使用变量作为R中的列名称来拟合glm

Question

我试图glm()使用变量而不是列名来适应R,但它不起作用.这可以帮助我自动生成glms.当我glm使用列名称时,程序运行正常,当我用包含列名的变量交换列名时,程序给出并出错.

这是我的命令的样子:

##The data
mydata <- structure(list(var1 = c(10L, 100L, 50L, 40L, 20L, 50L, 60L, 55L, 
45L), var2 = c(1.5, 1.2, 1, 1.4, 1.2, 1.4, 1.3, 1.4, 1.3), var3 = c(5L, 
3L, 4L, 1L, 5L, 2L, 7L, 5L, 4L), group = structure(c(1L, 1L, 
2L, 2L, 1L, 1L, 2L, 1L, 1L), .Label = c("A", "B"), class = "factor")), .Names = c("var1", 
"var2", "var3", "group"), class = "data.frame", row.names = c(NA, 
-9L))
## My variable
x <- c("var1+var2")
##Fitting the model
myglm <- glm(formula = group ~ var1+var2 , family = "binomial", data = mydata) ## works fine


myglm2 <- glm(formula = group ~ x , family = "binomial", data = mydata)
Error in model.frame.default(formula = group ~ x, data = mydata, drop.unused.levels = TRUE) : 
  variable lengths differ (found for 'x')

我试图使用paste(x)和cat(x)功能,但它没有用.可以在R中这样做吗？我需要使用它,因为我glm在for循环中制作了大约1000个.

Answer 1

编辑,更简单,as.formula:

valid.names <- names(mydata)[names(mydata) != "group"]  # all but group
for(i in 2:length(valid.names)) {
  frm <- as.formula(paste("group ~", valid.names[i - 1], "+" , valid.names[i]))
  myglm <- glm(formula = frm, family = "binomial", data = mydata) ## works fine
}

旧版

这是一个潜在的解决方案parse:

valid.names <- names(mydata[, -4])  # all but group
frm <- group ~ x
for(i in 2:length(valid.names)) {
  varplusvar <- parse(text=paste(valid.names[i - 1], "+" , valid.names[i]))[[1]]
  frm[[3]] <- varplusvar
  myglm <- glm(formula = frm, family = "binomial", data = mydata) ## works fine
}

Answer 2

reformulate当您想要基于字符串创建公式时,该功能非常有用.你不需要paste:

x <- c("var1+var2")

form <- reformulate(x, response = "group")
# group ~ var1 + var2

glm(formula = form , family = "binomial", data = mydata)