我的数据集在 R 中使用方差分析函数有什么问题？答案

【问题标题】：What's wrong with my dataset to use anova function in R?我的数据集在 R 中使用方差分析函数有什么问题？
【发布时间】：2021-01-07 13:12:23
【问题描述】：

我想使用 R 中的 anova 函数比较嵌套模型。我的数据集：

structure(list(Gene = c("ID-1", "ID-1", "ID-1", "ID-1", "ID-1", 
"ID-1", "ID-1", "ID-1", "ID-1", "ID-1", "ID-1", "ID-1", "ID-1", 
"ID-1", "ID-1", "ID-4", "ID-4", "ID-4", "ID-4", "ID-4", "ID-4", 
"ID-4", "ID-4", "ID-4", "ID-4", "ID-4", "ID-4", "ID-4", "ID-4", 
"ID-4", "ID-4", "ID-5", "ID-5", "ID-5", "ID-5", "ID-5", "ID-5", 
"ID-5", "ID-5", "ID-5", "ID-5", "ID-5", "ID-5", "ID-5", "ID-5", 
"ID-5", "ID-5", "ID-5", "ID-5", "ID-6", "ID-6", "ID-6", "ID-6", 
"ID-6", "ID-6", "ID-6", "ID-6", "ID-6", "ID-6", "ID-6", "ID-6", 
"ID-6", "ID-6", "ID-6", "ID-6", "ID-6", "ID-6", "ID-7", "ID-7", 
"ID-7", "ID-7", "ID-7", "ID-7", "ID-7", "ID-7", "ID-7", "ID-7", 
"ID-7", "ID-7", "ID-7", "ID-7", "ID-7", "ID-7", "ID-7", "ID-7", 
"ID-1", "ID-1", "ID-4", "ID-4", "ID-5", "ID-5", "ID-6", "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, -0.003983171, -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, 
0.349957464, 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, 0.206584568, 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, 0.102772933, 0.247872038, 0.216357646, 
0.126169901, -0.237278842, -0.066908278, 0.105082639, NA, -0.050061512, 
-0.143484352), Time = c(20L, 20L, 20L, 40L, 40L, 20L, 40L, 40L, 
60L, 60L, 60L, 60L, 120L, 120L, 120L, 20L, 20L, 20L, 40L, 40L, 
20L, 40L, 40L, 60L, 60L, 60L, 60L, 120L, 120L, 120L, 120L, 20L, 
20L, 20L, 0L, 0L, 0L, 40L, 40L, 20L, 40L, 40L, 60L, 60L, 60L, 
120L, 120L, 120L, 120L, 20L, 20L, 20L, 0L, 0L, 0L, 40L, 40L, 
20L, 40L, 40L, 60L, 60L, 60L, 60L, 120L, 120L, 120L, 20L, 20L, 
20L, 0L, 0L, 0L, 40L, 20L, 40L, 40L, 60L, 60L, 60L, 60L, 120L, 
120L, 120L, 120L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), Condition = c("Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "Irradiated", 
"Irradiated", "Irradiated", "Irradiated", "Irradiated", "reference", 
"reference", "reference", "reference", "reference", "reference", 
"reference", "reference", "reference", "reference")), class = "data.frame", row.names = c(NA, 
-95L))

还有我的代码：

model1 <- lm(mRNA ~ Time, data=GenemRNATimeCondition)
model2 <- lm(mRNA ~ Time + Gene , data=GenemRNATimeCondition)
model3 <- lm(mRNA ~ Time + Gene + Condition, data=GenemRNATimeCondition)
anova_df <- anova(model1,model2,model3)

anova_df[,"model"] <- c("Time","Time+Gene","Time+Gene+Condition")
anova_df
anova(model1,model2,model3)

当我运行 model3 时它给出了这个错误：

Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : 
  contrasts can only be applied to factor variables with 2 or more levels

当我跑步时

anova_df <- anova(model1,model2,model3)

我收到此错误：

Error in anova.lmlist(object, ...) : 
  the models were estimated for different sample sizes

我知道对于“条件”列中的参考值，我在“时间”列中有相应的 NA 值，但我不明白为什么这是一个问题（如果这是一个问题）。希望你能帮助我理解的通俗易懂（可能也从统计的角度）。

【问题讨论】：

标签： r model nested linear-regression anova

【解决方案1】：

对于第一个错误，它告诉您缺少因子，要么是因为您没有它们，要么是因为它们因缺少值而被删除。所以对于前。如果对于特定组合，您只有缺失值，则该组合的所有行都将被删除，并且不会估计此类项，这将引发错误。

第二个错误是相关的，因为您在每个模型中对数据进行不同的分组，所以会丢弃不同数量的行，这导致模型在不同的子样本上进行估计，这也是比较模型时的问题。

基本上这是因为缺少值，您应该在继续之前处理这些值，或者采用其他方法。

【讨论】：