R：简化长 ifelse 语句

Question

我正在尝试根据医疗数据集中具有 2500 多个值的过程代码变量创建新变量，以提取抗生素、它们的剂量和途径。我已经能够使用 ifelse 语句来做到这一点，但它很耗时，而且很难发现和纠正错误。有没有简化的方法来做到这一点？不幸的是，代码没有以任何逻辑方式组织。

vet <-mutate(vet, ab = ifelse(ProcedureCode=="6160"|ProcedureCode=="2028"|ProcedureCode=="6121"|ProcedureCode=="6130"|ProcedureCode=="6131"|ProcedureCode=="6132"|ProcedureCode=="6133" |ProcedureCode=="6134"|ProcedureCode=="6135"|ProcedureCode=="6136"|ProcedureCode=="6090" |ProcedureCode=="6137"|ProcedureCode=="6138"|ProcedureCode=="6139" |ProcedureCode=="6140" |ProcedureCode=="6510"|ProcedureCode=="680D" |ProcedureCode=="633E"|ProcedureCode=="661J"|ProcedureCode=="627I" |ProcedureCode=="6198"|ProcedureCode=="6199"|ProcedureCode=="6200" |ProcedureCode=="6201" |ProcedureCode=="6202"|ProcedureCode=="622G" |ProcedureCode=="697C" |ProcedureCode=="698C" |ProcedureCode=="6204"|ProcedureCode=="6775"| ProcedureCode=="6229" |ProcedureCode=="6207" |ProcedureCode=="6203" |ProcedureCode=="6205" |ProcedureCode=="6206" |ProcedureCode=="6212" |ProcedureCode=="6213" |ProcedureCode=="6214" |ProcedureCode=="6215" |ProcedureCode=="6216" |ProcedureCode=="6219" |ProcedureCode=="692C" |ProcedureCode=="643C" |ProcedureCode=="601E" |ProcedureCode=="629G" |ProcedureCode=="6234" |ProcedureCode=="6235" |ProcedureCode=="6236" |ProcedureCode=="6237" |ProcedureCode=="6238" |ProcedureCode=="615J" |ProcedureCode=="6242" |ProcedureCode=="6243" |ProcedureCode=="6244" |ProcedureCode=="6245" |ProcedureCode=="1193" |ProcedureCode=="652G" |ProcedureCode=="657G" |ProcedureCode=="697B"|ProcedureCode=="6336" |ProcedureCode=="6337" |ProcedureCode=="6338" |ProcedureCode=="6152" |ProcedureCode=="603C" |ProcedureCode=="655B" |ProcedureCode=="6357" |ProcedureCode=="6358" |ProcedureCode=="6399" |ProcedureCode=="666B" |ProcedureCode=="695D" |ProcedureCode=="699C" |ProcedureCode=="6365" |ProcedureCode=="6366" |ProcedureCode=="696F" |ProcedureCode=="6497" |ProcedureCode=="6613" |ProcedureCode=="6508" |ProcedureCode=="6509" |ProcedureCode=="617I" |ProcedureCode=="6506" |ProcedureCode=="2029" |ProcedureCode=="6538" |ProcedureCode=="671J" |ProcedureCode=="633H" |ProcedureCode=="621G" |ProcedureCode=="680J" |ProcedureCode=="672G" |ProcedureCode=="673G" |ProcedureCode=="6559" |ProcedureCode=="6652" |ProcedureCode=="6593" |ProcedureCode=="651C" |ProcedureCode=="633B" |ProcedureCode=="659E" |ProcedureCode=="676D" |ProcedureCode=="678D" |ProcedureCode=="620B" |ProcedureCode=="6562" |ProcedureCode=="6564" |ProcedureCode=="6585" |ProcedureCode=="6766" |ProcedureCode=="6595" |ProcedureCode=="6607" |ProcedureCode=="6608" |ProcedureCode=="627B" |ProcedureCode=="6653" |ProcedureCode=="6654" |ProcedureCode=="6655"|ProcedureCode=="6732" |ProcedureCode=="6733" |ProcedureCode=="6734"|ProcedureCode=="6735" |ProcedureCode=="6795"|ProcedureCode=="6745" |ProcedureCode=="6746" |ProcedureCode=="6748" |ProcedureCode=="6758" |ProcedureCode=="697E" |ProcedureCode=="6761" |ProcedureCode=="6032" |ProcedureCode=="6747" |ProcedureCode=="6749" |ProcedureCode=="668A" |ProcedureCode=="648A" |ProcedureCode=="649A" |ProcedureCode=="6765" |ProcedureCode=="6768" |ProcedureCode=="6771" |ProcedureCode=="637B"|ProcedureCode=="6894", 1,0))

