编辑：如何为PostHoc Tukey测试和多重比较编写Loop

Question

这是我的大数据矩阵示例＆amp;并且每列都以多个信息命名，并以下划线分隔。

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"))

我想跟踪Tukey测试并绘制每个基因的条形图（响应与时间的关系;由两种基因型填充）和多个比较字母。

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ANOVA

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

此后，我想进行Tukey测试（多重比较），例如基因“AGI4120.1_UBQ”，绘制响应与时间的关系，并查看每个基因型（WT＆amp; mut1）在每个时间点（0T，1T，3T和5T）的表现如何？如果响应明显不同或不同，并用图中的字母表示。

如下所示，lsmeans语法将所有Genes组合成一个并给出输出，如何让它分别为每个基因循环（即“AGI4120.1_UBQ”，“AGI570.1_Acin”）并获取字母以显示统计上不同的组（又名“紧凑字母显示”）

 lsmeans(fit1, pairwise ~ Genotype | Time) 

我的最终目标是在下图中绘制每个基因，并表示重要字母。

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，我将非常感谢您的帮助。

Answer 1

您的代码存在许多问题。我甚至说它对StackOverflow来说并不是一个合适的帖子，因为这里的问题是多种多样的，并且不能超出这个特定的错误和语法问题。但我会发布一些建议作为让你开始的答案 - 希望他们帮助。

<强> 1。 lsmeans()
此函数需要拟合模型（如lm()）或ref.grid对象。但是你传递的是一个数据框，没有任何计算最小二乘法所需的回归属性。 （想一想：lsmeans()当你要求与Genes作为自变量进行成对比较时，{1}}如何知道因变量应该是什么？）查看Using lsmeans documentation了解更多详情。

根据您的数据，我想您可能希望运行多级回归，使用lme4包，Gene和Genotype以及{{{} 1}}作为嵌套分组级别。

但是为了演示，我会用Time保持简单。将拟合的回归对象传递给lm()按预期工作：

lsmeans()

<强> 2。 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
您尚未在自己提供的代码中定义ggplot()或colours;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average_measure并允许df1进行分组，而不是ggplot进入group_by。然后使用df2和stat="summary"完成您在fun.y="mean"中执行的summarise()次计算。这样做还允许您使用df2函数作为误差线。像这样：

mean_se

最后请注意，在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') 中使用separate()会发出警告，但不会出现错误，因为在df1上拆分会产生额外的值。您可以通过添加一个级别来捕获最终值（这似乎是一个时间索引）来避免这种情况（如果它引起混淆）：

sep="_"