CODEWITHSUNDEEP
rdataframedplyrbioinformatics
Kenn
Kenn

Reputation: 83

Split dataframe into multiple dataframes by grouping columns

I have a dataframe of expression data where gene are rows and columns are samples. I also have a dataframe containing metadata for each sample in the expression dataframe. In reality my expr dataframe has 30,000+ rows and 100+ columns. However, below is an example with smaller data.

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"))

I want to find the mean for each gene per treatment. From the examples I've seen with split() or group_by() people group based on a column already present in the data.frame. However, I have a separate dataframe (meta) that classifies the grouping for the columns in another dataframe (expr).

I would like my output to be a dataframe with genes as rows, treatment as columns, and values as the mean.

#        control   treatment1   treatment2
#  gene1  mean        mean         mean
#  gene2  mean        mean         mean

Upvotes: 7

Views: 211

Answers (5)

M--
M--

Reputation: 28825

Here's another approach in tidyverse, replacing sample numbers with treatments from meta by creating a named vector instead of left_join(), and also using values_fn within pivot_wider() instead of summarise():

library(dplyr)
library(tidyr)

expr %>% 
  tibble::rownames_to_column("gene") %>% 
  pivot_longer(-gene) %>% 
  mutate(name = split(meta$treatment, meta$sample)[name]) %>%
  pivot_wider(values_fn = 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

Created on 2023-12-21 with reprex v2.0.2

Upvotes: 2

Onyambu
Onyambu

Reputation: 79228

A base R approach:

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

Upvotes: 4

TarJae
TarJae

Reputation: 78927

Here is a data.table approach with the same logic provided by @Gregor Thomas:

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

Upvotes: 2

Friede
Friede

Reputation: 7410

An approach in base R which works for the particular toy data example given:

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

Data

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"))

Upvotes: 6

Gregor Thomas
Gregor Thomas

Reputation: 145775

Something like this. It's not entirely clear what you want to group by in the last step, but you can adjust that easily.

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

Upvotes: 2

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