ddply总结比例计数

时间:2013-08-05 11:31:21

标签: r plyr

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)

这可以正确地给出X5employff每个实例的计数,但似乎正在计算每一行的比例而不是因子X5employf的每个级别,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

正如您所见,X5employf因子的每个等级的比例总和等于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。任何人都可以帮助解决这个以及我对总结功能如何运作的理解吗?!

很多,非常感谢

玛蒂

3 个答案:

答案 0 :(得分:6)

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)

修改:使用ddply一次调用,并在您暗示时使用table

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

答案 1 :(得分:1)

我在这里添加一个dplyr示例,它可以很容易地在一个步骤中使用短代码和易于阅读的语法。

d是您的data.frame

library(dplyr)
d%.%
  dplyr:::group_by(X5employf, X5employff) %.%
  dplyr:::summarise(n = length(X5employff)) %.%
  dplyr:::mutate(ngr = sum(n)) %.% 
  dplyr:::mutate(prop = n/ngr*100)

将导致

Source: local data frame [15 x 5]
Groups: X5employf

   X5employf X5employff  n ngr      prop
1   increase          1 26  44 59.090909
2   increase          2  1  44  2.272727
3   increase          3 15  44 34.090909
4   increase    1 and 8  1  44  2.272727
5   increase         NA  1  44  2.272727
6   decrease          4  1  10 10.000000
7   decrease          5  5  10 50.000000
8   decrease          6  2  10 20.000000
9   decrease          7  1  10 10.000000
10  decrease          8  1  10 10.000000
11      same          4  4  19 21.052632
12      same          5  6  19 31.578947
13      same          6  5  19 26.315789
14      same    6 and 7  3  19 15.789474
15      same          7  1  19  5.263158

答案 2 :(得分:0)

您显然想要查找X5employff的每个值的X5employff比例。但是,你没有告诉ddply X5employf和X5employff是不同的;对于ddply来说,这两个变量只是两个分开数据的变量。此外,由于每行有一个观察点,即每个数据行的count = 1,每个(X5employf,X5employff)组合的长度等于每个(X5employf,X5employff)组合的总和。

我能想到的解决问题的最简单的“plyr方式”如下:

result <- ddply(kano_final, .(X5employf, X5employff), summarise, n=length(X5employff), drop=FALSE)
n <- result$n
n2 <- ddply(kano_final, .(X5employf), summarise, n=length(X5employff))$n
result <- data.frame(result, prop=n/rep(n2, each=13)*100)

你也可以使用好的旧版xtabs:

a <- xtabs(~X5employf + X5employff, kano_final)
b <- xtabs(~X5employf, kano_final)
a/matrix(b, nrow=3, ncol=ncol(a))