Question

d = data.frame(
    Temperature = c(rep("Cool", 6), rep("Warm", 6)),
    Bact = c(rep("Bact 1", 2), rep("Bact 2", 2), rep("Bact 3", 2), rep("Bact 1", 2), rep("Bact 2", 2), rep("Bact 3", 2)),
    Time = c(15.23,14.32,14.77,15.12,14.05,15.48,14.13,16.13,16.44,14.82,17.96,16.65)
)

我为双向ANOVA自行创建了一个小数据框。我想通过

执行双向ANOVA模型

summary(aov(Time~Bact*Temperature, data=d))

时间是因变量，而Bact和温度是两个分类独立变量。

不是以ANOVA的方式进行，我想学习并证明ANOVA也可以用线性回归模型完成。我想将我的数据转换为虚拟变量并对其执行线性回归。我希望我能恢复相同的结果。虚拟变量还将包括Bact和温度之间的相互作用。

问题在于，我不知道如何将我的数据帧转换为虚拟变量，以便可以在lm（）函数中使用它。

Answer 1

我也和你做同样的事情。我希望能够控制住，所以每当我有空的时候，我都会用以下方法设计假人：

d = data.frame(
  Temperature = c(rep("Cool", 6), rep("Warm", 6)),
  Bact = c(rep("Bact 1", 2), rep("Bact 2", 2), rep("Bact 3", 2), rep("Bact 1", 2), rep("Bact 2", 2), rep("Bact 3", 2)),
  Time = c(15.23,14.32,14.77,15.12,14.05,15.48,14.13,16.13,16.44,14.82,17.96,16.65)
)

是：

> d
   Temperature   Bact  Time
1         Cool Bact 1 15.23
2         Cool Bact 1 14.32
3         Cool Bact 2 14.77
4         Cool Bact 2 15.12
5         Cool Bact 3 14.05
6         Cool Bact 3 15.48
7         Warm Bact 1 14.13
8         Warm Bact 1 16.13
9         Warm Bact 2 16.44
10        Warm Bact 2 14.82
11        Warm Bact 3 17.96
12        Warm Bact 3 16.65

因此，您只需要对因子（温度，温度）进行虚拟化，以便以下过程有效：

xfactors <- Filter(is.factor,d) #filter only the factors to dummify
b <- data.frame(matrix(NA,nrow=nrow(xfactors),ncol=1)) #make empty data.frame to initiate b
for ( i in 1:ncol(xfactors)) { #start loop
  a <- data.frame(model.matrix(~xfactors[,i])) #make dummies here
  b <- cbind(b, a[-1]) #remove intercept and combine dummies
}
b <- data.frame(b[-1]) #make a data.frame
#the reference dummy gets excluded automatically by model.matrix
colnames(b) <- c('warm' , 'bact2' , 'bact3') #you will probably want to change the names to sth smaller

> b
   warm bact2 bact3
1     0     0     0
2     0     0     0
3     0     1     0
4     0     1     0
5     0     0     1
6     0     0     1
7     1     0     0
8     1     0     0
9     1     1     0
10    1     1     0
11    1     0     1
12    1     0     1

然后运行模型：

new_data <- cbind(b, Time=d$Time) #add time to the data
mymod <- lm(Time ~ warm*bact2+warm*bact3, data=new_data) #compute lm with interactions
#you shouldn't compute the interactions between dummy variables because they come from the same variable

输出：

> summary(mymod)

Call:
lm(formula = Time ~ warm * bact2 + warm * bact3, data = new_data)

Residuals:
   Min     1Q Median     3Q    Max 
 -1.00  -0.67   0.00   0.67   1.00 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  14.7750     0.6873  21.498 6.61e-07 ***
warm          0.3550     0.9719   0.365    0.727    
bact2         0.1700     0.9719   0.175    0.867    
bact3        -0.0100     0.9719  -0.010    0.992    
warm:bact2    0.3300     1.3745   0.240    0.818    
warm:bact3    2.1850     1.3745   1.590    0.163    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9719 on 6 degrees of freedom
Multiple R-squared:  0.6264,    Adjusted R-squared:  0.3151 
F-statistic: 2.012 on 5 and 6 DF,  p-value: 0.2097

Answer 2

lm()将为您创建虚拟变量。无需自己创建它们：

m <- lm(Time ~ Bact*Temperature, data = d)
anova(m)

修改

如果你想在lm()的引导下进行对等，你可以看到设计矩阵model.matrix(m)

创建虚拟变量以进行双向ANOVA

2 个答案: