R 中的因子水平默认为 1 和 2 |虚拟变量 [英] Factor levels default to 1 and 2 in R | Dummy variable
问题描述
我正在从 Stata 过渡到 R.在 Stata 中，如果我将因子水平(比如-0 和 1)标记为(M 和 F)，则 0 和 1 将保持原样.此外，在大多数软件(包括 Excel 和 SPSS)中，这是虚拟变量线性回归所必需的.
I am transitioning from Stata to R. In Stata, if I label a factor levels (say--0 and 1) to (M and F), 0 and 1 would remain as they are. Moreover, this is required for dummy-variable linear regression in most software including Excel and SPSS.
但是，我注意到 R 默认因子级别为 1,2 而不是 0,1.我不知道为什么 R 会这样做，尽管内部回归(并且正确地)假定 0 和 1 作为因子变量.我将不胜感激任何帮助.
However, I've noticed that R defaults factor levels to 1,2 instead of 0,1. I don't know why R does this although regression internally (and correctly) assumes 0 and 1 as the factor variable. I would appreciate any help.
这是我所做的:
尝试 #1:
sex<-c(0,1,0,1,1)
sex<-factor(sex,levels = c(1,0),labels = c("F","M"))
str(sex)
Factor w/ 2 levels "F","M": 2 1 2 1 1
现在似乎因子水平被重置为 1 和 2.我相信 1 和 2s 是对此处因子水平的引用.但是，我丢失了原始值，即 0 和 1.
It seems that factor levels are now reset to 1 and 2. I believe 1 and 2s are references to the factor level here. However, I have lost the original values i.e. 0s and 1s.
尝试 2:
sex<-c(0,1,0,1,1)
sex<-factor(sex,levels = c(0,1),labels = c("F","M"))
str(sex)
Factor w/ 2 levels "F","M": 1 2 1 2 2
同上.我的 0 和 1 现在是 1 和 2.相当令人惊讶.为什么会发生这种情况.
Ditto. My 0's and 1's are now 1's and 2's. Quite Surprising. Why is this happening.
尝试3现在，我想看看 1s 和 2s 是否有任何不良影响回归.所以，这就是我所做的:
Try3 Now, I wanted to see whether 1s and 2s have any bad effect regression. So, here's what I did:
这是我的数据的样子:
> head(data.frame(sassign$total_,sassign$gender))
sassign.total_ sassign.gender
1 357 M
2 138 M
3 172 F
4 272 F
5 149 F
6 113 F
myfit<-lm(sassign$total_ ~ sassign$gender)
myfit$coefficients
(Intercept) sassign$genderM
200.63522 23.00606
所以，结果证明手段是正确的.在运行回归时，R 确实使用了 0 和 1 值作为虚拟变量.
So, it turns out that the means are correct. While running the regression, R did use 0 and 1 value as dummies.
我确实检查了 SO 上的其他线程，但他们大多谈论 R 如何编码因子变量而没有告诉我原因.Stata 和 SPSS 通常要求基变量为0".所以，我想问问这个.
I did check other threads on SO, but they mostly talk about how R codes factor variables without telling me why. Stata and SPSS generally require the base variable to be "0." So, I thought of asking about this.
我很感激任何想法.
推荐答案
简而言之，你只是混淆了两个不同的概念.我将在下面一一阐明.
你在str()
您从 str() 中看到的是因子变量的内部表示.一个因子在内部是一个整数，其中的数字给出了向量内水平的位置.例如:
What you see from str() is the internal representation of a factor variable. A factor is internally an integer, where the number gives the position of levels inside the vector. For example:
x <- gl(3, 2, labels = letters[1:3])
#[1] a a b b c c
#Levels: a b c
storage.mode(x) ## or `typeof(x)`
#[1] "integer"
str(x)
# Factor w/ 3 levels "a","b","c": 1 1 2 2 3 3
as.integer(x)
#[1] 1 1 2 2 3 3
levels(x)
#[1] "a" "b" "c"
此类位置的常见用途是以最有效的方式执行 as.character(x):
A common use of such positions, is to perform as.character(x) in the most efficient way:
levels(x)[x]
#[1] "a" "a" "b" "b" "c" "c"
<小时>
您对模型矩阵的误解
在我看来，您认为模型矩阵是通过
It seems to me that you thought a model matrix is obtained by
cbind(1L, as.integer(x))
# [,1] [,2]
#[1,] 1 1
#[2,] 1 1
#[3,] 1 2
#[4,] 1 2
#[5,] 1 3
#[6,] 1 3
这不是真的.在这种情况下，您只是将因子变量视为数值变量.
which is not true. In this fashion, you are just treating a factor variable as a numerical variable.
模型矩阵是这样构建的:
The model matrix is constructed this way:
xlevels <- levels(x)
cbind(1L, match(x, xlevels[2], nomatch=0), match(x, xlevels[3], nomatch=0))
# [,1] [,2] [,3]
#[1,] 1 0 0
#[2,] 1 0 0
#[3,] 1 1 0
#[4,] 1 1 0
#[5,] 1 0 1
#[6,] 1 0 1
1 和 0 分别表示匹配"/出现"和不匹配"/不出现".
The 1 and 0 implies "match" / "occurrence" and "no-match" / "no-occurrence", respectively.
R 例程 model.matrix 将有效地为您完成此操作，并提供易于阅读的列名和行名:
The R routine model.matrix will do this for you efficiently, with easy-to-read column names and row names:
model.matrix(~x)
# (Intercept) xb xc
#1 1 0 0
#2 1 0 0
#3 1 1 0
#4 1 1 0
#5 1 0 1
#6 1 0 1
<小时>
自己编写一个R函数来生成模型矩阵
我们可以编写一个名义例程 mm 来生成模型矩阵.虽然它比 model.matrix 效率低得多，但它可能有助于更好地理解这一概念.
We could write a nominal routine mm to generate a model matrix. Though it is much less efficient than model.matrix, it may help one digest this concept better.
mm <- function (x, contrast = TRUE) {
xlevels <- levels(x)
lst <- lapply(xlevels, function (z) match(x, z, nomatch = 0L))
if (contrast) do.call("cbind", c(list(1L), lst[-1]))
else do.call("cbind", lst)
}
例如，如果我们有一个具有 5 个级别的因子 y:
For example, if we have a factor y with 5 levels:
set.seed(1); y <- factor(sample(1:5, 10, replace=TRUE), labels = letters[1:5])
y
# [1] b b c e b e e d d a
#Levels: a b c d e
str(y)
#Factor w/ 5 levels "a","b","c","d",..: 2 2 3 5 2 5 5 4 4 1
其有/无对比处理的模型矩阵分别为:
Its model matrix with / without contrast treatment is respectively:
mm(y, TRUE)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 0 0 0
# [2,] 1 1 0 0 0
# [3,] 1 0 1 0 0
# [4,] 1 0 0 0 1
# [5,] 1 1 0 0 0
# [6,] 1 0 0 0 1
# [7,] 1 0 0 0 1
# [8,] 1 0 0 1 0
# [9,] 1 0 0 1 0
#[10,] 1 0 0 0 0
mm(y, FALSE)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 1 0 0 0
# [2,] 0 1 0 0 0
# [3,] 0 0 1 0 0
# [4,] 0 0 0 0 1
# [5,] 0 1 0 0 0
# [6,] 0 0 0 0 1
# [7,] 0 0 0 0 1
# [8,] 0 0 0 1 0
# [9,] 0 0 0 1 0
#[10,] 1 0 0 0 0
对应的model.matrix调用将分别为:
model.matrix(~ y)
model.matrix(~ y - 1)
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