在R语言中计算分组求和、总计及排名并生成新列

Got it, let's break this down so it works even with your 200-column dataset—no need to manually list columns! Here's a clean, scalable approach using both tidyverse (dplyr) and base R, whichever you prefer:

Using dplyr (Tidyverse) #

This method is super readable, especially for large datasets:

# Load the package (install first if needed: install.packages("dplyr"))
library(dplyr)

# Your sample data
df <- read.table(text="Q1a Q2a Q3b Q4c Q5a Q6c Q7b 1 2 4 2 2 0 1 3 2 1 2 2 1 1 4 3 2 1 1 1 1", h=T)

# Add grouped sums, total, and rank
result_df <- df %>%
  # Calculate row sums for each group (a, b, c) by column suffix
  mutate(
    a = rowSums(select(., ends_with("a"))),
    b = rowSums(select(., ends_with("b"))),
    c = rowSums(select(., ends_with("c"))),
    # Total is sum of a, b, c
    Total = a + b + c,
    # Rank by Total (descending, same totals get same rank)
    Rank = min_rank(desc(Total))
  )

# View the result (matches your expected output)
result_df[, c(colnames(df), "a", "b", "c", "Total", "Rank")]

Using Base R (No Packages Needed) #

If you prefer not to load external packages, this works just as well:

# Your sample data
df <- read.table(text="Q1a Q2a Q3b Q4c Q5a Q6c Q7b 1 2 4 2 2 0 1 3 2 1 2 2 1 1 4 3 2 1 1 1 1", h=T)

# Calculate grouped row sums using regex to match column suffixes
df$a <- rowSums(df[, grepl("a$", colnames(df))])
df$b <- rowSums(df[, grepl("b$", colnames(df))])
df$c <- rowSums(df[, grepl("c$", colnames(df))])

# Add Total and Rank
df$Total <- df$a + df$b + df$c
# Use rank() with ties.method="min" to get same rank for equal totals
df$Rank <- rank(-df$Total, ties.method = "min")

# View the final table
df

Key Notes: #

• Both methods scale perfectly to 200 columns—no need to adjust code for more columns, since we're using pattern matching (ends_with or regex) instead of hardcoding column names.
• The ranking uses min rank (same totals get the same rank, which matches your sample output). If you want different ranking behavior (like dense rank), just swap min_rank() for dense_rank() in dplyr, or adjust the ties.method in base R.

内容的提问来源于stack exchange，提问作者user9285150