在R中按列统计层级计数及矩阵等位基因列频的实现方法

Hey there! Let's break down your two R questions step by step—first the general case of counting levels per column, then the specific allele matrix scenario.

There are a few solid approaches depending on whether you prefer base R or the tidyverse ecosystem. Let's start with a sample dataset to demo:

# Create a sample data frame with categorical columns
set.seed(123)
sample_df <- data.frame(
  Group = sample(c("Control", "Treatment"), 15, replace = TRUE),
  Status = sample(c("Positive", "Negative", "Neutral"), 15, replace = TRUE),
  Category = sample(c("X", "Y"), 15, replace = TRUE)
)

Option 1: Base R with apply() + table() #

This is quick and doesn't require any extra packages:

# Count levels for each column (returns a list of tables)
column_level_counts <- apply(sample_df, 2, table)

# View the result
column_level_counts

The output is a list where each element corresponds to a column, showing the count of each level present.

Option 2: Tidyverse (dplyr + tidyr) for a structured data frame #

If you want the results in a clean, tabular format (easier to work with for further analysis), use this workflow:

library(tidyverse)

sample_df %>%
  # Convert wide data to long format
  pivot_longer(cols = everything(), names_to = "Column", values_to = "Level") %>%
  # Count occurrences of each level per column
  count(Column, Level) %>%
  # Convert back to wide format, filling missing levels with 0
  pivot_wider(names_from = Level, values_from = n, values_fill = 0)

This gives you a data frame where each row is an original column, and each column is a level with its count.

Option 3: Janitor package for simplified tabulation #

The janitor package has a handy tabyl() function that makes this even easier:

library(janitor)
library(purrr)

# Generate a tabyl for each column and combine into a single data frame
map(sample_df, tabyl) %>%
  map_dfr(~ as.data.frame(.x), .id = "Column")

This includes both counts and percentages, which can be useful for quick summaries.

For your specific case with an A/T/C/G allele matrix, we can adapt the above methods to ensure we always include all four alleles (even if one isn't present in a column). Let's start with a sample allele matrix:

# Simulate an allele matrix (rows = samples, columns = loci)
set.seed(456)
allele_matrix <- matrix(
  sample(c("A", "T", "C", "G"), 50, replace = TRUE),
  nrow = 10, ncol = 5,
  dimnames = list(paste0("Sample", 1:10), paste0("Locus", 1:5))
)

Method 1: Tidyverse Workflow (Recommended for Readability) #

library(tidyverse)

allele_counts <- allele_matrix %>%
  # Convert matrix to data frame
  as.data.frame() %>%
  # Reshape to long format
  pivot_longer(cols = everything(), names_to = "Locus", values_to = "Allele") %>%
  # Count each allele per locus
  count(Locus, Allele) %>%
  # Ensure all four alleles are included (fill missing with 0)
  pivot_wider(
    names_from = Allele,
    values_from = n,
    values_fill = 0,
    names_expand = TRUE # Forces inclusion of all possible alleles
  )

# View the final dataset
allele_counts

This produces a clean data frame where each row is a locus (original column), and columns show the count of A, T, C, and G.

Method 2: Base R for No Extra Dependencies #

If you prefer sticking to base R, use this approach to force all four alleles to appear:

# Define the four possible alleles
alleles <- c("A", "T", "C", "G")

# Count alleles per column, ensuring all four are included
count_list <- apply(allele_matrix, 2, function(col) {
  table(factor(col, levels = alleles))
})

# Convert the list of tables to a data frame
allele_counts_df <- as.data.frame(t(count_list))

# View the result
allele_counts_df

The factor() function ensures that even if an allele is missing from a column, it still shows up with a count of 0.

内容的提问来源于stack exchange，提问作者user11709754