Question

我正在尝试根据医疗数据集中具有2500+值的程序代码变量创建新变量，以提取抗生素，其剂量和途径。我已经能够使用ifelse语句执行此操作，但这非常耗时，并且很难找到并纠正错误。有一种简化的方法吗？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[是/否]，剂量，途径），我觉得有一种更好的方法，我错过了不涉及剪切和粘贴变量和每次引号。是否有可能建立数据帧并使用ifelse将任何也在该数据帧中的代码分配为1，将其他代码分配为0？

很抱歉，如果这是重复的，我对R来说比较新，我很难找到词汇来搜索我需要的东西。我环顾四周（比如Nested ifelse statement，但还没找到我需要的东西。

Answer 1

两种替代方法，都使用合并/连接。这种方法的一个优点是维护起来要容易得多：您拥有结构良好且易于管理的过程表，而不是ifelse语句的（可能非常长的）代码行。建议%in%的评论也可以减少这个问题，尽管你会处理可管理的向量而不是可管理的帧。

虚假数据：

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

每个程序类型一帧。这是可以管理的，但如果你有很多不同的类型，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

使用ab=TRUE（etc）是为了合并一列。不匹配的行将有NA，这要求!is.na(.)将T,NA,T转换为T,F,T。

您甚至可以使用过程代码的向量，例如：

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

虽然这只有在你已经将代码作为矢量的情况下才有用，否则它似乎只是为了让你更容易维护。

包含所有程序的一个框架，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

你可以保持原样（如果以这种方式存储它是有意义的）或spread它就像其他人一样：

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

（此处加入/合并后的顺序不同，但数据保持不变。）

（我使用logical s，很容易将它们转换为1和0，可能是mutate(ab=1L*ab)或mutate(ab=as.integer(ab))。）

Answer 2

基本R方法的简单选项：

# 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，这是足够的，但要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

R：简化长篇ifelse陈述

2 个答案: