R中的数据字典链接

Question

我有两份数据如下。

• I想要更新patients dataframe
• ，并将Option_name与Option_num交换为all.

## dataframe
patid = c(1, 2, 3, 4, 5, 6, 7, 8, 9) 
technique = c("VMAT", "IMRT", "ELEC", "VMAT", "VMAT", "VMAT", "ELEC", "VMAT", "IMRT")
intent = c("Radical", "Curative", "Adjuvant", "Curative", "Curative", "Curative", "Curative", "Curative", "Adjuvant")
patients = data.frame(patid, technique, intent)

## data dictionary
variable = c("technique", "technique", "technique", "intent", "intent", "intent")
Option_name = c("VMAT", "IMRT", "ELEC", "Radical", "Curative", "Adjuvant")
Option_num = c(1, 2, 3, 1, 2, 3)
dict = data.frame(variable, Option_name, Option_num)

因此，patients数据将更改。

从…

​

至

​

我知道如何通过一个非常手动的过程来做到这一点：

patients$technique <- replace(as.character(patients$technique), patients$technique == "VMAT",1)

但我相信会有一个更有效的方法。

任何帮助都将不胜感激！

Answer 1

你可以试试这个：

library(dplyr)
library(purrr)
patients %>%
    mutate(across(c(technique, intent), 
                  ~map(as.character(.), 
                       ~dict$Option_num[which(dict$variable == cur_column() & dict$Option_name == .)])))

或完全tidyverse选项：

library(dplyr)
library(purrr)
patients %>%
    mutate(across(c(technique, intent), 
                  ~map(as.character(.), 
                       ~pull(filter(dict, 
                                    variable == cur_column(), 
                                    Option_name == .x), 
                             Option_num))))

Answer 2

这肯定不是最优雅的解决方案，但我想不出其他的解决方案。

for (a in c("technique", "intent")) {
  # go through the columns
  for (b in unique(patients[, a])) {
    # go through the Option_name
    patients[patients[, a] == b, a] <- dict[dict[, "variable"] == a & dict[, "Option_name"] == b, "Option_num"]
  }
}