model.matrix() 与 na.action=NULL？

发布于 2024-10-31 12:07:43 字数 1268 浏览 3 评论 0原文

我有一个公式和一个数据框，我想提取 model.matrix()。但是，我需要生成的矩阵包含在原始数据集中找到的 NA。如果我要使用 model.frame() 来执行此操作，我只需传递 na.action=NULL 即可。但是，我需要的输出是 model.matrix() 格式。具体来说，我只需要右侧变量，我需要输出是一个矩阵（而不是数据框），并且我需要将因子转换为一系列虚拟变量。

我确信我可以使用循环或其他东西将某些东西组合在一起，但我想知道是否有人可以建议一种更干净、更有效的解决方法。非常感谢您抽出时间！

这是一个例子：

dat <- data.frame(matrix(rnorm(20),5,4), gl(5,2))
dat[3,5] <- NA
names(dat) <- c(letters[1:4], 'fact')
ff <- a ~ b + fact

# This omits the row with a missing observation on the factor
model.matrix(ff, dat) 

# This keeps the NA, but it gives me a data frame and does not dichotomize the factor
model.frame(ff, dat, na.action=NULL)

这是我想要获得的：

   (Intercept)          b fact2 fact3 fact4 fact5
1            1  0.7266086     0     0     0     0
2            1 -0.6088697     0     0     0     0
3            NA 0.4643360     NA    NA    NA    NA
4            1 -1.1666248     1     0     0     0
5            1 -0.7577394     0     1     0     0
6            1  0.7266086     0     1     0     0
7            1 -0.6088697     0     0     1     0
8            1  0.4643360     0     0     1     0
9            1 -1.1666248     0     0     0     1
10           1 -0.7577394     0     0     0     1

原文

I have a formula and a data frame, and I want to extract the model.matrix(). However, I need the resulting matrix to include the NAs that were found in the original dataset. If I were to use model.frame() to do this, I would simply pass it na.action=NULL. However, the output I need is of the model.matrix() format. Specifically, I need only the right-hand side variables, I need the output to be a matrix (not a data frame), and I need factors to be converted to a series of dummy variables.

I'm sure I could hack something together using loops or something, but I was wondering if anyone could suggest a cleaner and more efficient workaround. Thanks a lot for your time!

And here's an example:

dat <- data.frame(matrix(rnorm(20),5,4), gl(5,2))
dat[3,5] <- NA
names(dat) <- c(letters[1:4], 'fact')
ff <- a ~ b + fact

# This omits the row with a missing observation on the factor
model.matrix(ff, dat) 

# This keeps the NA, but it gives me a data frame and does not dichotomize the factor
model.frame(ff, dat, na.action=NULL)

Here is what I would like to obtain:

   (Intercept)          b fact2 fact3 fact4 fact5
1            1  0.7266086     0     0     0     0
2            1 -0.6088697     0     0     0     0
3            NA 0.4643360     NA    NA    NA    NA
4            1 -1.1666248     1     0     0     0
5            1 -0.7577394     0     1     0     0
6            1  0.7266086     0     1     0     0
7            1 -0.6088697     0     0     1     0
8            1  0.4643360     0     0     1     0
9            1 -1.1666248     0     0     0     1
10           1 -0.7577394     0     0     0     1

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温柔一刀 2024-11-07 12:07:43

Joris 的建议有效，但更快、更干净的方法是通过全局 na.action 设置。 “通过”选项实现了我们从原始数据集中保留 NA 的目标。

选项 1：通过

结果矩阵将在与原始数据集对应的行中包含 NA。

options(na.action='na.pass')
model.matrix(ff, dat)

选项 2：省略

结果矩阵将跳过包含 NA 的行。

options(na.action='na.omit')
model.matrix(ff, dat)

选项 3：失败

如果原始数据包含 NA，则会发生错误。

options(na.action='na.fail')
model.matrix(ff, dat)

当然，更改全局选项时请务必小心，因为它们可能会改变代码其他部分的行为。谨慎的人可能会使用 current.na.action <- options('na.action') 之类的内容存储原始设置，然后在创建 model.matrix 后将其更改回来。

Joris's suggestion works, but a quicker and cleaner way to do this is via the global na.action setting. The 'Pass' option achieves our goal of preserving NA's from the original dataset.

Option 1: Pass

Resulting matrix will contain NA's in rows corresponding to the original dataset.

options(na.action='na.pass')
model.matrix(ff, dat)

Option 2: Omit

Resulting matrix will skip rows containing NA's.

options(na.action='na.omit')
model.matrix(ff, dat)

Option 3: Fail

An error will occur if the original data contains NA's.

options(na.action='na.fail')
model.matrix(ff, dat)

Of course, always be careful when changing global options because they can alter behavior of other parts of your code. A cautious person might store the original setting with something like current.na.action <- options('na.action'), and then change it back after making the model.matrix.

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千纸鹤带着心事 2024-11-07 12:07:43

另一种方法是使用带有参数 na.action=na.pass 的 model.frame 函数作为 model.matrix 的第二个参数：

> model.matrix(ff, model.frame(~ ., dat, na.action=na.pass))
   (Intercept)          b fact2 fact3 fact4 fact5
1            1 -1.3560754     0     0     0     0
2            1  2.5476965     0     0     0     0
3            1  0.4635628    NA    NA    NA    NA
4            1 -0.2871379     1     0     0     0
5            1  2.2684958     0     1     0     0
6            1 -1.3560754     0     1     0     0
7            1  2.5476965     0     0     1     0
8            1  0.4635628     0     0     1     0
9            1 -0.2871379     0     0     0     1
10           1  2.2684958     0     0     0     1

< code>model.frame 允许您为 na.action 设置适当的操作，该操作在调用 model.matrix 时维护。

Another way is to use the model.frame function with argument na.action=na.pass as your second argument to model.matrix:

> model.matrix(ff, model.frame(~ ., dat, na.action=na.pass))
   (Intercept)          b fact2 fact3 fact4 fact5
1            1 -1.3560754     0     0     0     0
2            1  2.5476965     0     0     0     0
3            1  0.4635628    NA    NA    NA    NA
4            1 -0.2871379     1     0     0     0
5            1  2.2684958     0     1     0     0
6            1 -1.3560754     0     1     0     0
7            1  2.5476965     0     0     1     0
8            1  0.4635628     0     0     1     0
9            1 -0.2871379     0     0     0     1
10           1  2.2684958     0     0     0     1

model.frame allows you to set the appropriate action for na.action which is maintained when model.matrix is called.

