通过对列进行分组将数据帧拆分为多个数据帧

问题描述 投票:0回答:4

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
r dplyr bioinformatics rna-seq
4个回答
2
投票

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  

2
投票

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"))

1
投票

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

0
投票

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
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