如何將公式作為參數傳遞給r中的函數？

Question

如何在R中將公式作為參數傳遞？

下面的代碼適用於前兩種情況，但是當我傳遞公式時，會出現錯誤： Error in model.frame.default(formula = formula, weights = weights, na.action = na.omit, : invalid type (closure) for variable '(weights)'

makeModel<-function(formula,weights) {
    m <- lm(formula, na.action = na.omit, weights = weights)
    return(m);
}
run<-function(t) {
    f<-formula(t$y~t$x+t$r)
    m <- lm(t$y~t$x+t$r, na.action = na.omit, weights = t$size)
    m <- lm(f, na.action = na.omit, weights = t$size)
    m <- makeModels(f,t$size)    
}
l<-20
x<-seq(0,1,1/l)
y<-sqrt(x)
r=round(runif(n=length(x),min=0,max=.8))
n<-1:(l+1)
size=n/sum(n)
t<-data.frame(x,y,r,n,size)
run(t)

編輯1：此代碼：

makeModel<-function(formula,weights,t) {
    print(class(weights))
    m <- lm(formula, na.action = na.omit, weights = weights,data=t)
    return(m);
}
run<-function(t) {
    f<-formula(y~x+r)
    f <- as.formula("t$y~t$x+t$r")
    m <- lm(y~x+r, na.action = na.omit, weights = t$size,data=t)
    m <- lm(f, na.action = na.omit, weights = t$size,data=t)
    m <- makeModel(f,t$size,t)    
}

生產：

model.frame.default中的錯誤（公式=公式，數據= t，權重=權重，：變量“（權重）”的無效類型（閉包）

編輯2：作品：

makeModel <- function(formula, data) {
    # size is looked in data first, which is why this works
    m <- lm(formula, na.action = na.omit, weights = size, data =  data) # works
    #m <- lm(formula, na.action = na.omit, weights = data$size, data =  data) # fails!
    return(m)
}

r很奇怪！

有誰知道為什么weights = data $ size行失敗？

編輯3：得到了：weights = data $ size起作用。

makeModel<-function(formula,w,data) {
    print(class(weights))
    m <- lm(formula, na.action = na.omit, weights = size, data =  data) # works
    m <- lm(formula, na.action = na.omit, weights = data$size, data =  data) #works
    m <- lm(formula, na.action = na.omit, weights = w,data=data) # fails
    return(m);
}
run<-function(data) {
    f<-formula(y~x+r)
    #f <- as.formula("t$y~t$x+t$r")
    m <- lm(y~x+r, na.action = na.omit, weights = data$size,data=data)
    m <- lm(f, na.action = na.omit, weights = data$size,data=data)
    m <- makeModel(f,data$size,data)    
}

最后一個失敗，並顯示以下錯誤：eval（extras，data，env）中的錯誤：找不到對象“ w”

Answer 1

避免分配與轉置函數一致的名為t的對象。 看回溯率

makeModel<-function(formula,weights) {
  m <- lm(formula, na.action = na.omit, weights = weights)
  return(m)
}
run<-function(x) {
  f<-formula(x$y~x$x+x$r)
  m <- lm(x$y~x$x+x$r, na.action = na.omit, weights = x$size)
  m <- lm(f, na.action = na.omit, weights = x$size)
  m <- makeModel(f,x$size)    
}
l<-20
x<-seq(0,1,1/l)
y<-sqrt(x)
r=round(runif(n=length(x),min=0,max=.8))
n<-1:(l+1)
size=n/sum(n)
x<-data.frame(x,y,r,n,size)
run(x)
#R Error in model.frame.default(formula = formula, weights = weights, na.action = na.omit,  : 
#R    invalid type (closure) for variable '(weights)'
traceback()
#R 7: model.frame.default(formula = formula, weights = weights, na.action = na.omit, 
#R                        drop.unused.levels = TRUE)
#R 6: stats::model.frame(formula = formula, weights = weights, na.action = na.omit, 
#R                       drop.unused.levels = TRUE)
#R 5: eval(mf, parent.frame())
#R 4: eval(mf, parent.frame())
#R 3: lm(formula, na.action = na.omit, weights = weights) at #3
#R 2: makeModel(f, x$size) at #5
#R 1: run(t)

現在， debug(model.frame.default)顯示該行是由於這些行和這一行而出錯的地方。 原因是它調用

eval(list(weights = weights), environment(formula), environment(formula))

並且在run環境（分配了公式的環境）中未分配weights對象，因此它產生stats::weights 。 三種解決方案是

makeModel <- function(formula, weights) {
  environment(formula) <- environment()
  lm(formula, na.action = na.omit, weights = weights)
}
run<-function(x) {
  f <- x$y ~ x$x + x$r
  makeModel(f, x$size)  
}
x1 <- run(x)

makeModel <- function(formula, weights) {
  cl <- match.call()
  cl[[1L]] <- quote(lm)
  cl$na.action <- quote(na.omit)
  eval(cl, parent.frame())
}
run<-function(x) {
  f <- x$y ~ x$x + x$r
  makeModel(f, x$size)  
}
x2 <- run(x)

makeModel <- function(formula, weights, x) {
  cl <- match.call()
  cl[[1]] <- quote(lm)
  cl$x <- NULL
  cl[c("data", "formula", "na.action")] <- 
    list(quote(x), formula, quote(na.omit))
  eval(cl)
}
run<-function(x) {
  f <- y ~ x + r
  makeModel(f, size, x)  
}
x3 <- run(x)

stopifnot(all.equal(coef(x1), coef(x2)))
stopifnot(all.equal(coef(x1), coef(x3), check.attributes = FALSE))

例如，上面的第一個解決方案意味着

eval(list(weights = weights), environment(formula), environment(formula))

成功，因為在formula環境中分配了一個weights對象。 第二種解決方案在run環境中使用weights = x$size進行調用，從而成功。 如果您知道weights參數始終是size列，那么第三個問題就像RomanLuštrik的答案，盡管他的解決方案比我建議的第三個問題更為清晰。 這里的電話是

eval(list(weights = size), data, environment(formula))

因為size是data一列，所以它起作用。

Answer 2

請參閱?as.formula示例。 您不應從變量名中顯式調用變量。 該公式應該是抽象的，並且lm將知道您應該從data提取哪些變量，您應該指定哪些變量。

makeModels <- function(formula, data) {
  # size is looked in data first, which is why this works
  m <- lm(formula, na.action = na.omit, weights = size, data =  data)
  return(m)
}

run <- function(t) {
  f <- formula(y ~ x + r)
  m1 <- lm(formula = f, na.action = na.omit, weights = size, data = t)
  m2 <- makeModels(formula = f, data = t)
  return(list(m1, m2))
}

l<-20
x<-seq(0,1,1/l)
y<-sqrt(x)
r=round(runif(n = length(x), min = 0, max = 0.8))
n<-1:(l+1)
size=n/sum(n)
t<-data.frame(x,y,r,n,size)
run(t)

[[1]]

Call:
lm(formula = f, data = t, weights = t$size, na.action = na.omit)

Coefficients:
(Intercept)            x            r  
   0.327154     0.706553    -0.008167  


[[2]]

Call:
lm(formula = formula, data = data, weights = size, na.action = na.omit)

Coefficients:
(Intercept)            x            r  
   0.327154     0.706553    -0.008167