将系数名称转换为R中的公式

Question

当使用具有因子的公式时,拟合模型将系数命名为XY,其中X是因子的名称,Y是其特定级别.我希望能够从这些系数的名称创建一个公式.

原因是:如果我将套索适合稀疏设计矩阵(如下所示),我想创建一个新的公式对象,它只包含非零系数的项.

require("MatrixModels")
require("glmnet")
set.seed(1)
n <- 200
Z <- data.frame(letter=factor(sample(letters,n,replace=T),letters),
                x=sample(1:20,200,replace=T))
f <- ~ letter + x:letter + I(x>5):letter
X <- sparse.model.matrix(f, Z)
beta <- matrix(rnorm(dim(X)[2],0,5),dim(X)[2],1)
y <- X %*% beta + rnorm(n)

myfit <- glmnet(X,as.vector(y),lambda=.05)
fnew <- rownames(myfit$beta)[which(myfit$beta != 0)]
 [1] "letterb"              "letterc"              "lettere"             
 [4] "letterf"              "letterg"              "letterh"             
 [7] "letterj"              "letterm"              "lettern"             
[10] "lettero"              "letterp"              "letterr"             
[13] "letters"              "lettert"              "letteru"             
[16] "letterw"              "lettery"              "letterz"             
[19] "lettera:x"            "letterb:x"            "letterc:x"           
[22] "letterd:x"            "lettere:x"            "letterf:x"           
[25] "letterg:x"            "letterh:x"            "letteri:x"           
[28] "letterj:x"            "letterk:x"            "letterl:x"           
[31] "letterm:x"            "lettern:x"            "lettero:x"           
[34] "letterp:x"            "letterq:x"            "letterr:x"           
[37] "letters:x"            "lettert:x"            "letteru:x"           
[40] "letterv:x"            "letterw:x"            "letterx:x"           
[43] "lettery:x"            "letterz:x"            "letterb:I(x > 5)TRUE"
[46] "letterc:I(x > 5)TRUE" "letterd:I(x > 5)TRUE" "lettere:I(x > 5)TRUE"
[49] "letteri:I(x > 5)TRUE" "letterj:I(x > 5)TRUE" "letterl:I(x > 5)TRUE"
[52] "letterm:I(x > 5)TRUE" "letterp:I(x > 5)TRUE" "letterq:I(x > 5)TRUE"
[55] "letterr:I(x > 5)TRUE" "letteru:I(x > 5)TRUE" "letterv:I(x > 5)TRUE"
[58] "letterx:I(x > 5)TRUE" "lettery:I(x > 5)TRUE" "letterz:I(x > 5)TRUE"

从这个我想有一个公式

~ I(letter=="d") + I(letter=="e") + ...(etc)

我检查了公式()和all.vars()无济于事.此外,编写一个函数来解析这个问题有点痛苦,因为可能出现的术语类型不同.例如,对于x:字母,当x是数值而字母是因子,或I(x> 5):字母作为另一个烦人的情况.

所以我不知道在公式和它的字符表示之间转换并返回的某些函数？

Answer 1

当我运行代码时，我得到了一些不同的东西，因为 set.seed() 尚未指定。我没有使用变量名“letter”，而是使用“letter_”作为方便的分割标记：

> fnew <- rownames(myfit$beta)[which(myfit$beta != 0)]

> fnew
 [1] "letter_c" "letter_d" "letter_e" "letter_f" "letter_h" "letter_k" "letter_l"
 [8] "letter_o" "letter_q" "letter_r" "letter_s" "letter_t" "letter_u" "letter_v"
[15] "letter_w"

然后进行分割并打包成字符矩阵：

> fnewmtx <- cbind( lapply(sapply(fnew, strsplit, split="_"), "[[", 2),
+ lapply(sapply(fnew, strsplit, split="_"), "[[", 1))

fnewmtx [,1] [,2]
letter_c “c” “字母” letter_d “d” “字母” letter_e “e” “字母” letter_f “f” “字母” 剪掉其余部分

并将粘贴函数输出包装在 as.formula() 中，这是如何“在公式及其字符表示形式之间进行转换”的答案的一半。另一半是as.character()

form <- as.formula( paste("~", 
             paste( 
               paste(" I(", fnewmtx[,2], "_ ==", "'",fnewmtx[,1],"') ", sep="") , 
             sep="", collapse="+")
                 ) 
           )  # edit: needed to add back the underscore

现在输出是一个适当的类对象：

> class(form)
[1] "formula"
> form
~I(letter_ == "c") + I(letter_ == "d") + I(letter_ == "e") + 
    I(letter_ == "f") + I(letter_ == "h") + I(letter_ == "k") + 
    I(letter_ == "l") + I(letter_ == "o") + I(letter_ == "q") + 
    I(letter_ == "r") + I(letter_ == "s") + I(letter_ == "t") + 
    I(letter_ == "u") + I(letter_ == "v") + I(letter_ == "w")

我发现有趣的是 as.formula 转换使字母周围的单引号变成了双引号。

编辑：既然问题有一个或两个额外的维度，我的建议是跳过公式的重新创建。请注意，myfit$beta 的行名称与 X 的列名称完全相同，因此请使用非零行名称作为索引来选择 X 矩阵中的列：

> str(X[ , which( colnames(X) %in% rownames(myfit$beta)[which(myfit$beta != 0)] )] )
Formal class 'dgCMatrix' [package "Matrix"] with 6 slots
  ..@ i       : int [1:429] 9 54 91 157 166 37 55 68 117 131 ...
  ..@ p       : int [1:61] 0 5 13 20 28 36 42 50 60 68 ...
  ..@ Dim     : int [1:2] 200 60
  ..@ Dimnames:List of 2
  .. ..$ : chr [1:200] "1" "2" "3" "4" ...
  .. ..$ : chr [1:60] "letter_b" "letter_c" "letter_e" "letter_f" ...
  ..@ x       : num [1:429] 1 1 1 1 1 1 1 1 1 1 ...
  ..@ factors : list()

> myfit2 <- glmnet(X[ , which( colnames(X) %in% rownames(myfit$beta)[which(myfit$beta != 0)] )] ,as.vector(y),lambda=.05)
> myfit2

Call:  glmnet(x = X[, which(colnames(X) %in% rownames(myfit$beta)[
                                           which(myfit$beta != 0)])], 
              y = as.vector(y), lambda = 0.05) 

     Df   %Dev Lambda
[1,] 60 0.9996   0.05