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[英]Condition on the existence of the object in the User-defined function, not in the Global environment in R
[英]R Step function looks for data in global environment, not inside defined function
我有逐步回歸問題,我的理解是我沒有正確傳遞參數Data
。
我有這個功能:
ForwardStep <- function(df,yName, Xs, XsMin) {
Data <- df[, c(yName,Xs)]
fit <- glm(formula = paste(yName, " ~ ", paste0(XsMin, collapse = " + ")),
data = Data, family = binomial(link = "logit") )
ScopeFormula <- list(lower = paste(yName, " ~ ", paste0(XsMin, collapse = " + ")),
upper = paste(yName, " ~ ", paste0(Xs, collapse = " + ")))
result <- step(fit, direction = "forward", scope = ScopeFormula, trace = 1 )
return(result)
}
當我嘗試使用以下參數運行它時
df <- data.frame(Y= rep(c(0,1),25),time = rpois(50,2), x1 = rnorm(50, 0,1),
x2 = rnorm(50,.5,2), x3 = rnorm(50,0,1))
yName = "Y"
Xs <- c("x1","x2","x3")
XsMin <- 1
res <- ForwardStep(df,Yname,Xs,XsMin)
我收到一個錯誤: is.data.frame(data) 中的錯誤:找不到對象“數據”
但是如果我首先在 Global Env 中定義Data
,它就可以正常工作。
Data <- df[, c(yName,Xs)]
res <- ForwardStep(df,Yname,Xs,XsMin)
我想我對函數 step 的實現有誤,但是我不知道如何以正確的方式執行。
您需要意識到公式始終具有關聯的環境,請參閱help("formula")
。 永遠不要將文本傳遞給模型函數的formula
參數,永遠不要。 如果這樣做,遲早會遇到范圍界定問題。 通常,我建議改為使用語言進行計算,但您也可以從正確范圍內的文本創建公式:
ForwardStep <- function(df,Yname, Xs, XsMin) {
Data <- df[, c(Yname,Xs)]
f1 <- as.formula(paste(Yname, " ~ ", paste0(XsMin, collapse = " + ")))
fit <- glm(formula = f1,
data = Data, family = binomial(link = "logit") )
f2 <- as.formula(paste(Yname, " ~ ", paste0(XsMin, collapse = " + ")))
f3 <- as.formula(paste(Yname, " ~ ", paste0(Xs, collapse = " + ")))
ScopeFormula <- list(lower = f2,
upper = f3)
step(fit, direction = "forward", scope = ScopeFormula, trace = 1)
}
df <- data.frame(Y= rep(c(0,1),25),time = rpois(50,2), x1 = rnorm(50, 0,1),
x2 = rnorm(50,.5,2), x3 = rnorm(50,0,1))
YName = "Y"
Xs <- c("x1","x2","x3")
XsMin <- 1
res <- ForwardStep(df,YName,Xs,XsMin)
#Start: AIC=71.31
#Y ~ 1
#
# Df Deviance AIC
#<none> 69.315 71.315
#+ x1 1 68.661 72.661
#+ x3 1 68.797 72.797
#+ x2 1 69.277 73.277
(公共服務公告:逐步回歸是垃圾生成器。有更好的統計技術可用。)
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