代码之家 › 专栏 › 技术社区 › Laura

如何将变量名和另一个变量名的数据框与回归数据相匹配?

lm linear-regression formula regression r

Laura · 技术社区 · 6 年前

我有两个数据帧:

x = data.frame(Var1= c("A", "B", "C", "D","E"),Var2=c("F","G","H","I","J"),
    Value= c(11, 12, 13, 14,18))

y = data.frame(A= c(11, 12, 13, 14,18), B= c(15, 16, 17, 14,18),C= c(17, 22, 23, 24,18), D= c(11, 12, 13, 34,18),E= c(11, 5, 13, 55,18),  F= c(8, 12, 13, 14,18),G= c(7, 5, 13, 14,18),
    H= c(8, 12, 13, 14,18), I= c(9, 5, 13, 14,18), J= c(11, 12, 13, 14,18))

Var3 <- rep("time", each=length(x$Var1))

x=cbind(x,Var3)

time=seq(1:length(y[,1]))
y=cbind(y,time)

> x
  Var1 Var2 Value Var3
1    A    F    11 time
2    B    G    12 time
3    C    H    13 time
4    D    I    14 time
5    E    J    18 time
> y
   A  B  C  D  E  F  G  H  I  J time
1 11 15 17 11 11  8  7  8  9 11    1
2 12 16 22 12  5 12  5 12  5 12    2
3 13 17 23 13 13 13 13 13 13 13    3
4 14 14 24 34 55 14 14 14 14 14    4
5 18 18 18 18 18 18 18 18 18 18    5

x DF,我有变量 A 和 F y DF并实现一个简单的回归: lm(A ~ F, data = y) ,并将结果保存在列表的第一个位置。我也会对第二排的 DF实现回归 lm(B ~ G, data = y)

如何匹配中的变量名数据输入 是的 为了回归?

更复杂的回归如何 Var1 ~ Var2 + Var3

1 回复 | 直到 6 年前

Zheyuan Li 6 年前

x = data.frame(Var1= c("A", "B", "C", "D","E"),
               Var2=c("F","G","H","I","J"),
               Value= c(11, 12, 13, 14,18))

y = data.frame(A= c(11, 12, 13, 14,18),
               B= c(15, 16, 17, 14,18),
               C= c(17, 22, 23, 24,18),
               D= c(11, 12, 13, 34,18),
               E= c(11, 5, 13, 55,18),
               F= c(8, 12, 13, 14,18),
               G= c(7, 5, 13, 14,18),
               H= c(8, 12, 13, 14,18), 
               I= c(9, 5, 13, 14,18),
               J= c(11, 12, 13, 14,18))

我们可以利用

fitmodel <- function (RHS, LHS) do.call("lm", list(formula = reformulate(RHS, LHS),
                                              data = quote(y)))

modList <- Map(fitmodel, as.character(x$Var2), as.character(x$Var1))

modList[[1]]  ## for example
#Call:
#lm(formula = A ~ F, data = y)
#
#Coefficients:
#(Intercept)            F  
#     4.3500       0.7115

评论:

使用 do.call reformulate 传递给时计算 lm . 这是需要的,因为它允许像 update Showing string in formula and not as variable in lm fit . 作为比较:

oo <- Map(function (RHS, LHS) lm(reformulate(RHS, LHS), data = y),
          as.character(x$Var2), as.character(x$Var1))
oo[[1]]
#Call:
#lm(formula = reformulate(RHS, LHS), data = y)
#
#Coefficients:
#(Intercept)            F  
#     4.3500       0.7115

这个 as.character 在 x$Var1 和 x$Var2 重新制定 不能用。如果你把 stringsAsFactors = FALSE 在里面 data.frame 当你建立你的 x

对你有用吗?它不应该有一个“for”循环?

这个 Map 函数隐藏“for”循环。它是 mapply *apply R中的族函数是 a syntactic sugar .

你最初的问题是构造一个模型公式 Var1 ~ Var2 .

你的新问题是 Var1 ~ Var2 + Var3 .

x$Var3 <- rep("time", each=length(x$Var1))
y$time <- seq(1:length(y[,1]))

## collect multiple RHS variables (using concatenation function `c`)
RHS <- Map(base::c, as.character(x$Var2), as.character(x$Var3))
#str(RHS)
#List of 5  ## oh this list has names! annoying!!
# $ F: chr [1:2] "F" "time"
# $ G: chr [1:2] "G" "time"
# $ H: chr [1:2] "H" "time"
# $ I: chr [1:2] "I" "time"
# $ J: chr [1:2] "J" "time"
LHS <- as.character(x$Var1)
modList <- Map(fitmodel, RHS, LHS)  ## `fitmodel` function unchanged
modList[[1]]  ## for example
#Call:
#lm(formula = A ~ F + time, data = y)
#
#Coefficients:
#(Intercept)            F         time  
#        5.6          0.5          0.5