在R数据框中生成按Subject分组的多变量组合均值数据框

Hey there! Let's work through this problem together—since you're new to R, I'll keep things clear and provide code you can test right away.

First, let's start with a simulated version of your data frame so you can follow along and verify the output:

# Simulate sample data (matches your structure)
set.seed(123) # Makes results reproducible
df <- data.frame(
  Subject = rep(1:6, each = 10), # 6 subjects, 10 rows each
  T = sample(0:1, 60, replace = TRUE),
  O = sample(0:1, 60, replace = TRUE),
  E = sample(0:1, 60, replace = TRUE),
  P = sample(0:1, 60, replace = TRUE),
  Score = rnorm(60, mean = 50, sd = 10) # Random numeric scores
)

Approach 1: Using dplyr (Tidyverse, Recommended for Readability) #

This method is intuitive and easy to modify if you need to add more combinations later. First, install and load the dplyr package if you haven't already:

install.packages("dplyr")
library(dplyr)

Now, group the data by Subject and calculate the mean Score for each of your desired combinations:

result_df <- df %>%
  group_by(Subject) %>%
  summarise(
    # Single variables (mean Score where the variable equals 1)
    T = mean(Score[T == 1], na.rm = TRUE),
    O = mean(Score[O == 1], na.rm = TRUE),
    E = mean(Score[E == 1], na.rm = TRUE),
    P = mean(Score[P == 1], na.rm = TRUE),
    
    # Two-variable combinations (mean Score where both equal 1)
    TO = mean(Score[T == 1 & O == 1], na.rm = TRUE),
    TE = mean(Score[T == 1 & E == 1], na.rm = TRUE),
    TP = mean(Score[T == 1 & P == 1], na.rm = TRUE),
    OE = mean(Score[O == 1 & E == 1], na.rm = TRUE),
    OP = mean(Score[O == 1 & P == 1], na.rm = TRUE),
    PE = mean(Score[P == 1 & E == 1], na.rm = TRUE),
    
    # Three-variable combinations (mean Score where all equal 1)
    TOP = mean(Score[T == 1 & O == 1 & P == 1], na.rm = TRUE),
    TOE = mean(Score[T == 1 & O == 1 & E == 1], na.rm = TRUE),
    POE = mean(Score[P == 1 & O == 1 & E == 1], na.rm = TRUE),
    PET = mean(Score[P == 1 & E == 1 & T == 1], na.rm = TRUE)
  )

# View the result
print(result_df)

The na.rm = TRUE argument ensures that if a subject has no rows matching a combination (e.g., no rows where T=1 and O=1), we get NA instead of an error—you can replace these NAs later with replace_na() from tidyr if needed.

Approach 2: Base R (No Extra Packages Needed) #

If you prefer not to install new packages, you can use base R's by() function to process each subject separately, then combine the results:

# Process each subject and calculate means
subject_results <- by(df, df$Subject, function(subject_data) {
  data.frame(
    Subject = unique(subject_data$Subject),
    T = mean(subject_data$Score[subject_data$T == 1], na.rm = TRUE),
    O = mean(subject_data$Score[subject_data$O == 1], na.rm = TRUE),
    E = mean(subject_data$Score[subject_data$E == 1], na.rm = TRUE),
    P = mean(subject_data$Score[subject_data$P == 1], na.rm = TRUE),
    TO = mean(subject_data$Score[subject_data$T == 1 & subject_data$O == 1], na.rm = TRUE),
    TE = mean(subject_data$Score[subject_data$T == 1 & subject_data$E == 1], na.rm = TRUE),
    TP = mean(subject_data$Score[subject_data$T == 1 & subject_data$P == 1], na.rm = TRUE),
    OE = mean(subject_data$Score[subject_data$O == 1 & subject_data$E == 1], na.rm = TRUE),
    OP = mean(subject_data$Score[subject_data$O == 1 & subject_data$P == 1], na.rm = TRUE),
    PE = mean(subject_data$Score[subject_data$P == 1 & subject_data$E == 1], na.rm = TRUE),
    TOP = mean(subject_data$Score[subject_data$T == 1 & subject_data$O == 1 & subject_data$P == 1], na.rm = TRUE),
    TOE = mean(subject_data$Score[subject_data$T == 1 & subject_data$O == 1 & subject_data$E == 1], na.rm = TRUE),
    POE = mean(subject_data$Score[subject_data$P == 1 & subject_data$O == 1 & subject_data$E == 1], na.rm = TRUE),
    PET = mean(subject_data$Score[subject_data$P == 1 & subject_data$E == 1 & subject_data$T == 1], na.rm = TRUE)
  )
})

# Combine all subject results into one data frame
result_base <- do.call(rbind, subject_results)

# View the result
print(result_base)

Both methods will give you a 6-row data frame (one per subject) with all the mean values you need. The dplyr approach is cleaner for beginners, but the base R method works if you want to avoid package installs.

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