在用户定义的函数中调用 glm() 函数

Question

我一直在尝试创建一个在其中使用 glm() 的函数。但我总是收到一条错误消息。看起来函数没有检索变量的值。

set.seed(234)
sex <- sample(c("M", "F"), size=100, replace=TRUE)
age <- rnorm(n=100, mean=20 + 4*(sex=="F"), sd=0.1)
dsn <- data.frame(sex, age)
rm(sex, age) #remove sex and age from the global environment for reproducibility

to_analyze <- function(dep, indep, data){
  glm(dep~factor(indep), data=data)
}

to_analyze(dep=age, indep=sex, data=dsn)
#> Error in eval(predvars, data, env): object 'age' not found

Answer 1

在函数中创建一个"formula"对象并传递给glm。

要在不报错的情况下获取变量，标准技巧是 deparse(substitute(.))。
然后用paste.

组成公式

to_analyze <- function(dep, indep, data){
  dep <- deparse(substitute(dep))
  indep <- deparse(substitute(indep))
  indep <- paste0("factor(", indep, ")")
  fmla <- paste(dep, indep, sep = " ~ ")
  fmla <- as.formula(fmla)
  glm(fmla, data = data)
}

to_analyze(dep=age, indep=sex, data=dsn)
#
#Call:  glm(formula = fmla, data = data)
#
#Coefficients:
# (Intercept)  factor(sex)M  
#      23.984        -3.984  
#
#Degrees of Freedom: 99 Total (i.e. Null);  98 Residual
#Null Deviance:     396.2 
#Residual Deviance: 0.837   AIC: -188.5

Answer 2

您可以使用以下任何一项：

Using substitute:

to_analyze <- function(dep, indep, data){
  glm(substitute(dep ~ factor(indep)), data=data)
}

to_analyze(dep=age, indep=sex, data=dsn)

优点：可以独立的写成公式

例如

 to_analyze(Petal.Width, Sepal.Length + Sepal.Width, data = iris)

Using reformulate as stated by

to_analyze <- function(dep, indep, data){ 
  glm(reformulate(sprintf("factor(%s)",indep), dep),  data = data) 
 }

注意调用此函数，变量必须是字符类型

 to_analyze(dep= "age", indep="sex", data=dsn)

Recall glm can also take a string that can be parsed to a formula:

to_analyze <- function(dep, indep, data){ 
  glm(sprintf("%s~factor(%s)", dep, indep),  data = data) 
}

to_analyze("age", "sex", data=dsn)

甚至：

to_analyze <- function(dep, indep, data){ 
  glm(paste(dep,"~ factor(",indep,")"),  data = data) 
}

to_analyze("age", "sex", data=dsn)

LASTLY: to combine both the substitute and paste:

to_analyze <- function(dep, indep, data){ 
  glm(paste(substitute(dep),"~ factor(",substitute(indep),")"),  data = data) 
}

将适用于符号和字符。例如：

to_analyze(age, sex, data=dsn)
to_analyze("age", "sex", data=dsn)

Answer 3

@Onyambu 等。 substitute 命令似乎只适用于一次调用，因为它适用于 to_analyze()。但是，当我在其中调用另一个函数时，它再次抱怨。任何帮助将不胜感激

to_analyze <- function(dep, indep, data){
  glm(substitute(dep ~ factor(indep)), data=data)
}

to_analyze(dep=age, indep=sex, data=dsn)
#> 
#> Call:  glm(formula = substitute(dep ~ factor(indep)), data = data)
#> 
#> Coefficients:
#>  (Intercept)  factor(sex)M  
#>       24.006        -4.034  
#> 
#> Degrees of Freedom: 99 Total (i.e. Null);  98 Residual
#> Null Deviance:       397.3 
#> Residual Deviance: 0.8152    AIC: -191.2

但是，我再次陷入困境，因为我试图在 lsmeans::lsmeans() 中调用此模型的输出来预测边际均值和 return 输出，但它给了我一个错误。虽然它不需要偏移量，但我将其包含在此处以便获得更通用的代码以便稍后修改。任何帮助将不胜感激

to_predict_lsmeans <- function(dep, indep, data){
  model <- glm(substitute(dep ~ factor(indep)), data=data)
  pred <- lsmeans:: lsmeans(model, substitute(~ factor(indep)), offset=substitute(data)$log(age), type ="response" )
  return(pred)
}

pred <- to_predict_lsmeans(dep=age, indep=sex, data=dsn)
#> Error in ref_grid(object, ...): We are unable to reconstruct the data.
#> The variables needed are:
#>  sex
#> Are any of these actually constants? (specify via 'params = ')
#> The dataset name is:
#>  data
#> Does the data still exist? Or you can specify a dataset via 'data = '
pred
#> Error in eval(expr, envir, enclos): object 'pred' not found

在用户定义的函数中调用 glm() 函数

calling the glm() function within a user-defined function

r

function

glm