Hey there, let's break down what's happening here and how to fix it.

Why the Error Happens #

The error Error in matrix(0, 2^nvar, nvar) : invalid 'nrow' value (too large or NA) comes down to a core design choice in countpattern: it tries to create a matrix with 2^nvar rows to enumerate every possible binary pattern for your columns. When you have more than 19 columns, 2^20 hits 1,048,576 rows—and it only gets worse from there (25 columns would need 33 million rows!). This either exceeds R's ability to handle such a large matrix, or causes integer overflow that makes the row count invalid.

Can You Fix This With countpattern Itself? #

Unfortunately, no—there's no way to tweak the function to get around this limitation. That matrix creation is baked into how countpattern works, so once your column count passes ~19, this method just isn't feasible in R, even with lots of memory.

Better, More Efficient Alternatives #

Since your data is a large 114x9800 matrix, we should stop trying to enumerate every possible pattern (most of which won't even exist in your data!) and instead count only the patterns that actually appear. Here are a few solid approaches:

1. Use data.table to Count Row Strings

Convert each row to a single string of 0s and 1s, then count how often each string appears. This is fast and straightforward for 9800 rows:

library(data.table)

# Replace 'your_matrix' with your actual matrix name
row_strings <- apply(your_matrix, 1, paste, collapse = "")
pattern_counts <- data.table(pattern = row_strings)[, .N, by = pattern]

# Optional: Convert strings back to binary matrix format
pattern_matrix <- do.call(rbind, strsplit(pattern_counts$pattern, ""))
pattern_matrix <- matrix(as.integer(pattern_matrix), ncol = ncol(your_matrix))
final_result <- cbind(pattern_matrix, count = pattern_counts$N)

2. Tidyverse Approach with dplyr

If you prefer the tidyverse workflow, group by all columns and count occurrences:

library(dplyr)

# Convert matrix to data frame
mat_df <- as.data.frame(your_matrix)

# Count unique row patterns
pattern_counts <- mat_df %>%
  group_by(across(everything())) %>%
  summarise(count = n(), .groups = "drop")

3. Hash Table for Ultra-Large Datasets

If you ever work with even bigger datasets, a hash table can be more efficient than string conversion:

library(hash)

row_keys <- apply(your_matrix, 1, function(row) paste(row, collapse = ","))
count_hash <- hash()

for (key in row_keys) {
  count_hash[[key]] <- if (has.key(key, count_hash)) count_hash[[key]] + 1 else 1
}

# Convert to a readable data frame
pattern_counts <- data.frame(
  pattern = names(count_hash),
  count = unlist(count_hash),
  stringsAsFactors = FALSE
)

Testing with Your Sample Data #

Using your provided data_sample:

data_sample <- rbind(c(1,1,1,0,1,0,1,1,0,1,0), c(1,0,0,1,1,1,9,1,0,0,1), c(1,0,0,0,0,1,0,1,1,0,0), c(0,1,1,0,0,0,0,0,1,1,1), c(1,1,1,0,0,1,1,0,1,1,0))

# Test the dplyr method
as.data.frame(data_sample) %>%
  group_by(across(everything())) %>%
  summarise(count = n(), .groups = "drop")

Since all rows in your sample are unique, you'll get 5 rows each with a count of 1—exactly what you'd expect!

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