Question

我正在尝试将两种方法结合起来：

Bootstrapping multiple columns in data.table in a scalable fashion

带

Bootstrap weighted mean in R

以下是一些随机数据：

## Generate sample data

# Function to randomly generate weights
set.seed(7)
rtnorm <- function(n, mean, sd, a = -Inf, b = Inf){
qnorm(runif(n, pnorm(a, mean, sd), pnorm(b, mean, sd)), mean, sd)
}

# Generate variables
nps    <- round(runif(3500, min=-1, max=1), 0) # nps value which takes 1, 0 or -1
group  <- sample(letters[1:11], 3500, TRUE) # groups
weight <- rtnorm(n=3500, mean=1, sd=1, a=0.04, b=16) # weights between 0.04 and 16

# Build data frame
df = data.frame(group, nps, weight)

# The following packages / libraries are required:
require("data.table")
require("boot")

这是上面提到的加权平均值的第一篇文章中的代码：

samplewmean <- function(d, i, j) {
  d <- d[i, ]
  w <- j[i, ]
  return(weighted.mean(d, w))   
}

results_qsec <- boot(data= df[, 2, drop = FALSE], 
                     statistic = samplewmean, 
                     R=10000, 
                     j = df[, 3 , drop = FALSE])

这完全没问题。

下面是第二篇文章中的代码，用数据表中的组引导平均值：

dt = data.table(df)
stat <- function(x, i) {x[i, (m=mean(nps))]}
dt[, list(list(boot(.SD, stat, R = 100))), by = group]$V1

这也可以。

我无法结合两种方法：

跑步......

dt[, list(list(boot(.SD, samplewmean, R = 5000, j = dt[, 3 , drop = FALSE]))), by = group]$V1

...显示错误消息：

Error in weighted.mean.default(d, w) : 
  'x' and 'w' must have the same length

跑步......

dt[, list(list(boot(dt[, 2 , drop = FALSE], samplewmean, R = 5000, j = dt[, 3 , drop = FALSE]))), by = group]$V1

......提出了一个不同的错误：

Error in weighted.mean.default(d, w) : 
  (list) object cannot be coerced to type 'double'

我仍然无法理解data.table中的参数以及如何组合运行data.table的函数。

我将不胜感激。

Answer 1

它与data.table在函数范围内的行为有关。即使在使用data.table进行子集化之后，d仍然是samplewmean i，而weighted.mean期待权重和值的数字向量。如果您在致电unlist之前weighted.mean，则可以修复此错误

weighted.mean.default（d，w）出错：（list）对象无法强制键入＆＃39; double＆＃39;

在传递到weighted.mean之前要取消列出的代码：

samplewmean <- function(d, i, j) {
  d <- d[i, ]
  w <- j[i, ]
  return(weighted.mean(unlist(d), unlist(w)))   
}

dt[, list(list(boot(dt[, 2 , drop = FALSE], samplewmean, R = 5000, j = dt[, 3 , drop = FALSE]))), by = group]$V1

更多data.table - like（data.table version＆gt; = v1.10.2）语法可能如下：

#a variable named original is being passed in from somewhere and i am unable to figure out from where
samplewmean <- function(d, valCol, wgtCol, original) {
    weighted.mean(unlist(d[, ..valCol]), unlist(d[, ..wgtCol]))
}

dt[, list(list(boot(.SD, statistic=samplewmean, R=1, valCol="nps", wgtCol="weight"))), by=group]$V1

或另一种可能的语法是:(见data.table faq 1.6）

samplewmean <- function(d, valCol, wgtCol, original) {
    weighted.mean(unlist(d[, eval(substitute(valCol))]), unlist(d[, eval(substitute(wgtCol))]))
}

dt[, list(list(boot(.SD, statistic=samplewmean, R=1, valCol=nps, wgtCol=weight))), by=group]$V1

R引导加权意味着按组分组，数据表为

1 个答案: