方差分析:自由度几乎都等于 1 [英] ANOVA: Degrees of freedom almost all equal 1

问题描述

我有一个这样开头的数据集:

I have a data set that begins like this:

> d.weight
    R   N   P  C D.weight
1   1   0   0 GO     45.3
2   2   0   0 GO     34.0
3   3   0   0 GO     19.1
4   4   0   0 GO     26.6
5   5   0   0 GO     23.5
6   1  45   0 GO     22.1
7   2  45   0 GO     15.5
8   3  45   0 GO     23.4
9   4  45   0 GO     15.8
10  5  45   0 GO     42.9
...

等等.

• R 是复制的，有 5 个(1-5).
• N 是氮含量，也有 5 个(0、45、90、180、360).
• P 是磷含量，也有 5 个(0, 35, 70, 140, 280).
• C 是植物组合，有 4 个(GO、GB、LO、LB).
• D.weight 是以克为单位的干重.

• R is replicate and there are 5 of them (1-5).
• N is nitrogen level, and there are 5 as well (0, 45, 90, 180, 360).
• P is phosphorus level, and there are 5 as well (0, 35, 70, 140, 280).
• C is plant combination, and there are 4 (GO, GB, LO, LB).
• D.weight is dry weight in grams.

然而，当我做方差分析时，我得到了错误的自由度.我通常在完整数据集的子集上运行我的方差分析，但让我们做一个分析，否则我实际上不会做，这样你就可以看到几乎所有的 Df(自由度)都是错误的.

However, when I do an ANOVA I get the wrong degrees of freedom. I usually run my ANOVAs on subsets of that full set of data, but let's just do an analysis I wouldn't actually do otherwise, just so you can see that almost all of the Df (degrees of freedom) are wrong.

> example.aov=aov(D.weight ~ R+N+P+C, data=d.weight)
> summary(example.aov)
         Df Sum Sq Mean Sq F value  Pr(>F)    
R             1   1158    1158   9.484 0.00226 ** 
N             1    202     202   1.657 0.19900    
P             1  11040   11040  90.408 < 2e-16 ***
C             3  41032   13677 112.010 < 2e-16 ***
Residuals   313  38220     122

所以，基本上，唯一正确的是 C 因子.是不是因为它有字母而不是数字?

So, basically, the only one that's right is the C factor. Is it because it has letters instead of numbers?

我发现如果我为每个术语编写 interaction() ，我会得到正确的 Df，但我不知道这是否是整体上正确的做法.例如:

I found somewhere that if I write interaction() with each term, I get the right Df, but I don't know if that's the right thing to do overall. For example:

> example.aov2=aov(D.weight ~ interaction(R)+interaction(N)+interaction(P)+interaction(C), data=d.weight)
> summary(example.aov2)
                Df Sum Sq Mean Sq F value   Pr(>F)    
interaction(R)   4   7423    1856  19.544 2.51e-14 ***
interaction(N)   4    543     136   1.429    0.224    
interaction(P)   4  13788    3447  36.301  < 2e-16 ***
interaction(C)   3  41032   13677 144.042  < 2e-16 ***
Residuals      304  28866      95

我用 C 因子试了一下，只是想看看它是否搞砸了什么:

I tried it with the C factor only to see if it messed up anything:

> example.aov3=aov(D.weight ~ C, data=d.weight)
> summary(example.aov3)
             Df Sum Sq Mean Sq F value Pr(>F)    
C             3  41032   13677   85.38 <2e-16 ***
Residuals   316  50620     160                   
> 
> example.aov4=aov(D.weight ~ interaction(C), data=d.weight)
> summary(example.aov4)
                Df Sum Sq Mean Sq F value Pr(>F)    
interaction(C)   3  41032   13677   85.38 <2e-16 ***
Residuals      316  50620     160

它看起来一样.我应该在任何地方添加 interaction() 吗?

And it looks the same. Should I be adding interaction() everywhere?

推荐答案

R 通过检查变量是否为 numeric 或 factor 变量.最简单的是，您可以通过

R determines whether it should treat variables as categorical (ANOVA-type analysis) or continuous (regression-type analysis) by checking whether they are numeric or factor variables. Most simply, you can convert your predictor (independent) variables to factors via

facs <- c("R","N","P")
d.weight[facs] <- lapply(d.weight[facs],factor) 

如果你想创建辅助变量而不是覆盖你可以做类似的事情

If you want to create auxiliary variables instead of overwriting you could do something like

for (varname in facs) {
   d.weight[[paste0("f",varname)]] <- factor(d.weight[[varname]])
}

可能有一种更紧凑的方法来做到这一点，但应该可以...

There might be a more compact way to do it but that should serve ...

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