在R语言中根据列名匹配值创建新变量grade的实现方法

Absolutely! You can easily create the grade variable by pulling the test score specified in the select column for each row. Here are two practical approaches in R:

Base R Approach #

Using row-column indexing, we can directly fetch the correct score from the column named in select:

# Original data frame
students <- data.frame(name = c("student1", "student2", "student3", "student4"), 
                       test1 = c(50, 30, 20, 6), 
                       test2 = c(30, 20, 15, 10), 
                       select = c("test2", "test1", "test2", "test1"))

# Add the grade variable
students$grade <- students[cbind(seq(nrow(students)), match(students$select, colnames(students)))]

# View the result
students

This works because cbind(seq(nrow(students)), match(students$select, colnames(students))) creates a matrix of (row, column) pairs. For each row, we use the position of the select value in the column names to grab the corresponding test score.

Tidyverse (dplyr) Approach #

If you prefer a more readable, pipe-based workflow:

library(dplyr)

students <- students %>%
  rowwise() %>%
  mutate(grade = pull(cur_data(), select)) %>%
  ungroup()

# Check the output
students

rowwise() ensures we process each row individually, and pull(cur_data(), select) extracts the value from the column specified in the select column for that row.

Final Output #

Both methods will produce your desired result:

> students
       name test1 test2 select grade
1 student1    50    30  test2    30
2 student2    30    20  test1    30
3 student3    20    15  test2    15
4 student4     6    10  test1     6

内容的提问来源于stack exchange，提问作者imprela