问题还在于我需要创建多个组（例如：抗生素 [yes/no]、剂量、路线）而且我觉得我缺少一种不涉及切割的更好方法并每次粘贴变量和引号。是否有可能制作数据框并使用 ifelse 将也在该数据框中的任何代码分配为 1，将其他代码分配为 0？

抱歉，如果这是重复的，我是 R 的新手，很难找到词汇表来搜索我需要的内容。我环顾四周（比如 Nested ifelse statement ，但还没有找到我需要的东西。

Answer 1

两种替代方法，均使用 merges/joins。这种方法的一个优点是它更容易维护：您拥有结构良好且易于管理的过程表，而不是使用 ifelse 语句的（可能非常长的）代码行。建议 %in% 的评论也减少了这个问题，尽管您将处理可管理的矢量而不是可管理的框架。

虚假数据：

library(dplyr)
library(tidyr)
vet <- data_frame(ProcedureCode = c('6160', '2028', '2029'))

每个过程类型一帧。这是可管理的，但如果您有很多不同的类型，可能会很烦人。对每种类型重复 left_join。

abs <- data_frame(ab=TRUE, ProcedureCode = c('6160', '2028'))
antis <- data_frame(antibiotic=TRUE, ProcedureCode = c('2029'))
vet %>%
  left_join(abs, by = "ProcedureCode") %>%
  left_join(antis, by = "ProcedureCode") %>%
  mutate_at(vars(ab, antibiotic), funs(!is.na(.)))
# # A tibble: 3 × 3
#   ProcedureCode    ab antibiotic
#           <chr> <lgl>      <lgl>
# 1          6160  TRUE      FALSE
# 2          2028  TRUE      FALSE
# 3          2029 FALSE       TRUE

ab=TRUE（等）的使用是为了有一个列要合并。不匹配的行将有一个 NA，这要求 !is.na(.) 将 T,NA,T 转换为 T,F,T。

您甚至可以改用过程代码向量，例如：

vet %>%
  left_join(data_frame(ab=TRUE, ProcedureCode=vector_of_abs), by = "ProcedureCode") %>%
  ...

尽管只有当您已经将代码作为向量时才真正有用，否则它似乎只是您更容易维护的那个。

一个框架包含所有程序，只需要一个框架用于类型和一个left_join.

procedures <- tibble::tribble(
  ~ProcedureCode, ~procedure,
  '6160'        , 'ab',
  '2028'        , 'ab',
  '2029'        , 'antibiotic'
)
left_join(vet, procedures, by = "ProcedureCode")
# # A tibble: 3 × 2
#   ProcedureCode  procedure
#           <chr>      <chr>
# 1          6160         ab
# 2          2028         ab
# 3          2029 antibiotic

您可以保持原样（如果以这种方式存储它有意义）或spread它与其他的一样：

left_join(vet, procedures, by = "ProcedureCode") %>%
  mutate(ignore=TRUE) %>%
  spread(procedure, ignore) %>%
  mutate_at(vars(ab, antibiotic), funs(!is.na(.)))
# # A tibble: 3 × 3
#   ProcedureCode    ab antibiotic
#           <chr> <lgl>      <lgl>
# 1          2028  TRUE      FALSE
# 2          2029 FALSE       TRUE
# 3          6160  TRUE      FALSE

（此处join/merge后的顺序不同，但数据保持不变。）

（我用的是logicals，很容易将它们转换为1s和0s，也许是mutate(ab=1L*ab)或mutate(ab=as.integer(ab))。）

Answer 2

一个简单选项的基础 R 方法：

# my dummy data
df1 <- data.frame("v1" = c(LETTERS[1:10]), "v2" = rep(NA, 10))

# step 1, fill the column with 0 (the else part of your code)
df1[,'v2'] <- 0

# step 2, create a vector containing ids you want to change
change_vec <- c("A", "C", "D", "F")

# step 3, use %in% to index and replace with 1
df1[,'v2'][df1[,'v1'] %in% change_vec] <- 1

在大多数情况下这就足够了，但请注意使用包含数值的索引向量的风险。

https://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f

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