我有一个表达数据的数据框,其中基因是行,列是样本。我还有一个数据框,其中包含表达式数据框中每个样本的元数据。实际上,我的 expr 数据框有 30,000 多行和 100 多列。不过,下面是一个数据较小的示例。
expr <- data.frame(sample1 = c(1,2,2,0,0),
sample2 = c(5,2,4,4,0),
sample3 = c(1,2,1,0,1),
sample4 = c(6,5,6,6,7),
sample5 = c(0,0,0,1,1))
rownames(expr) <- paste0("gene",1:5)
meta <- data.frame(sample = paste0("sample",1:5),
treatment = c("control","control",
"treatment1",
"treatment2", "treatment2"))
我想找到每次治疗中每个基因的平均值。从我看到的 split() 或 group_by() 示例中,人们根据 data.frame 中已存在的列进行分组。但是,我有一个单独的数据框(元),用于对另一个数据框(expr)中的列进行分组。
我希望我的输出是一个数据框,其中基因作为行,治疗作为列,值作为平均值。
# control treatment1 treatment2
# gene1 mean mean mean
# gene2 mean mean mean
类似这样的事情。目前尚不完全清楚您要在最后一步中按什么进行分组,但您可以轻松调整。
library(dplyr)
library(tidyr)
expr |>
mutate(gene = row.names(expr)) |>
pivot_longer(-gene, names_to = "sample") |>
left_join(meta, by = "sample") |>
summarize(mean = mean(value), .by = c(gene, treatment)) |>
pivot_wider(names_from = treatment, values_from = mean)
# # A tibble: 5 × 4
# gene control treatment1 treatment2
# <chr> <dbl> <dbl> <dbl>
# 1 gene1 3 1 3
# 2 gene2 2 2 2.5
# 3 gene3 3 1 3
# 4 gene4 2 0 3.5
# 5 gene5 0 1 4
基础 R 中的一种方法适用于给定的特定玩具数据示例:
colnames(expr) = paste0(colnames(expr), "_",
meta$treatment[match(colnames(expr), meta$sample)])
vapply(unique(meta$treatment),
\(i) rowMeans(expr[grepl(i, colnames(expr))]), numeric(nrow(expr)))
#> control treatment1 treatment2
#> gene1 3 1 3.0
#> gene2 2 2 2.5
#> gene3 3 1 3.0
#> gene4 2 0 3.5
#> gene5 0 1 4.0
数据
expr <- data.frame(sample1 = c(1,2,2,0,0),
sample2 = c(5,2,4,4,0),
sample3 = c(1,2,1,0,1),
sample4 = c(6,5,6,6,7),
sample5 = c(0,0,0,1,1))
rownames(expr) <- paste0("gene",1:5)
meta <- data.frame(sample = paste0("sample",1:5),
treatment = c("control","control",
"treatment1",
"treatment2", "treatment2"))
基本 R 方法:
expr|>
split.default(with(meta, treatment[match(names(expr), sample)]))|>
lapply(rowMeans)|>
structure(dim=3)|>
array2DF()
Var1 gene1 gene2 gene3 gene4 gene5
1 control 3 2.0 3 2.0 0
2 treatment1 1 2.0 1 0.0 1
3 treatment2 3 2.5 3 3.5 4
这是一个
data.table
方法,与 @Gregor Thomas 提供的逻辑相同:
library(data.table)
expr_dt <- setDT(expr)
expr_dt[, gene := rownames(expr)]
meta_dt <- setDT(meta)
melt(expr_dt, id.vars = "gene", variable.name = "sample", value.name = "expression")[
meta_dt, on = .(sample)][
, .(mean = mean(expression)), by = .(gene, treatment)][
, dcast(.SD, gene ~ treatment, value.var = "mean")]
gene control treatment1 treatment2
1: 1 3 1 3.0
2: 2 2 2 2.5
3: 3 3 1 3.0
4: 4 2 0 3.5
5: 5 0 1 4.0