R 使用apply函数根据条件简化替换向量元素

我有一个数据集，它有一个名为

Region

的变量，代表澳大利亚的不同地区。以下是数据中的25行：

> head(sample.2013$Region, n = 25)
 [1] QLD major urban - capital city   VIC rural                        NSW regional - low urbanisation 
 [4] SA regional - low urbanisation   NSW regional - low urbanisation  Tas rural                       
 [7] ACT major urban - capital city   QLD rural                        ACT major urban - capital city  
[10] NT regional - low urbanisation   NSW other                        QLD rural                       
[13] ACT major urban - capital city   VIC regional - high urbanisation Tas rural                       
[16] QLD major urban - capital city   Tas rural                        VIC regional - high urbanisation
[19] QLD rural                        Tas rural                        VIC rural                       
[22] QLD other urban                  Tas rural                        VIC rural                       
[25] ACT major urban - capital city
36 Levels: ACT major urban - capital city NSW major urban - capital city NSW other urban ... ?

原始解决方案 我需要根据此列中的变量创建另一个名为

state

的变量。目前，我正在使用蛮力方法创建一个新向量，如下所示：

add_states <- function(sample.2013) {
    # Add states from the region variable
    sample.2013$State[grepl('NSW', sample.2013$Region) == TRUE] <- 'NSW'
    sample.2013$State[grepl('VIC', sample.2013$Region) == TRUE] <- 'VIC'
    sample.2013$State[grepl('QLD', sample.2013$Region) == TRUE] <- 'QLD'
    sample.2013$State[grepl('WA', sample.2013$Region) == TRUE] <- 'WA'
    sample.2013$State[grepl('SA', sample.2013$Region) == TRUE] <- 'SA'
    sample.2013$State[grepl('Tas', sample.2013$Region) == TRUE] <- 'TAS'
    sample.2013$State[grepl('TAS', sample.2013$Region) == TRUE] <- 'TAS'
    sample.2013$State[grepl('ACT', sample.2013$Region) == TRUE] <- 'ACT'
    sample.2013$State[grepl('NT', sample.2013$Region) == TRUE] <- 'NT'
    return(sample.2013)
}

add_state <- function(input, output, state) {
    # Change the variable y in place, prevents duplication
    output <- replace(output, grepl(state, input, ignore.case = TRUE), state)
    output
}

state_codes <- c('NSW', 'VIC', 'QLD', 'WA', 'SA', 'TAS', 'ACT', 'NT')
test_vector <- head(sample.2013$Region, n = 500)

y = vector('character', length = length(test_vector))

for (i in 1:length(state_codes)) {
    y <- add_state(test_vector, y, state_codes[i])
}

    table(y)
y
    ACT NSW  NT QLD  SA TAS VIC  WA 
 14  99  50  42  49  98  92  45  11 

但这也相当冗长，for循环不是惯用的R。我无法用apply函数替换这段代码，并让它替换向量中的值，而不是创建一堆其他向量

拉普拉 这是我用

lappy

管理过的最好的一次：

add_state3 <- function(x, state) {
    x <- replace(x, grepl(state, x, ignore.case = TRUE), state)
    x
}

test_vector_short <- c("NSW 1", "NSW 2", "Vic", "Goo")

> output <- lapply(state_codes, add_state3, x = test_vector_short)
> output
[[1]]
[1] "NSW" "NSW" "Vic" "Goo"

[[2]]
[1] "NSW 1" "NSW 2" "VIC"   "Goo"  

[[3]]
[1] "NSW 1" "NSW 2" "Vic"   "Goo"  

[[4]]
[1] "NSW 1" "NSW 2" "Vic"   "Goo"  

[[5]]
[1] "NSW 1" "NSW 2" "Vic"   "Goo"  

[[6]]
[1] "NSW 1" "NSW 2" "Vic"   "Goo"  

[[7]]
[1] "NSW 1" "NSW 2" "Vic"   "Goo"  

[[8]]
[1] "NSW 1" "NSW 2" "Vic"   "Goo"  

---

add_state3您可以使用
gsub
组合您的搜索和替换，例如
gsub（'^.*\\bNT\\b.*'，'NT'）
将替换所有匹配的NT字符串（
\\b
以避免类似“pint”匹配“NT”的事情）

如果将正则表达式设置为类似于
'^.*\\b（NSW | NT | QLD |…）\b.*
，然后替换为
\\1
（捕获的匹配项），则可以执行以下操作：

state.regex <- sprintf('^.*\\b(%s)\\b.*$', paste(state_codes, collapse='|'))
# "^.*\\b(NSW|VIC|QLD|WA|SA|TAS|ACT|NT)\\b.*$"
gsub(state.regex, '\\1', test_vector_short, ignore.case=T)
# [1] "NSW" "NSW" "Vic" "Goo"

state.regex似乎有一种模式是
STATECODE-other-stuff
，因此您可以
strsplit
并获取第一个元素

使用
测试
：

test <- c(
  "QLD major urban - capital city",
  "Vic rural",
  "NSW regional - low urbanisation", 
  "SA regional - low urbanisation",
  "NSW regional - low urbanisation",
  "guff and goo"
)

result <- toupper(sapply(strsplit(test," "),`[`,1))
replace(result, !result %in% state_codes, NA)
#[1] "QLD" "VIC" "NSW" "SA"  "NSW" NA   

test由于每个
区域的第一个单词是状态代码，您可以去掉其余的单词，并将结果用作新的
state
变量：

sample.2013 <- data.frame(Region=c('QLD major urban - capital city','VIC rural','NSW regional - low urbanisation','SA regional - low urbanisation','NSW regional - low urbanisation  Tas rural','ACT major urban - capital city','QLD rural','ACT major urban - capital city','NT regional - low urbanisation','NSW other','QLD rural','ACT major urban - capital city','VIC regional - high urbanisation Tas rural','QLD major urban - capital city','Tas rural','VIC regional - high urbanisation','QLD rural','Tas rural','VIC rural','QLD other urban','Tas rural','VIC rural','ACT major urban - capital city'));
sample.2013$state <- toupper(sub(' .*','',sample.2013$Region));
sample.2013;
##                                        Region state
## 1              QLD major urban - capital city   QLD
## 2                                   VIC rural   VIC
## 3             NSW regional - low urbanisation   NSW
## 4              SA regional - low urbanisation    SA
## 5  NSW regional - low urbanisation  Tas rural   NSW
## 6              ACT major urban - capital city   ACT
## 7                                   QLD rural   QLD
## 8              ACT major urban - capital city   ACT
## 9              NT regional - low urbanisation    NT
## 10                                  NSW other   NSW
## 11                                  QLD rural   QLD
## 12             ACT major urban - capital city   ACT
## 13 VIC regional - high urbanisation Tas rural   VIC
## 14             QLD major urban - capital city   QLD
## 15                                  Tas rural   TAS
## 16           VIC regional - high urbanisation   VIC
## 17                                  QLD rural   QLD
## 18                                  Tas rural   TAS
## 19                                  VIC rural   VIC
## 20                            QLD other urban   QLD
## 21                                  Tas rural   TAS
## 22                                  VIC rural   VIC
## 23             ACT major urban - capital city   ACT

sample.2013次要要点：你不需要
grepl（…）==TRUE
中的
==TRUE，因为
grepl（）
已经返回了一个逻辑向量（正如测试本身所示）。@bgoldst是的，我在发布之后注意到了这一点。干杯