dummy.coef() 用于多元模型

发布于 2024-12-10 14:37:45 字数 1123 浏览 2 评论 0原文

我想知道 dummy.coef() 中的一些内容，它将方差分析模型中的估计参数（对比度）转换为原始参数。它仅适用于单变量模型，但使其适用于多变量模型所需的更改似乎很小。在 dummy.coef.lm() 中：

第 52 行 coef <- object$coefficients 必须是 coef <- as.matrix(object$coefficients ) 以适应单变量和多变量模型（coef(object) 在第一种情况下是一个向量，在第二种情况下是一个矩阵）
第 60 行ans <- drop(mm[rn == tl[j], keep, drop = FALSE] %*% coef[keep]) 必须是 ans <- drop( mm[rn == tl[j], keep, drop = FALSE] %*% coef[keep, ]) 保留 coef
第 61 行中的所有列names(ans) <- rnn[rn == tl[j]] 可以是 names(ans) <-rep(rnn[rn == tl[j]], ncol (coef)) 为所有列的行命名

打印方法需要一些更改，但似乎就是这样。 有人知道为什么 dummy.coef() 不是为处理多元模型而设计的吗？

我偶然发现的另一件事：第 20-22 行

for (i in vars) args[[i]] <- if (nxl[[i]] == 1)
    rep.int(1, nl)
else factor(rep.int(xl[[i]][1L], nl), levels = xl[[i]])

安全吗？即，如果 if() 子句为 TRUE，是否会出现意外的 else？我本来期望类似的东西

for (i in vars) args[[i]] <- if (nxl[[i]] == 1) {
    rep.int(1, nl)
} else { factor(rep.int(xl[[i]][1L], nl), levels = xl[[i]]) }

原文

I was wondering about some things in dummy.coef() which converts the estimated parameters (contrasts) in ANOVA models to the original ones. It only works for univariate models, but the changes required to also make it work for multivariate models seem minor. In dummy.coef.lm():

line 52 coef <- object$coefficients would have to be coef <- as.matrix(object$coefficients) to accomodate univariate and multivariate models (coef(object) is a vector in the 1st case, and a matrix in the 2nd)
line 60 ans <- drop(mm[rn == tl[j], keep, drop = FALSE] %*% coef[keep]) would have to be ans <- drop(mm[rn == tl[j], keep, drop = FALSE] %*% coef[keep, ]) to keep all columns in coef
line 61 names(ans) <- rnn[rn == tl[j]] could be names(ans) <- rep(rnn[rn == tl[j]], ncol(coef)) to give names to the rows of all columns

The printing method would need some changes, but that seems to be it. Does anybody know why dummy.coef() was not designed to handle multivariate models?

Another thing I stumbled upon: Lines 20-22 are

for (i in vars) args[[i]] <- if (nxl[[i]] == 1)
    rep.int(1, nl)
else factor(rep.int(xl[[i]][1L], nl), levels = xl[[i]])

Is that safe? I.e., if the if() clause is TRUE, wouldn't there be an unexpected else? I would have expected something like

for (i in vars) args[[i]] <- if (nxl[[i]] == 1) {
    rep.int(1, nl)
} else { factor(rep.int(xl[[i]][1L], nl), levels = xl[[i]]) }

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离笑几人歌 2024-12-17 14:37:45

这仅解决了您的第二个问题（关于代码中第 20-22 行的 if() x else y 语句的问题）。

首先，尝试将这两个块剪切并粘贴到 R 会话中：

test <- TRUE

# Block 1 -- Doesn't work
if(test) 
    cat("test is TRUE\n")
    else
    cat("test is FALSE\n")

# Block 2 -- Works
{
if(test) 
    cat("test is TRUE\n")
    else
    cat("test is FALSE\n")
}

发生了什么？ {} 在这里发挥了重要作用。块 1 逐行“读取-解析-评估”代码，导致出现您所期望的问题。另一方面，在进行任何评估之前会读取并完全解析块 2。这是 {} 指示 R 执行的操作的一部分。当解析器接收到整个代码块时，它会明确地将 if() x else y 块解析为一个表达式。

您引用的代码取自函数体内（即 {} 对内部）。在这种情况下，它被正确处理（即像块 2 一样）。

华泰

This only addresses your second question (the one about the if() x else y statement at lines 20-22 in the code).

To start with, try cut-and-pasting these two blocks into an R session:

test <- TRUE

# Block 1 -- Doesn't work
if(test) 
    cat("test is TRUE\n")
    else
    cat("test is FALSE\n")

# Block 2 -- Works
{
if(test) 
    cat("test is TRUE\n")
    else
    cat("test is FALSE\n")
}

What's going on? The {} make all the difference here. Block 1 'read-parse-evaluates' the code line-by-line, causing just the problem you'd expect. Block 2, on the other hand, is read and completely parsed before any evaluation takes place. That's part of what {} directs R to do. When it receives the block of code as a whole, the parser clearly parses the if() x else y block as one expression.

The code you quoted was taken from inside the body of a function (i.e. inside of a {} pair). In that context, it is handled correctly (i.e. like Block 2).

HTH

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