From two dataframes with single expression values (rows) per sample (cols) of different groups, I want to calculate the mean and median per group. My solution seems a bit verbose and I wonder if there is a more elegant solution.
# expression values
genes <- paste("gene",1:1000,sep="")
x <- list(
A = sample(genes,300),
B = sample(genes,525),
C = sample(genes,440),
D = sample(genes,350)
)
# expression dataframe
crete_exp_df <- function(gene_nr, sample_nr){
df <- replicate(sample_nr, rnorm(gene_nr))
rownames(df) <- paste("Gene", c(1:nrow(df)))
colnames(df) <- paste("Sample", c(1:ncol(df)))
return(df)
}
exp1 <- crete_exp_df(50, 20)
exp2 <- crete_exp_df(50, 20)
# sample annotation
san <- data.frame(
id = colnames(exp1),
group = sample(1:4, 20, replace = TRUE))
# get ids of samples per group
ids_1 <- san %>% filter(group == 1) %>% pull(id)
ids_2 <- san %>% filter(group == 2) %>% pull(id)
ids_3 <- san %>% filter(group == 3) %>% pull(id)
ids_4 <- san %>% filter(group == 4) %>% pull(id)
id_list <- list(group1 = ids_1, group2 = ids_2, group3 = ids_3, group4 = ids_4)
# fct means df1
get_means_exp1 <- function(id){
apply(exp1[, id], 1, mean, na.rm = T)
}
# fct means df2
get_means_exp2 <- function(id){
apply(exp2[, id], 1, mean, na.rm = T)
}
# lapply on df1
list_means_exp1 <- lapply(id_list, get_means_exp1)
means_exp1 <- as.data.frame(list_means_exp1)
# lapply on df2
list_means_exp2 <- lapply(id_list, get_means_exp2)
means_exp2 <- as.data.frame(list_means_exp2)
I suppose this can be solved much more elegant. Specifically, how to get the ids per group and write a function that works for both df. Looking forwards to learning from your solutions, thanks a lot!
So, I worked with the data generation process you provided and came up with a more simple solution. I changed exp1 into a dataframe, brought it in tidy format (pivot_longer()), added the groups from the san dataframe and finally applied the simple dplyr syntax to summarise your data.
library(tidyverse)
as.data.frame(exp1) %>%
rownames_to_column("Gene") %>%
pivot_longer(cols= 2:21, names_to = "id", values_to = "Values") %>%
left_join(., san) %>%
group_by(group) %>%
summarise(mean= mean(Values),
median= median(Values))
#> Joining with `by = join_by(id)`
#> # A tibble: 4 × 3
#> group mean median
#> <int> <dbl> <dbl>
#> 1 1 0.0803 0.0568
#> 2 2 -0.0383 -0.0387
#> 3 3 -0.00929 0.0356
#> 4 4 -0.0840 -0.0306
Considering your comment, simply also group by gene and that gets you the expected output.
library(tidyverse)
as.data.frame(exp1) %>%
rownames_to_column("Gene") %>%
pivot_longer(cols= 2:21, names_to = "id", values_to = "Values") %>%
left_join(., san) %>%
group_by(group, Gene) %>%
summarise(mean= mean(Values),
median= median(Values))
#> Joining with `by = join_by(id)`
#> `summarise()` has grouped output by 'group'. You can override using the
#> `.groups` argument.
#> # A tibble: 200 × 4
#> # Groups: group [4]
#> group Gene mean median
#> <int> <chr> <dbl> <dbl>
#> 1 1 Gene 1 -0.0642 -0.122
#> 2 1 Gene 10 0.0151 0.563
#> 3 1 Gene 11 -0.0585 -0.0367
#> 4 1 Gene 12 -0.978 -0.917
#> 5 1 Gene 13 -1.01 -1.37
#> 6 1 Gene 14 0.160 -0.394
#> 7 1 Gene 15 -0.295 -0.689
#> 8 1 Gene 16 0.774 0.729
#> 9 1 Gene 17 -0.356 -0.336
#> 10 1 Gene 18 -0.741 -0.103
#> # … with 190 more rows
Created on 2023-04-13 with reprex v2.0.2