R ddply行汇总统计

Question

对于下面数据框中的每一行（由 FID_Bounda、NAME、DESCRIPTIO 和 SOVEREIGNT 定义）我正在尝试计算平均值、标准值以 crN 开头的每列中所有值的偏差和变异系数。

structure(list(FID_Bounda = 0:7, NAME = c("Bedfordshire", "Berkshire", 
"Bristol", "Buckinghamshire", "Cambridgeshire", "Cheshire", "Derbyshire", 
"Devon"), DESCRIPTIO = c("Ceremonial County", "Ceremonial County", 
"Ceremonial County", "Ceremonial County", "Ceremonial County", 
"Ceremonial County", "Ceremonial County", "Ceremonial County"
), SOVEREIGNT = c("England", "England", "England", "England", 
"England", "England", "England", "England"), crN1 = c(61.944107636, 
38.769347117, 0.810167027, 63.721241962, 191.046323469, 81.467146994, 
61.65529268, 288.751788714), crN10 = c(60.33595964, 38.326639788, 
0.834289164, 63.009539538, 185.25772542, 82.936101454, 61.985178493, 
304.951827268), crN100 = c(53.385110882, 33.530058107, 0.739041324, 
55.601839364, 165.604271128, 76.386014559, 55.591194915, 284.739586188
), crN1000 = c(58.397452282, 37.277298648, 0.820739862, 61.716749153, 
175.436497697, 82.461823706, 61.762203751, 321.414544333)), .Names = c("FID_Bounda", 
"NAME", "DESCRIPTIO", "SOVEREIGNT", "crN1", "crN10", "crN100", 
"crN1000"), row.names = c(NA, 8L), class = "data.frame")

我尝试使用 cookbook-r 中概述的代码来导出这些值：

cdata <- ddply(uadt, c("FID_Bounda","NAME","DESCRIPTIO","SOVEREIGNT"), summarise,
               N    = length(grep("crN", names(uadt), value = T)),
               mean = mean(grep("crN", names(uadt), value = F)),
               sd   = sd(grep("crN", names(uadt), value = F)),
               se   = sd / sqrt(N)
)
cdata

它正确地计算了 crN 列的总数，但它为每一行提供了相同的均值、sd 和 se。对于问题所在的任何帮助将不胜感激，因为真实数据集有 1000 列，所有列都具有相同的命名模式 crNnumber.

Answer 1

我知道这不是完美的答案，但可能值得使用更多最新的工具（同样我知道这句话的讽刺意味，因为我的答案没有使用 tidyr）。但我会采取的方法是：

library(reshape2)
madt <- melt(uadt, 
             id.vars = c("FID_Bounda", "NAME", 
                         "DESCRIPTIO", "SOVEREIGNT"))
library(dplyr)
cdata <- summarise(group_by(madt,
                            FID_Bounda, NAME, 
                            DESCRIPTIO, SOVEREIGNT), 
                   N = n_distinct(variable), 
                   mean = mean(value), 
                   sd = sd(value), 
                   se = sd / sqrt(N))

这确实产生了正确的输出

Answer 2

食谱中的示例正在计算均值，其他函数沿列向下而不是跨行，这就是您想要的。

使用基数 R 实现此目的的一种方法是：

functions <- list(length, mean, sd)

d <- lapply(functions, function(y) {
  apply(uadt, 1, function(x) y(as.numeric(x[5:8])))
})

calc <- as.data.frame(do.call(cbind, d))
names(calc) <- c("N", "mean", "sd")

cdata <- cbind(uadt[1:4], calc)
cdata$se <- cdata$sd / sqrt(cdata$N)

如果您有更多数字列，只需适当更改间隔 5:8。

R ddply行汇总统计

R ddply row summary statistics

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