如何在R中为data.frame各列提取首值>1的指定行至新表

Here's a clean implementation to achieve your goal:

library(data.table)

# Convert your data.frame to data.table (if not already done)
setDT(dt)

# Add a column for the actual row numbers (since your rows start at 52)
dt[, row_id := as.integer(rownames(dt))]

# Function to get the row range for a single column
get_target_rows <- function(col_name) {
  # Find the first row number where the column value is >1
  first_row <- dt[get(col_name) > 1, min(row_id)]
  # Return the range if such a row exists, else NULL
  if (!is.na(first_row)) {
    return(seq(from = first_row - 1, to = first_row + 25))
  } else {
    return(NULL)
  }
}

# Apply the function to all columns (excluding the new row_id column)
all_target_rows <- unique(unlist(lapply(names(dt)[names(dt) != "row_id"], get_target_rows)))

# Subset the original data to these rows, removing the row_id column
result_dt <- dt[row_id %in% all_target_rows, .SD, .SDcols = names(dt)[names(dt) != "row_id"]]

Key Step Explanations: #

• Add row_id column: Captures your actual row numbers (52, 53, ...) instead of the default 1-based data frame indices, which is critical for accurate range calculation.
• get_target_rows function: For each column, finds the first row where values exceed 1, then generates the required range from n-1 to n+25.
• Combine ranges: Flattens all column-specific ranges and keeps only unique rows to avoid duplicates.
• Subset data: Extracts the relevant rows from the original data, excluding the helper row_id column.

Why Your Previous Attempts Failed: #

1. Your first code used i>1 which checks if the column index is greater than 1, not if values in the column are greater than 1. You should reference column values with df[[i]] > 1 instead.
2. Your second code used min(which(df[i]>1)) which returns the 1-based index of the row (e.g., 3 for row 54) instead of the actual row number (54), leading to incorrect range calculation.

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