Question

当我使用standardize()定义arm对象并在formula中使用它时，as.formula包中的

lm(formula, data = df)失败了。

选项A（我不想要）标准化lm之外的输入。选项B尝试（并且失败）标准化lm对象。

（注意：保持我的循环结构，因为我的实际用例有点复杂）

# create data
  library(arm)
  set.seed(324)
  df <- data.frame(y=sample(0:50, 100, replace=T),
                   x1=sample(0:1, 100, replace=T),
                   x2=sample(0:50, 100, replace=T))

# rescale outside of lm for comparison
  df$x1Z <- rescale(df$x1, binary.inputs = "0/1")
  df$x2Z <- rescale(df$x2, binary.inputs = "0/1")

# actual use case has more vars
  var <- c("x1", "x2")
  varZ <- c("x1Z", "x2Z")

# Option A: lm on rescaled
  a <- data.frame(matrix(NA, nrow = 0, ncol = 6))
  for (i in 1:length(var)) {
    formula <- as.formula(paste("y ~", varZ[i])) # use standardized
    m1 <- lm(formula, data = df)
    ms1 <- summary(m1)
    a[i, 1] <- var[i]
    a[i, 2] <- coefficients(ms1)[1,1]  
    a[i, 3] <- coefficients(ms1)[2,1]  
    a[i, 4] <- coefficients(ms1)[2,4]  
    a[i, 5] <- confint(m1)[2,1]        
    a[i, 6] <- confint(m1)[2,2]        
  }
  names(a) <- c("predictor", "intercept", "est", "p", "L95CI", "U95CI")

# Option B: lm, rescaling within lm
  b <- data.frame(matrix(NA, nrow = 0, ncol = 6))
  for (i in 1:length(var)) {
    formula <- as.formula(paste("y ~", var[i]))  # use raw
    m2 <- lm(formula, data = df)
    m2Z <- standardize(m2, binary.inputs="0/1")  # error
    ms2 <- summary(m2Z)
    b[i, 1] <- var[i]
    b[i, 2] <- coefficients(ms2)[1,1]  
    b[i, 3] <- coefficients(ms2)[2,1]  
    b[i, 4] <- coefficients(ms2)[2,4]  
    b[i, 5] <- confint(m2)[2,1]        
    b[i, 6] <- confint(m2)[2,2]        
  }
  names(b) <- c("predictor", "intercept", "est", "p", "L95CI", "U95CI")

仅显示standardize有效：

m3 <- lm(y ~ x2, data=df)
standardize(m3, binary.inputs="0/1")

Answer 1

使用

m2 <- do.call("lm", list(formula = formula, data = quote(df)))

在选项B的循环中。

您的问题或多或少类似于此问题：Showing string in formula and not as variable in lm fit。你想在m2$call中保留一个不错的公式。

如果您想知道为什么这很重要，请参阅standardize的源代码：

getMethod("standardize", "lm")

此功能的工作原理是提取和分析$call对象的lm。

为什么arm :: standardize（）无法在循环中的lm对象上工作？

1 个答案: