r - 在 R 中解析数据帧以查找值频率的更快方法？

Question

我在 R 中有一个数据帧，其中包含一列距离(第 12 列)以及该距离是否存在匹配项(第 13 列)。 1 代表匹配，0 代表不匹配。例如:

distance    match
1           0
1           1
1           1
2           1
2           0
3           1
4           0
4           0

我想找到每个距离值的频率，并找到每个值的匹配百分比。

例如，对于上面的表格，我想要得到这样的东西:

distance    frequency    matches
1           3            2
2           2            1
3           1            1
4           2            0

我当前的代码如下所示:

  #Create a new list with unique distance values
  distance <- unique(methyl_data[,12])

  #Count how many of each distance and how many matches there are
  total = c()
  matches = c()

  dl = length(distance)
  ml = length(methyl_data[,12])

  match = FALSE
  tcounter = 0
  mcounter = 0
  for (d in 1:dl) {
    for (x in 1:ml){
      if (distance[d] == methyl_data[x, 12]) {
        match = TRUE
        tcounter <- tcounter + 1
        if (methyl_data[x, 13] == 1) {
          mcounter <- mcounter + 1
        }
      }

      #Add the frequency and number of matches for the last data entry
      if(d== dl && x ==ml) {
        total = c(total, tcounter)
        matches = c(matches, mcounter)
      }
      if((distance[d] != methyl_data[x, 12]) && match == TRUE) {
        match = FALSE
        total = c(total, tcounter)
        matches = c(matches, mcounter)
        tcounter =0
        mcounter =0
      }

    }



  }

  View(distance)
  #Create a table with the frequency of distances and matches and percentage of matches
  percentage = (matches/total)
  table = cbind(distance, total, matches, percentage)

但是，该数据帧有近 200 万行，因此该循环效率低下。有没有内置的 R 函数可以简化我的代码？我的最终目标是查看距离和匹配之间是否存在关系，那么对于非常大的数据集是否有更简单的方法来做到这一点？

提前致谢。

Answer 1

有多种方法可以做到这一点。

方法1:使用dplyr包:

library(dplyr)
df %>% 
    group_by(distance) %>% 
    mutate(frequency = n(), matches = sum(match)) %>% 
    select(distance, frequency, matches) %>% 
    distinct()

print(df)

  distance frequency matches
     <int>     <int>   <int>
1        1         3       2
2        2         2       1
3        3         1       1
4        4         2       0

方法2:使用data.table包(如果您的数据量很大，则首选此方法)

library(data.table)
setDT(df)
df[,':='(frequency = .N, matches = sum(match)), .(distance)]
df <- unique(df[,.(distance, frequency, matches)])

print(df)
       distance frequency matches
1:        1         3       2
2:        2         2       1
3:        3         1       1
4:        4         2       0