model.matrix() 与 na.action=NULL? [英] model.matrix() with na.action=NULL?

问题描述

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() 来做到这一点，我会简单地传递它 na.action=NULL.但是，我需要的输出是 model.matrix() 格式.具体来说，我只需要右边的变量，我需要输出是一个矩阵(不是数据框)，我需要将因子转换为一系列虚拟变量.

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.

我确信我可以使用循环或其他方式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清洁、更有效的解决方法.非常感谢您的时间！

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!

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

推荐答案

您可以根据行名对 model.matrix 对象进行一些处理:

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)

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() 为您完成这项工作.手动构建矩阵将是:

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)

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

无论如何，这两种方法都可以合并到一个可以处理更复杂公式的函数中.我将练习留给读者(当我在论文中遇到这句话时，我有什么不喜欢的；-))

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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