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r - 计算所有数值列的绝对值的平均值

问题描述

我想计算示例数据集的所有数字列的绝对值的平均值DT：

library(data.table)
set.seed(1)
DT <- data.table(panelID = sample(50,50),                                                    # Creates a panel ID
                      Country = c(rep("Albania",30),rep("Belarus",50), rep("Chilipepper",20)),       
                      some_NA = sample(0:5, 6),                                             
                      some_NA_factor = sample(0:5, 6),         
                      Group = c(rep(1,20),rep(2,20),rep(3,20),rep(4,20),rep(5,20)),
                      Time = rep(seq(as.Date("2010-01-03"), length=20, by="1 month") - 1,5),
                      norm = round(runif(100)/10,2),
                      Income = round(rnorm(10,-5,5),2),
                      Happiness = sample(10,10),
                      Sex = round(rnorm(10,0.75,0.3),2),
                      Age = sample(100,100),
                      Educ = round(rnorm(10,0.75,0.3),2))           
DT [, uniqueID := .I]                                                                        # Creates a unique ID     
DT[DT == 0] <- NA                                                                            # https://stackoverflow.com/questions/11036989/replace-all-0-values-to-na
DT$some_NA_factor <- factor(DT$some_NA_factor)

我试图计算平均值和绝对平均值如下：

mean_of_differences <- DT[,lapply(Filter(is.numeric,.SD),mean, na.rm=TRUE)]
mean_of_differences <- as.data.frame(t(mean_of_differences))
mean_of_differences <- round(mean_of_differences, digits=2)
mean_of_absolute_diff <- DT[,lapply(Filter(is.numeric,.SD),function(x) mean(abs(x),na.rm=TRUE))]
mean_of_absolute_diff <- as.data.frame(t(mean_of_absolute_diff))
mean_of_absolute_diff <- round(mean_of_differences, digits=2)

然而，绝对差异的收入平均值是负数（就像正常平均值一样），这显然是不可能的。如果我查看我的代码，我不明白我做错了什么。我在看什么？

标签： rdata.tablelapplymeanabsolute-value

解决方案

这是使用 data.table 的解决方案。它 (i) 识别数字列，​​并且 (ii) 获得每个数字列的绝对值的平均值。

数据

dt = data.table(
num1 = rnorm(100),
num2 = rnorm(100),
strv = sample(LETTERS, 100, replace = T)
)

代码

numcols = colnames(dt)[unlist(lapply(dt, is.numeric))] # Which columns are numeric?

# > numcols
# [1] "num1" "num2"

meandt = dt[, lapply(.SD, function(x) mean(abs(x))), .SDcols = numcols]
newcols = paste('mean_abs_', numcols, sep = ''); colnames(meandt) = newcols

# > meandt
#        mean_abs_num1 mean_abs_num2
# 1:     0.8287523     0.8325123

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