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我有一个包含三个数据框的列表，并希望生成另一个包含三个数据框的列表，其行由分组变量 (g1) 的每个值和 g1 变量的六个变量的平均值组成。扭曲的是，我只想在相应的虚拟变量的值等于 1 时计算三个连续变量的均值。

可重现的例子：

    a <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),c(1,1,1,1,0,0,0,1,0,0),c(0,0,1,0,1,0,0,1,0,1),c(0,0,0,1,0,0,1,1,0,0),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
b <- data.frame(c("fj","a","fj","a","fj","fj","fj","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
c <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
u <- list(a,b,c)
u <- lapply(u, setNames, nm = c('g1','dummy1','dummy2','dummy3','contin1','contin2','contin3'))

u[[1]]

> u
[[1]]
   g1 dummy1 dummy2 dummy3  contin1 contin2 contin3
1  fj      1      0      0       199      18      61
2  fj      1      0      0        91     158      28
3  fj      1      1      0       147      67     190
4   a      1      0      1       181     105      22
5  fj      0      1      0        14      16     156
6   a      0      0      0       178      14      98
7   g      0      0      1       116      97      30
8   g      1      1      1        48      31     144
9   g      0      0      0        60      21     112
10  g      0      1      0        95     145     199

我想仅在 dummy1 = 1 时计算 contin1 的平均值，仅在 dummy2 = 1 时计算 contin2 的平均值，仅在 dummy3 = 1 时计算 contin3 的平均值

我想要的第一个列表的输出：

> rates
[[1]]
  x[, 1]   V1  V2  V3 x[, 1] x[, 6] x[, 1] x[, 7] x[, 1] x[, 8]
1      a 0.50 0.0 0.5      a 181         a  NA         a  22
2     fj 0.75 0.5 0.0     fj 145.67     fj  41.5      fj  NA
3      g 0.25 0.5 0.5      g  48         g  88         g  87

我试过的：

rates <- lapply(u, function(x) {
    cbind(aggregate(cbind(x[,2],x[,3],x[,4]) ~ x[,1], FUN = mean, na.action = NULL),
    aggregate(x[,6] ~ x[,1], FUN = mean, na.action = NULL, subset = (x[,2] == 1)),
    aggregate(x[,7] ~ x[,1], FUN = mean, na.action = NULL, subset = (x[,3] == 1)),
    aggregate(x[,8] ~ x[,1], FUN = mean, na.action = NULL, subset = (x[,4] == 1)))
    })
Error in data.frame(..., check.names = FALSE) : 
  arguments imply differing number of rows: 3, 2

我知道此错误来自 cbind，因为每当您尝试 cbind 具有不同行数的对象时，cbind 都会失败。（列 x[, 6] 有三行，而 x[, 7] 和 x[, 8] 有两行。）我想我希望聚合有某种方法可以为每个分组变量保留一行，这意味着我将拥有相同数量的行并且 cbind 会起作用。根据 R 文档，这可能是不可能的吗？：“结果中将省略任何 by 变量中缺少值的行。”

我已经轻快地阅读了聚合的文档。以下两篇文章解决了类似的问题，但没有使用不同的数据子集来计算均值。

R：计算组子集的均值 和 R中数据帧列表的均值

任何建议将不胜感激。

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2 回答 2

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如果您安装了 dplyr，以下代码似乎可以解决您的问题。

library(dplyr)

set.seed(1234)

a <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),c(1,1,1,1,0,0,0,1,0,0),c(0,0,1,0,1,0,0,1,0,1),c(0,0,0,1,0,0,1,1,0,0),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
b <- data.frame(c("fj","a","fj","a","fj","fj","fj","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
c <- data.frame(c("fj","fj","fj","a","fj","a","g","g","g","g"),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 0, max = 2)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)),floor(runif(10, min = 10, max = 200)))
u <- list(a,b,c)
u <- lapply(u, setNames, nm = c('g1','dummy1','dummy2','dummy3','contin1','contin2','contin3'))


