R中每次输出一个变量的输出回归统计量 [英] Output Regression statistics for each variable one at a time in R

问题描述

我有一个看起来像这样的数据框.名称和列数将不一致(有时不会出现"C"，有时可能会出现"D"，"E"，"F"，等等.)唯一一致的变量将始终是Y，并且我想对Y退缩.

I have a data frame that looks like this. names and number of columns will NOT be consistent (sometimes 'C' will not be present, other times "D', 'E', 'F' may be present, etc.). The only consistent variable will always be Y, and I want to regress against Y.

# name and number of columns varies...so need flexible process
Y <- c(4, 4, 3, 4, 3, 2, 3, 2, 2, 3, 4, 4, 3, 4, 8, 6, 5, 4, 3, 6)
A <- c(1, 2, 1, 2, 3, 2, 1, 1, 1, 2, 1, 4, 3, 1, 2, 2, 1, 2, 4, 8)
B <- c(5, 6, 6, 5, 3, 7, 2, 1, 1, 2, 7, 4, 7, 8, 5, 7, 6, 6, 4, 7)
C <- c(9, 1, 2, 2, 1, 4, 5, 6, 7, 8, 89, 9, 7, 6, 5, 6, 8, 9 , 67, 6)
YABC <- data.frame(Y, A, B, C)

我想遍历每个变量并收集回归模型的输出.

I want to loop through each variable and collect output from regression model.

此过程创建所需的输出，但仅针对此特定迭代.

This process creates the desired output, but only for this specific iteration.

model_A <- lm(Y ~ A, YABC)

ID <- 'A'
rsq <- summary(model_A)$r.squared
adj_rsq <- summary(model_A)$adj.r.squared
sig <- summary(model_A)$sigma

datA <- data.frame(ID, rsq, adj_rsq, sig)

model_B <- lm(Y ~ B, YABC)

ID <- 'B'
rsq <- summary(model_B)$r.squared
adj_rsq <- summary(model_B)$adj.r.squared
sig <- summary(model_B)$sigma

datB <- data.frame(ID, rsq, adj_rsq, sig)

model_C <- lm(Y ~ C, YABC)

ID <- 'C'
rsq <- summary(model_C)$r.squared
adj_rsq <- summary(model_C)$adj.r.squared
sig <- summary(model_C)$sigma

datC <- data.frame(ID, rsq, adj_rsq, sig)

output <- rbind(datA, datB, datC)

如何将其包装在循环中或其他可以解释列数和列名不同的过程中?这是我的尝试...是的，我知道这是不对的，只是我将自己想要的能力概念化.

How can I wrap this in a loop or some other process that will account for varied number and name of columns? Here is my attempt...yes I know it's not right, just me conceptualizing the kind of capability I'd like.

# initialize data frame
output__ <- data.frame(ID__ = as.character(),
                     rsq__ = as.numeric(),
                     adj_rsq__ = as.numeric(),
                     sig__ = as.numeric())

# loop through A, then B, then C
for(i in A:C) {
  model_[i] <- lm(Y ~ [i], YABC)

  ID <- '[i]'
  rsq <- summary(model_[i])$r.squared
  adj_rsq <- summary(model_[i])$adj.r.squared
  sig <- summary(model_[i])$sigma
  data__temp <- (ID__, rsq__, adj_rsq__, sig__)
  data__ <- rbind(data__, data__temp)
}

使用@BigDataScientist方法...这是我使用的解决方案.

Using @BigDataScientist approach...here is the solution I went with.

# initialize data frame
data__ <- data.frame(ID__ = as.character(),
                     rsq__ = as.numeric(),
                     adj_rsq__ = as.numeric(),
                     sig__ = as.numeric())

# loop through A, then B, then C
for(char in names(YABC)[-1]){
  model <- lm(as.formula(paste("Y ~ ", char)), YABC)
  ID__ <- paste(char)
  rsq__ <- summary(model)$r.squared
  adj_rsq__ <- summary(model)$adj.r.squared
  sig__ <- summary(model)$sigma
  data__temp <- data.frame(ID__, rsq__, adj_rsq__, sig__)
  data__ <- rbind(data__, data__temp)

}

推荐答案

如注释中所述:?as.formula()是一种解决方案. 您可以像这样:

As written in the comment: ?as.formula() is one solution. You could do sthg like:

model = list()
for(char in names(YABC)[-1]) {
  model[[char]] <- lm(as.formula(paste("Y ~ ", char)), YABC)
}
model

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