多个barplot和t检验

Question

我想要一个基于r中数据集中特定列中字符串的出现数量。

同时，我想运行一个t检验，并使用杆顶部的恒星绘制重要的p值。非重要性可以表示为ns。

我的尝试是：

barplot(prop.table(table(ttcluster_dataset$Phenotype)),col=clustercolor,border="black",xlab="Phenotypes",ylab="Percentage of Samples expressed",main="Sample wise Phenotype distribution",cex.names = 0.8)

数据集列是：

ttcluster_dataset$Phenotype<- 
structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 
7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L), .Label = c("Proneural (Cluster 1)", "Proneural (Cluster 2)", "Neural (Cluster 1)", "Neural (Cluster 2)", 
"Classical (Cluster 1)", "Classical (Cluster 2)", "Mesenchymal (Cluster 1)", 
"Mesenchymal (Cluster 2)"), class = "factor")

应批准所有建议。

Answer 1

t检验可能不是您想要的，因为您正在研究两个群集之间的计数和比例。您的数据实际上并不是要进行的，因此首先需要将两个变量拆分：

Pheno.splt <- strsplit(as.character(ttcluster_dataset$Phenotype), " ")
Pheno.mat <- do.call(rbind, x)[, c(1, 3)]
ttclust <- data.frame(Phenotype=Pheno.mat[, 1], Cluster=gsub(")", "", Pheno.mat[, 2]))
str(ttclust)
# 'data.frame': 171 obs. of  2 variables:
#  $ Phenotype: chr  "Proneural" "Proneural" "Proneural" "Proneural" ...
#  $ Cluster  : chr  "1" "1" "1" "1" ...

现在的表型和群集是数据框中的单独列。有多种方法可以做到这一点，但是在这里，我们将您的表型分为三个部分，通过在它们之间的空间上分开。现在ttclust是具有两个变量的数据框架。现在一个摘要表和条图：

tbl <- xtabs(~Phenotype+Cluster, ttclust)
tbl
#              Cluster
# Phenotype      1  2
#   Classical   32  6
#   Mesenchymal 44 10
#   Neural      26  0
#   Proneural   45  8
tbl.row <- prop.table(tbl, 1)
barplot(t(tbl.row), beside=TRUE)

​

prop.test(tbl)

4-sample test for equality of proportions without continuity correction

data:  tbl
X-squared = 5.2908, df = 3, p-value = 0.1517
alternative hypothesis: two.sided
sample estimates:
   prop 1    prop 2    prop 3    prop 4 
0.8421053 0.8148148 1.0000000 0.8490566 

​在每种情况下，与群集2的差异都显着差异：

for(i in 1:4) print(prop.test(t(tbl[i, ])))

# First test
# 
#   1-sample proportions test with continuity correction
# 
# data:  t(tbl[i, ]), null probability 0.5
# X-squared = 16.447, df = 1, p-value = 5.002e-05
# alternative hypothesis: true p is not equal to 0.5
# 95 percent confidence interval:
#  0.6807208 0.9341311
# sample estimates:
#         p 
# 0.8421053 
    . . . .