R语言新手求助：基于因子级别及相同ID创建条件列f

Hey there! No worries at all—we all start somewhere with R, and it’s totally okay to ask basic questions. Let’s figure this out together.

Looking at your data and desired output, the rule for column f is clear: for each group of rows sharing the same id, if the group contains both a row where b=0 and c=1 and a row where b=1 and c=0, set f=1 for all rows in that group; otherwise, set f=0. And crucially, we’ll keep all your original columns (like e) intact.

Solution using dplyr (tidyverse) #

First, install the dplyr package if you haven’t already with install.packages("dplyr"), then use this code:

library(dplyr)

# Your sample data frame (I’ve structured it properly here)
df <- tibble(
  b = c(0, 1, 0, 1),
  c = c(1, 0, 1, 0),
  id = c(45, 45, 48, 46),
  e = c(5, 7, 5, 7)
)

# Add the new column f without altering existing columns
df_with_f <- df %>%
  group_by(id) %>%
  mutate(
    # Check if both required (b,c) pairs exist in the group
    has_b0_c1 = any(b == 0 & c == 1),
    has_b1_c0 = any(b == 1 & c == 0),
    # Convert the combined condition to 1/0 for column f
    f = as.integer(has_b0_c1 & has_b1_c0)
  ) %>%
  # Remove helper columns (optional but keeps data clean)
  select(-has_b0_c1, -has_b1_c0) %>%
  ungroup()

# View the final result
df_with_f

How this works: #

• group_by(id): Clusters rows by their id value so we can check conditions per group
• mutate(): Creates new columns within each group:
  • has_b0_c1 and has_b1_c0 are helper flags to check if each required (b,c) pair exists in the group
  • f is set to 1 only if both flags are TRUE, converted to an integer for your desired 1/0 format
• select(-...): Cleans up the helper columns we used to calculate f
• ungroup(): Returns the data to a regular ungrouped data frame

Base R alternative (no packages needed) #

If you prefer to stick to base R without installing extra packages, use this approach with ave():

# Your sample data frame
df <- data.frame(
  b = c(0, 1, 0, 1),
  c = c(1, 0, 1, 0),
  id = c(45, 45, 48, 46),
  e = c(5, 7, 5, 7)
)

# Add column f directly to the original data frame
df$f <- with(df, ave(
  x = 1:nrow(df),
  by = id,
  FUN = function(group_rows) {
    # Subset the data to the current id group
    group_data <- df[group_rows, ]
    # Check if both (b,c) pairs exist, return 1 or 0
    as.integer(any(group_data$b == 0 & group_data$c == 1) & any(group_data$b == 1 & group_data$c == 0))
  }
))

# View the result
df

Both methods will preserve all your original columns while adding the f column exactly as you need it. Let me know if you hit any snags getting this to work!

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