为什么 arm::standardize() 无法在循环中处理 lm 对象？

Question

当我使用 as.formula 定义 formula 对象并在 lm(formula, data = df).[=19 中使用它时，arm 包中的

standardize() 对我来说失败了=]

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

（注意：保留我的循环结构，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")

只是为了展示 standardize 作品：

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。

您的问题与此问题或多或少相似：。你想在 m2$call.

中保留一个像样的公式

如果您想知道为什么这很重要，请参阅 standardize 的源代码：

getMethod("standardize", "lm")

此函数通过提取和分析 lm 对象的 $call 来工作。

为什么 arm::standardize() 无法在循环中处理 lm 对象？

Why does arm::standardize() fail to work on a lm object in a loop?

r

regression

formula

linear-regression

lm