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旧街凉风 2024-11-07 12:07:43

在查看 mattdevlin 和 Nathan Gould 的回答：

 model.matrix.lm(ff, dat, na.action = "na.pass")

model.matrix.default 可能不支持 na.action 参数，但是 model.matrix.lm 确实如此！

（我从 Rstudio 的自动完成建议中找到了 model.matrix.lm - 如果您尚未加载任何模型，它似乎是 model.matrix 的唯一非默认方法然后我猜测它可能支持 na.action 参数。）

I half-stumbled across a simpler solution after looking at mattdevlin and Nathan Gould's answers:

 model.matrix.lm(ff, dat, na.action = "na.pass")

model.matrix.default may not support the na.action argument, but model.matrix.lm does!

(I found model.matrix.lm from Rstudio's auto-complete suggestions — it appears to be the only non-default method for model.matrix if you haven't loaded any libraries that add others. Then I just guessed it might support the na.action argument.)

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月下伊人醉 2024-11-07 12:07:43

您可以根据 rownames : 使用 model.matrix 对象进行一些修改，

MM <- model.matrix(ff,dat)
MM <- MM[match(rownames(dat),rownames(MM)),]
MM[,"b"] <- dat$b
rownames(MM) <- rownames(dat)

它给出：

> MM
     (Intercept)         b fact2 fact3 fact4 fact5
1              1 0.9583010     0     0     0     0
2              1 0.3266986     0     0     0     0
3             NA 1.4992358    NA    NA    NA    NA
4              1 1.2867461     1     0     0     0
5              1 0.5024700     0     1     0     0
6              1 0.9583010     0     1     0     0
7              1 0.3266986     0     0     1     0
8              1 1.4992358     0     0     1     0
9              1 1.2867461     0     0     0     1
10             1 0.5024700     0     0     0     1

或者，您可以使用 contrasts() 来为您完成这项工作。手动构造矩阵将是：

cont <- contrasts(dat$fact)[as.numeric(dat$fact),]
colnames(cont) <- paste("fact",colnames(cont),sep="")
out <- cbind(1,dat$b,cont)
out[is.na(dat$fact),1] <- NA
colnames(out)[1:2]<- c("Intercept","b")
rownames(out) <- rownames(dat)

给出：

> out
     Intercept          b fact2 fact3 fact4 fact5
1            1  0.2534288     0     0     0     0
2            1  0.2697760     0     0     0     0
3           NA -0.8236879    NA    NA    NA    NA
4            1 -0.6053445     1     0     0     0
5            1  0.4608907     0     1     0     0
6            1  0.2534288     0     1     0     0
7            1  0.2697760     0     0     1     0
8            1 -0.8236879     0     0     1     0
9            1 -0.6053445     0     0     0     1
10           1  0.4608907     0     0     0     1

在任何情况下，这两种方法都可以合并到可以处理更复杂公式的函数中。我把练习留给读者（当我在论文中遇到这句话时，我讨厌什么;-)）

You can mess around a little with the model.matrix object, based on the rownames :

MM <- model.matrix(ff,dat)
MM <- MM[match(rownames(dat),rownames(MM)),]
MM[,"b"] <- dat$b
rownames(MM) <- rownames(dat)

which gives :

> MM
     (Intercept)         b fact2 fact3 fact4 fact5
1              1 0.9583010     0     0     0     0
2              1 0.3266986     0     0     0     0
3             NA 1.4992358    NA    NA    NA    NA
4              1 1.2867461     1     0     0     0
5              1 0.5024700     0     1     0     0
6              1 0.9583010     0     1     0     0
7              1 0.3266986     0     0     1     0
8              1 1.4992358     0     0     1     0
9              1 1.2867461     0     0     0     1
10             1 0.5024700     0     0     0     1

Alternatively, you can use contrasts() to do the work for you. Constructing the matrix by hand would be :

cont <- contrasts(dat$fact)[as.numeric(dat$fact),]
colnames(cont) <- paste("fact",colnames(cont),sep="")
out <- cbind(1,dat$b,cont)
out[is.na(dat$fact),1] <- NA
colnames(out)[1:2]<- c("Intercept","b")
rownames(out) <- rownames(dat)

which gives :

> out
     Intercept          b fact2 fact3 fact4 fact5
1            1  0.2534288     0     0     0     0
2            1  0.2697760     0     0     0     0
3           NA -0.8236879    NA    NA    NA    NA
4            1 -0.6053445     1     0     0     0
5            1  0.4608907     0     1     0     0
6            1  0.2534288     0     1     0     0
7            1  0.2697760     0     0     1     0
8            1 -0.8236879     0     0     1     0
9            1 -0.6053445     0     0     0     1
10           1  0.4608907     0     0     0     1

In any case, both methods can be incorporated in a function that can deal with more complex formulae. I leave the exercise to the reader (what do I loath that sentence when I meet it in a paper ;-) )

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