问题：如果对样本之间的差异感兴趣，则需要对表达矩阵（行为基因，列为样本）按行进行标准化。

解答：对的，按行标准化，尝试3种方法，第1和3结果一致。scale函数是按列归一化。

mat <- read.table("expression_data.txt", header = TRUE, row.names = 1)
 #行基因，列为样本

## 用标准化的自定义函数
scale_rows<- function(x){
  m = apply(x, 1, mean, na.rm = T)
  s = apply(x, 1, sd, na.rm = T)
  return((x - m)/s)
}


## 测试1:用标准化的自定义函数
a <- scale_rows(mat)

## 测试2:用scale函数
b <- scale(mat,scale = TRUE,center = TRUE)

bk <- c(seq(-4,-0.1,by=0.01),seq(0,4,by=0.01))
p1 <- pheatmap(a,scale = 'none',
               color = c(colorRampPalette(colors = c("blue","white"))(length(bk)/2),colorRampPalette(colors = c("white","red"))(length(bk)/2)),
               legend_breaks=seq(-4,4,2),
               breaks=bk)

p2 <- pheatmap(b,scale = 'none',
               color = c(colorRampPalette(colors = c("blue","white"))(length(bk)/2),colorRampPalette(colors = c("white","red"))(length(bk)/2)),
               legend_breaks=seq(-4,4,2),
               breaks=bk)
## 测试3:直接在pheatmap中设置scale=‘row'
p3 <- pheatmap(mat,scale = 'row',
               color = c(colorRampPalette(colors = c("blue","white"))(length(bk)/2),colorRampPalette(colors = c("white","red"))(length(bk)/2)),
               legend_breaks=seq(-4,4,2),
               breaks=bk)