在plyr调用中使用svyglm

问题描述

这显然与R的调查包特有.我正在尝试使用plyr包中的llply进行列表的svyglm模型.这是一个示例:

This is clearly something idiosyncratic to R's survey package. I'm trying to use llply from the plyr package to make a list of svyglm models. Here's an example:

library(survey)
library(plyr)

foo <- data.frame(y1 = rbinom(50, size = 1, prob=.25),
                  y2 = rbinom(50, size = 1, prob=.5),
                  y3 = rbinom(50, size = 1, prob=.75),
                  x1 = rnorm(50, 0, 2),
                  x2 = rnorm(50, 0, 2),
                  x3 = rnorm(50, 0, 2),
                  weights = runif(50, .5, 1.5))

我的因变量列号列表

dvnum <- 1:3

在此样本中未显示任何聚类或地层

Indicating no clusters or strata in this sample

wd <- svydesign(ids= ~0, strata= NULL, weights= ~weights, data = foo)

一个svyglm通话有效

A single svyglm call works

svyglm(y1 ~ x1 + x2 + x3, design= wd)

然后llply将列出基本的R glm模型

And llply will make a list of base R glm models

llply(dvnum, function(i) glm(foo[,i] ~ x1 + x2 + x3, data = foo))

但是当我尝试将此方法调整为svyglm

But llply throws the following error when I try to adapt this method to svyglm

llply(dvnum, function(i) svyglm(foo[,i] ~ x1 + x2 + x3, design= wd))

Error in svyglm.survey.design(foo[, i] ~ x1 + x2 + x3, design = wd) : 
all variables must be in design= argument

所以我的问题是:如何使用llply和svyglm?

So my question is: how do I use llply and svyglm?

推荐答案

DWin谈到了他对正确公式的评论.

DWin was on to something with his comment about correct formula.

reformulate将执行此操作.

dvnum <- names(foo)[1:3]

llply(dvnum, function(i) {
    svyglm(reformulate(c('x1', 'x2', 'x3'),response = i), design = wd)})

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