问题描述
如果我考虑相同的线性模型，我不明白为什么aov和anova会产生不同的结果。首先，我要使用 aov 函数：
res.aov <- aov(mRNA ~ Time,data= ID1.4.5.6.7)
summary(res.aov)
，输出为：
Df Sum Sq Mean Sq F value Pr(>F)
Time 1 0.263 0.26332 6.228 0.0146 *
Residuals 83 3.509 0.04228
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
然后，我使用 anova 函数执行此操作：
model0 <- lm(mRNA ~ 1,data=ID1.4.5.6.7)
model1 <- lm(mRNA ~ Time,data=ID1.4.5.6.7)
model3 <- lm(mRNA ~ Time + Gene,data=ID1.4.5.6.7)
anova_df <- anova(model0,model1,model3)
anova_df[,"model"] <- c("Intercept","Time","Time+Gene")
anova_df
anova(model0,model3)
，输出如下：
Analysis of Variance Table
Model 1: mRNA ~ 1
Model 2: mRNA ~ Time
Model 3: mRNA ~ Time + Gene
Res.Df RSS Df Sum of Sq F Pr(>F) model
1 84 3.7727 1
2 83 3.5094 1 0.26332 7.2152 0.0088083 2
3 79 2.8831 4 0.62629 4.2902 0.0034166 3
所以首先我不明白为什么对于模型mRNA〜Time，我们同时获得不同的 F 值和不同的 p值？ （aov函数分别给出6.228和0.0146，而方差分析分别为7.2152和0.0088083）。
第二，如果我编写这段代码：
anova_df[,"Time+Gene")
为什么不在输出数据框的列模型下打印名称“ Intercept”，“ Time”和“ Time + Gene”？
我的数据集：
structure(list(Gene = c("ID-1","ID-1","ID-4","ID-5","ID-6","ID-7","ID-7"
),mRNA = c(-0.181385669,-0.059647494,0.104476117,-0.052190978,-0.040484945,0.194226742,-0.501601326,0.102342605,-0.127143845,-0.008523742,-0.102946211,-0.042894028,0.002922923,-0.134394347,-0.214204393,-0.138122686,0.203242361,0.097935502,0.147068146,-0.089430917,0.331565412,-0.034572422,-0.129896329,0.324191,0.470108479,-0.027268223,0.232304713,0.090348708,0.070848402,0.181540708,-0.502255367,-0.267631441,-0.368647839,-0.040910404,-0.003983171,-0.14980589,-0.119449612,-0.309154214,-0.487589361,0.272803506,-0.421733575,-0.467108567,0.024868338,-0.156025729,-0.044680175,-0.206716896,-0.272014193,-0.230499883,-0.238597397,-0.118130949,0.349957464,0.172048587,-0.186226994,0.16113822,-0.293029136,-0.111636253,-0.044189887,0.081555274,-0.048106079,-0.05853566,0.010407814,-0.066981809,-0.09828484,-0.315190986,-0.005102456,0.221556197,0.206584568,0.102649006,-0.011777384,-0.36963487,-0.054853074,-0.230240699,-0.210508323,-0.208889919,-0.050763372,0.023073782,-0.095118984,-0.091076071,-0.330257395),Time = c(2,2,3,4,5,1,5),predicted_mRNA = c(-0.00551000342030954,-0.00551000342030954,-0.0302695238715682,-0.0550290443228268,-0.129307605676603,0.0192495170309491,-0.129307605676603)),row.names = c(NA,-85L
),class = "data.frame")
解决方法
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