R有效地计算三位数字组合的频率

Question

我有一个 data.frame，其中每个 ID 都有 3 个属性。为了简化，我只放了 100 行，尽管在我的真实数据集中它大约是 1.000.000。大约有 50 种不同的可能属性。属性是数字和字符的混合体。

data <- data.frame(id = 1:100,
               a1 = sample(letters,100,replace = T),
               a2 = sample(letters,100,replace = T),
               a3 = sample(letters,100,replace = T),
               stringsAsFactors=FALSE) %>% 
               as_tibble()

我想知道最频繁的组合是什么(顺序无所谓)

所以结果应该是这样的

pattern | frequency
a,a,a   |  10
A,b,c   |  5
a,e,c   |  4
...     |  ....

首先我开始创建一个包含所有可能组合的向量:

possible_combinations <- combn(c(letters,LETTERS),3) %>% 
   t() %>% 
   as_tibble() %>%
   unite("combination",sep="") %>% 
   pull()

然后我写了这个嵌套循环来计算频率:

 counter = 0
 inner_counter = 0
 combination_counter = vector(mode = "numeric",length = length (possible_combinations))

  for (j in 1:length(possible_combinations)){
    for (i in 1:nrow(data)){

        # inner Counter Counts when Attribute of one ID is in one combination
        inner_counter = inner_counter + str_count(possible_combinations[j] , data[[i,2]] )
        inner_counter = inner_counter + str_count(possible_combinations[j] , data[[i,3]] )
        inner_counter = inner_counter + str_count(possible_combinations[j] , data[[i,4]] )

      # if all three attributes are in a combination, then the Counter increases by one 
    if(inner_counter == 3) {
       counter = counter + 1 }
       inner_counter = 0
                            }

  # combination_counter is a vector which saves the frequency with 
  # which a combination ocurred in all different ids

  combination_counter[[j]] = inner_counter
  inner_counter = 0 
 }

我知道这真的不是很喜欢 R，但我不知道如何以不同的方式做到这一点。运行时对于我的小玩具示例来说甚至很糟糕，而且对于我的真实数据来说几乎是不可行的。

Answer 1

你也可以用 base r 做这个:

table(apply(data[,2:4], 1, function(x) paste0(sort(x), collapse = ",")))