1.利用箱線圖比較兩類樣本的某個(gè)細(xì)胞比例差異

比較直觀，但是缺點(diǎn)在于如果單細(xì)胞樣本個(gè)數(shù)過少且異質(zhì)性大，導(dǎo)致很難有統(tǒng)計(jì)學(xué)顯著意義

library(ggpubr)
data <- data.frame(Cancer = c(0.5, 0.6, 0.8, 0.2),
                   Normal = c(0.2, 0.3, 0.7, 0.4),
                   Celltype = "T cells")
mydata <- reshape2::melt(data,id.vars=c("Celltype"))
ggboxplot(mydata, x = "Celltype", y = "value",
          color = "variable", palette = "jama",
          add = "jitter") + stat_compare_means(aes(color=variable))

2.R o/e 比值

好多文章都有用這個(gè)，我的理解是四格表卡方檢驗(yàn)計(jì)算出來的觀測(cè)除以期望

例如上述數(shù)據(jù)，一開始有三類細(xì)胞，分別在癌和正常的個(gè)數(shù)如表所示，那么計(jì)算R
o/e 的時(shí)候就要構(gòu)建四格表，以T細(xì)胞為例

##計(jì)算卡方值以及期望和觀測(cè)值
x <- chisq.test(matrix(c(80,300,200,220),ncol = 2))
Roe <- x$observed / x$expected
Roe

##           [,1]      [,2]
## [1,] 0.6015038 1.3605442
## [2,] 1.2145749 0.8058608

#p值
paste0("P-value = ",x$p.value)

## [1] "P-value = 6.55065992002061e-15"

可以看出Normal組Roe>1,說明T細(xì)胞比例在Normal組相對(duì)Cancer組會(huì)更多一些

3.OR指數(shù)

這個(gè)其實(shí)跟Roe是等同效果，只不過它是利用fisher檢驗(yàn)來計(jì)算OR值

###T細(xì)胞在Cancer組OR值
fisher.test(matrix(c(80,300,200,220),ncol = 2))

## 
##  Fisher's Exact Test for Count Data
## 
## data:  matrix(c(80, 300, 200, 220), ncol = 2)
## p-value = 2.184e-15
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##  0.2117305 0.4052156
## sample estimates:
## odds ratio 
##  0.2938061

###T細(xì)胞在Normal組OR值
fisher.test(matrix(c(200,220,80,300),ncol = 2))

## 
##  Fisher's Exact Test for Count Data
## 
## data:  matrix(c(200, 220, 80, 300), ncol = 2)
## p-value = 2.184e-15
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##  2.467822 4.722986
## sample estimates:
## odds ratio 
##   3.403605

終： 寫這個(gè)單純記錄一下過程，避免后面自己忘記了，僅為拙見