编辑:如何为PostHoc Tukey测试和多重比较编写循环

Question

这是我的大数据矩阵示例&每一列都用多个信息命名，并用下划线分隔。

structure(list(Gene = c("AGI4120.1_UBQ", "AGI570.1_Acin"), WT_Tissue_0T_1 = c(0.886461437, 1.093164915), WT_Tissue_0T_2 = c(1.075140682, 1.229862834), WT_Tissue_0T_3 = c(0.632903012, 1.094003128), WT_Tissue_1T_1 = c(0.883151274, 1.26322126), WT_Tissue_1T_2 = c(1.005627276, 0.962729188), WT_Tissue_1T_3 = c(0.87123469, 0.968078993), WT_Tissue_3T_1 = c(0.723601456, 0.633890322), WT_Tissue_3T_2 = c(0.392585237, 0.534819363), WT_Tissue_3T_3 = c(0.640185369, 1.021934772), WT_Tissue_5T_1 = c(0.720291294, 0.589244505), WT_Tissue_5T_2 = c(0.362131744, 0.475251717), WT_Tissue_5T_3 = c(0.549486925, 0.618177919), mut1_Tissue_0T_1 = c(1.464415756, 1.130533457), mut1_Tissue_0T_2 = c(1.01489573, 1.114915728), mut1_Tissue_0T_3 = c(1.171797418, 1.399956009), mut1_Tissue_1T_1 = c(0.927507448, 1.231911575), mut1_Tissue_1T_2 = c(1.089705396, 1.256782289 ), mut1_Tissue_1T_3 = c(0.993048659, 0.999044465), mut1_Tissue_3T_1 = c(1.000993049, 1.103486794), mut1_Tissue_3T_2 = c(1.062562066, 0.883617224 ), mut1_Tissue_3T_3 = c(1.037404833, 0.851875438), mut1_Tissue_5T_1 = c(0.730883813, 0.437440083), mut1_Tissue_5T_2 = c(0.480635551, 0.298762126 ), mut1_Tissue_5T_3 = c(0.85468388, 0.614923997)), row.names = c(NA, -2L), class = c("tbl_df", "tbl", "data.frame"), spec = structure(list( cols = list(Gene = structure(list(), class = c("collector_character", "collector")), WT_Tissue_0T_1 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_0T_2 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_0T_3 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_1T_1 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_1T_2 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_1T_3 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_3T_1 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_3T_2 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_3T_3 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_5T_1 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_5T_2 = structure(list(), class = c("collector_double", "collector")), WT_Tissue_5T_3 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_0T_1 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_0T_2 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_0T_3 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_1T_1 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_1T_2 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_1T_3 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_3T_1 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_3T_2 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_3T_3 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_5T_1 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_5T_2 = structure(list(), class = c("collector_double", "collector")), mut1_Tissue_5T_3 = structure(list(), class = c("collector_double", "collector"))), default = structure(list(), class = c("collector_guess", "collector"))), class = "col_spec"))

我想遵循Tukey测试，并用多个比较字母绘制每个基因的条形图(响应与时间；由两种基因型填充)。

语法

df1 <- df %>% 
         gather(var, response, WT_Tissue_0T_1:mut1_Tissue_5T_3) %>% 
         separate(var, c("Genotype", "Tissue", "Time"), sep = "_") %>% 
         arrange(desc(Gene))
df2 <- df1 %>% 
         group_by(`Gene`,Genotype,Tissue,Time) %>% 
         mutate(Response=mean(response),n=n(),se=sd(response)/sqrt(n))

双向方差分析

fit1 <- aov(Response ~ Genotype*Time, df2) 

此后，我想进行图基测试(多重比较)，例如基因"AGI4120.1_UBQ"，绘制响应与时间的关系图，并查看每个基因型(WT和mut1)在每个时间点(0T、1T、3T和5T)的表现。答案是否有显著不同，并在图中用字母表示。

如下所示，lsmeans语法将所有基因组合为一个并给出输出，我如何才能使其分别为每个基因循环(即"AGI4120.1_UBQ"，"AGI570.1_Acin")和获取字母以显示统计上不同的组(也称为“紧凑字母显示”)

 lsmeans(fit1, pairwise ~ Genotype | Time) 

我的最终目标是在下面的图表中绘制每个基因，并表示有意义的字母。

df2$genotype <- factor(df2$genotype, levels = c("WT","mut1")) colours <- c('#336600','#ffcc00')library(ggplot2)ggplot(df2,aes( x=Time, y=Response, fill=Genotype))+geom_bar(stat='identity', position='dodge')+scale_fill_manual(values=colours)+geom_errorbar(aes(ymin=average_measure-se, ymax=average_measure+se)+facet_wrap(~`Gene`)+labs(x='Time', y='Response')

Expecting Graph for denoting compact letter display

如果可能的话，我将非常感谢您的帮助。

Answer 1

你的代码有很多问题。但我会发布一些建议，作为你入门的答案--希望它们能有所帮助。

1. lsmeans()**:**

此函数需要一个拟合的模型(如来自lm()的模型)或ref.grid对象。但是您传递给它的是一个数据框，没有任何计算最小二乘均值所需的回归属性。(想想看:当您要求将Genes作为自变量进行成对比较时，lsmeans()如何知道因变量应该是什么？)有关更多详细信息，请查看Using lsmeans documentation。

但为了演示，我将使用lm()保持简单。将拟合的回归对象传递给lsmeans()可以按预期工作：

fit <- lm(Response ~ Gene + Genotype + Time, data=df2)

lsmeans(fit, pairwise ~ Gene)[[2]] 

# Output
contrast                        estimate        SE df t.ratio p.value
AGI4120.1_UBQ - AGI570.1_Acin -0.0515123 0.0299492 42   -1.72  0.0928

Results are averaged over the levels of: Genotype, Time 

2. ggplot()**:**

您没有在所提供的代码中定义colours或average_measure；调用这些未声明的变量将导致失败。

从结构上讲，我建议您使用df1并允许ggplot进行分组，而不是group_by到df2中。然后使用stat="summary"和fun.y="mean"完成您在df2中进行的summarise()计算。这样做还允许您对错误条使用mean_se函数。如下所示：

ggplot(df1,aes( x=Time, y=response, fill=Genotype))+ 
  geom_bar(stat='summary', fun.y='mean', position=position_dodge(0.9))+
    stat_summary(fun.data = mean_se, geom = "errorbar", 
                 color="gray60", width=.1, position=position_dodge(0.9)) +
  scale_fill_manual(values=c("steelblue","orange"))+
  facet_wrap(~`Gene`)+ 
  labs(x='Time', y='Response')

​

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separate(var, c("Genotype", "Tissue", "Time", "Index"), sep = "_")