我们只想要names，

n <- lapply(split(m <- factor(apply(d2, 2, toString)), m), names)

这实际上是结果，我们rbind与他们的lengths。

res <- do.call(rbind, lapply(n, function(x) cbind(toString(x), length(x))))
res
#     [,1]                  [,2]
# [1,] "col_2"               "1" 
# [2,] "col_4"               "1" 
# [3,] "col_1, col_3, col_5" "3" 

最后，我们可能想给矩阵一些有意义的dimnames。

dimnames(res) <- list(paste("Lineup", LETTERS[1:nrow(res)]), c("col", "n"))
res
#            col                   n  
# Lineup A "col_2"               "1"
# Lineup B "col_4"               "1"
# Lineup C "col_1, col_3, col_5" "3"

注意：如果你有超过26个阵容，你可能只想做1:nrow(res)而不是LETTERS[1:nrow(res)]。