rates <- lapply(u, function(x)
  x %>% 
    mutate( contin1_ = ifelse(dummy1==1, contin1, NA) ) %>%
    mutate( contin2_ = ifelse(dummy2==1, contin2, NA) ) %>%
    mutate( contin3_ = ifelse(dummy3==1, contin3, NA) ) %>%
    group_by(g1) %>%
    summarize( 
              V1 = mean(dummy1, na.rm=TRUE),
              V2 = mean(dummy2, na.rm=TRUE),
              V3 = mean(dummy3, na.rm=TRUE),
              mean1 = mean(contin1_, na.rm=TRUE),
              mean2 = mean(contin2_, na.rm=TRUE),
              mean3 = mean(contin3_, na.rm=TRUE)
               )
)

print(rates[[1]])

这给了我这个： 

Source: local data frame [3 x 7]

  g1   V1  V2  V3     mean1 mean2 mean3
1  a 0.50 0.0 0.5 128.00000   NaN    17
2 fj 0.75 0.5 0.0  94.66667    64   NaN
3  g 0.25 0.5 0.5  54.00000    57   146

我得到的数字似乎大致正确，NA 在所有正确的地方。不幸的是，您的示例不能完全重现，因为您没有指定生成随机变量的种子，因此，我的 runif 给了我与您不同的值。

于 2015-03-23T00:30:33.730 回答
1

另一种选择是将格式从“宽”更改为“长”，并在获得“平均”值后重新转换回“宽”。对于多值列，现在可以使用iemelt的dcast开发版本来实现。它可以从. （使用来自@akhmed 帖子的相同数据集）。data.tablev1.9.5here

我们可以melt通过将列（'dummy'和'contin'）的索引指定为列表来获取列表（'u'）中的数据集measure.vars。通过将 value.vars 指定为 'dummyMean' 和 'continMean' 来获取 按'g1' 和 'variable'（从 'melt' 创建）分组的 'dummy' 和 'contin' 列dcast的平均值。longwide

 res <-  lapply(u, function(x) {
   x1 <- melt(setDT(x), measure.vars=list(2:4,5:7),
                        value.name=c('dummy', 'contin'))
   x2 <- x1[, list(dummyMean = mean(dummy, na.rm=TRUE),
             continMean = mean(contin[dummy==1], na.rm=TRUE)), 
                           by=list(g1, variable)]

  dcast(x2, g1~variable, value.var=c('dummyMean', 'continMean'))})

 res[[1]]
 #   g1 1_dummyMean 2_dummyMean 3_dummyMean 1_continMean 2_continMean
 #1:  a        0.50         0.0         0.5    128.00000          NaN
 #2: fj        0.75         0.5         0.0     94.66667           64
 #3:  g        0.25         0.5         0.5     54.00000           57
 #    3_continMean
 #1:           17
 #2:          NaN
 #3:          146

或base R使用Map. 创建函数 'fdummy'、'fcontin' 来对 'dummy' 和 'contin' 列进行子集化。循环遍历“u”（lapply(...)）。用于Map获取 'dummy' 和 'contin' 的对应列，按 'g1' 列分组，使用 'dummy==1'获取mean'dummy' 和'contin' 列的结果。meantapplycbind

 fdummy <- function(x) x[grep('dummy', names(x))]
 fcontin <- function(x) x[grep('contin', names(x))]
 res2 <- lapply(u, function(x) {
        do.call(cbind.data.frame,
           Map(function(x,y,z) cbind(tapply(x,z, FUN=mean), 
                              tapply(y[x==1],z[x==1], FUN=mean)), 
                             fdummy(x), fcontin(x), x['g1']))})


lapply(res2, setNames, c(rbind(paste0('dummyMean', 1:3), 
                    paste0('continMean',1:3))))[[1]]
#    dummyMean1 continMean1 dummyMean2 continMean2 dummyMean3 continMean3
#a        0.50   128.00000        0.0          NA        0.5          17
#fj       0.75    94.66667        0.5          64        0.0          NA
#g        0.25    54.00000        0.5          57        0.5         146
于 2015-03-23T04:28:21.383 回答