ddply总结比例计数

Question

我在使用plyr包中的ddply函数时遇到了一些麻烦.我试图用每组中的计数和比例来总结以下数据.这是我的数据:

    structure(list(X5employf = structure(c(1L, 3L, 1L, 1L, 1L, 3L, 
1L, 1L, 1L, 3L, 1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 1L, 2L, 2L, 2L, 
2L, 2L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 1L, 1L, 3L, 1L, 
3L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 
3L, 3L, 1L), .Label = c("increase", "decrease", "same"), class = "factor"), 
    X5employff = structure(c(2L, 6L, NA, 2L, 4L, 6L, 5L, 2L, 
    2L, 8L, 2L, 2L, 2L, 7L, 7L, 8L, 11L, 7L, 2L, 8L, 8L, 11L, 
    7L, 6L, 2L, 5L, 2L, 8L, 7L, 7L, 7L, 8L, 6L, 7L, 5L, 5L, 7L, 
    2L, 6L, 7L, 2L, 2L, 2L, 2L, 2L, 5L, 5L, 5L, 2L, 5L, 2L, 2L, 
    2L, 5L, 12L, 2L, 2L, 2L, 2L, 5L, 5L, 5L, 5L, 2L, 5L, 2L, 
    13L, 9L, 9L, 9L, 7L, 8L, 5L), .Label = c("", "1", "1  and 8", 
    "2", "3", "4", "5", "6", "6 and 7", "6 and 7 ", "7", "8", 
    "1 and 8"), class = "factor")), .Names = c("X5employf", "X5employff"
), row.names = c(NA, 73L), class = "data.frame")

这是我使用ddply的电话:

ddply(kano_final, .(X5employf, X5employff), summarise, n=length(X5employff), prop=(n/sum(n))*100)

这给了我X5employff正确的每个实例的计数,但似乎比例是在每一行计算的,而不是在因子的每个级别内X5employf,如下所示:

   X5employf X5employff  n prop
1   increase          1 26  100
2   increase          2  1  100
3   increase          3 15  100
4   increase    1 and 8  1  100
5   increase       <NA>  1  100
6   decrease          4  1  100
7   decrease          5  5  100
8   decrease          6  2  100
9   decrease          7  1  100
10  decrease          8  1  100
11      same          4  4  100
12      same          5  6  100
13      same          6  5  100
14      same    6 and 7  3  100
15      same          7  1  100

当手动计算每个组中的比例时,我得到:

   X5employf X5employff  n prop
1   increase          1 26  59.09
2   increase          2  1  2.27
3   increase          3 15  34.09
4   increase    1 and 8  1  2.27
5   increase       <NA>  1  2.27
6   decrease          4  1  10.00
7   decrease          5  5  50.00
8   decrease          6  2  20.00
9   decrease          7  1  10.00
10  decrease          8  1  10.00
11      same          4  4  21.05
12      same          5  6  31.57
13      same          6  5  26.31
14      same    6 and 7  3  15.78
15      same          7  1  5.26

正如您所看到的,因子X5employf的每个级别的比例总和等于100.

我知道这可能是非常简单的,但尽管阅读了各种类似的帖子,但我似乎无法理解它.谁能帮助解决这个问题以及我对总结功能如何运作的理解？!

非常感谢

马蒂

Answer 1

您无法在一次ddply调用中执行此操作,因为传递给每个summarize调用的内容是您的组变量的特定组合的数据子集.在此最低级别,您无权访问该中间级别sum(n).相反,分两步完成:

kano_final <- ddply(kano_final, .(X5employf), transform,
                    sum.n = length(X5employf))

ddply(kano_final, .(X5employf, X5employff), summarise, 
      n = length(X5employff), prop = n / sum.n[1] * 100)

编辑:使用单个ddply呼叫并table按照您的暗示使用:

ddply(kano_final, .(X5employf), summarise,
      n          = Filter(function(x) x > 0, table(X5employff, useNA = "ifany")),
      prop       = 100* prop.table(n),
      X5employff = names(n))