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我正在定义一个函数来获取回归模型的预测值，其中包含不同亚组（亚群）的调查数据。我使用调查包中的 svyglm 函数。

我的问题涉及处理 svyglm 函数中的子集选项。由于它使用非标准评估，我理解这意味着它不会将列名作为字符串。我尝试只使用不带字符串的列名并引用（enquo）并取消引用它（!!）。但是，这两个选项都不起作用。我也玩过 ensym() 和 expr() 但没有得到任何结果。

资料库

library(dplyr)
library(survey)
library(srvyr)
library(purrr)
library(rlang)

mtcars <- read.table("https://forge.scilab.org/index.php/p/rdataset/source/file/master/csv/datasets/mtcars.csv",
                     sep=",", header=TRUE)

mtcars_cplx <- mtcars %>% as_survey_design(id = cyl, weights = qsec)

carb <- c(1:8)
cyl <- c(4:8)
new_data <- expand.grid(carb, cyl)
colnames(new_data) <- c("carb", "cyl")

有礼貌

功能和输入

subpop_pred <- function(formula, data, subpop, new_data) {
  
  subpop_quo <- enquo(subpop)
  subpop_txt <- data$variables %>% select(!!subpop_quo) %>% colnames()
  
  for(i in min(data$variables[subpop_txt]):max(data$variables[subpop_txt])){
    reg <- svyglm(formula, data, subset=!!subpop_quo==i)
    pred <- predict(reg, newdata=new_data)
    
    if(exists("reg_end")==TRUE){
      pred <- cbind(new_data, pred, confint(pred))
      pred[subpop_txt] <- i
      reg_end <- rbind(reg_end, pred)
    } else {
      reg_end <- cbind(new_data, pred, confint(pred))
      reg_end[subpop_txt] <- i
    }
  }
}

subpop_pred(mpg ~ carb + cyl + carb*cyl, 
            data=mtcars_cplx, 
            new_data=new_data,
            subpop=gear)

输出/错误

 Error: Base operators are not defined for quosures.
Do you need to unquote the quosure?

  # Bad:
  myquosure == rhs

  # Good:
  !!myquosure == rhs
Call `rlang::last_error()` to see a backtrace 
8. stop(cnd) 
7. abort(paste_line("Base operators are not defined for quosures.", 
    "Do you need to unquote the quosure?", "", "  # Bad:", bad, 
    "", "  # Good:", good, )) 
6. Ops.quosure(subpop_quo, i) 
5. eval(subset, model.frame(design), parent.frame()) 
4. eval(subset, model.frame(design), parent.frame()) 
3. svyglm.survey.design(formula, data, subset = !!subpop_quo == 
    i) 
2. svyglm(formula, data, subset = !!subpop_quo == i) 
1. subpop_pred(mpg ~ carb + cyl + carb * cyl, data = mtcars_cplx, 
    new_data = new_data, subpop = gear)

没有定论

功能和输入

subpop_pred <- function(formula, data, subpop, new_data) {
  
  subpop_quo <- enquo(subpop)
  subpop_txt <- data$variables %>% select(!!subpop_quo) %>% colnames()
  
  for(i in min(data$variables[subpop_txt]):max(data$variables[subpop_txt])){
    reg <- svyglm(formula, data, subset=subpop==i)
    pred <- predict(reg, newdata=new_data)
    
    if(exists("reg_end")==TRUE){
      pred <- cbind(new_data, pred, confint(pred))
      pred[subpop_txt] <- i
      reg_end <- rbind(reg_end, pred)
    } else {
      reg_end <- cbind(new_data, pred, confint(pred))
      reg_end[subpop_txt] <- i
    }
  }
}

subpop_pred(mpg ~ carb + cyl + carb*cyl, data=mtcars_cplx, new_data=new_data, subpop=gear)

输出

Error in eval(subset, model.frame(design), parent.frame()) : 
  object 'gear' not found 
5. eval(subset, model.frame(design), parent.frame()) 
4. eval(subset, model.frame(design), parent.frame()) 
3. svyglm.survey.design(formula, data, subset = subpop == i) 
2. svyglm(formula, data, subset = subpop == i) 
1. subpop_pred(mpg ~ carb + cyl + carb * cyl, data = mtcars_cplx, 
    new_data = new_data, subpop = gear)

您知道如何使该功能正常工作吗？

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2 回答 2

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我可以subset通过混合expr()和rlang::tidy_eval().

然后，您的函数中的模型行可以读取：

reg <- svyglm(formula, data = data, 
       subset = rlang::eval_tidy( expr( !!subpop_quo == i), data =  data) )

不过，我不知道这是否强大，或者是否有一些更直接的 tidyeval 方法。处理这个让我意识到subset()函数/参数很难在函数中使用。:-P

于 2019-06-17T13:45:00.460 回答
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不确定是否有更好的方法，因为svyby()似乎不支持svyglm(). 在这里，quo_squash()用于将表达式传递到subset(). 这可以扩展到进行预测。

gears = unique(mtcars$gear)
lapply(gears, function(x) {
  subset(mtcars_cplx, !!quo_squash(gear == x)) %>% 
    svyglm(mpg ~ carb + cyl + carb*cyl, design = .)
})
于 2019-06-17T05:33:53.923 回答