R 使用ifelse语句超出限制

问题：我写了一段包含100多条

ifelse

语句的巨大代码，但却发现

ifelse

语句的数量有限：超过50条会抛出错误。无论如何，我知道有一种更有效的方法来做我想做的事情

目标：尝试编写一个函数，将字符串的许多变体（参见下面的示例）重新编码为清晰的类别（例如下面的）。我使用

str\u detect

给出T/F，然后根据响应转换为正确的类别。如果没有超过100条

ifelse

语句（我有更多的类别），我怎么能做到这一点呢

例如：

mydf <- data_frame(answer = sample(1:5, 10, replace = T),
                location = c("at home", "home", "in a home", 
"school", "my school", "School", "Work", "work",
                         "working", "work usually"))

loc_function <- function(x) {
  home <- "home"
  school <- "school"
  work <- "work"
  ifelse(str_detect(x, regex(home, ignore_case = T)), "At Home",
     ifelse(str_detect(x, regex(school, ignore_case = T)), "At 
School",
            ifelse(str_detect(x, regex(work, ignore_case = T)), "At 
Work", x)))
}

### Using function to clean up messy strings (and recode first column too) into clean categories
mycleandf <- mydf %>%
  as_data_frame() %>%
  mutate(answer = ifelse(answer >= 2, 1, 0)) %>%
  mutate(location = loc_function(location)) %>%
  select(answer, location)

mycleandf

# A tibble: 10 x 2
   answer  location
    <dbl>     <chr>
 1      1   At Home
 2      1   At Home
 3      1   At Home
 4      1 At School
 5      1 At School
 6      1 At School
 7      1   At Work
 8      0   At Work
 9      1   At Work
10      0   At Work

---

mydf您可以将模式放入命名向量中，（注意
Other=”“
，当您的模式与字符串不匹配时，这是一种退步）：

模式而不是嵌套条件，您可以按顺序执行它们。对
循环使用
：

# Store the find-replace pairs in a data frame

word_map <- data.frame(pattern = c("home", "school", "work"),
                       replacement = c("At Home", "At School", "At Work"), 
                       stringsAsFactors = FALSE)

word_map
pattern replacement
1    home     At Home
2  school   At School
3    work     At Work

# Iterate through the pairs

for ( i in 1:nrow(word_map) ) {

    pattern     <- word_map$pattern[i]
    replacement <- word_map$replacement[i]

    mydf$location <- ifelse(grepl(pattern, mydf$location, ignore.case = TRUE), replacement, mydf$location)
}

mydf
   answer  location
1       4   At Home
2       4   At Home
3       1   At Home
4       5 At School
5       1 At School
6       2 At School
7       5   At Work
8       2   At Work
9       1   At Work
10      3   At Work

#将查找-替换对存储在数据帧中
当你发现自己需要更多的东西时，你应该开始考虑寻找更好的方法。当你达到两把时，你应该开始认为你做错了什么。如果你的手推车满了，要知道你完全搞砸了，需要寻求帮助。您已经达到了垃圾车级别。听起来您想使用case\u when语句，或者使用purr:map（）将函数映射到所有单词以使其成为标题case？非常有用。如果我有一个匹配这两个的字符串，我怎么能正确排序呢？例如：“在家然后在工作”我想分类为“在工作”。调整模式或匹配的逻辑顺序？你可以调整模式的顺序，把你想优先考虑的模式放在你不想优先考虑的模式之前。因此，如果你把
放在工作中
放在家中的
之前
，它会给你工作中的
。
match <- sapply(patterns, grepl, mydf$location, ignore.case = T)

mydf$clean_loc <- colnames(match)[max.col(match, ties.method = "first")]
mydf

# A tibble: 10 x 3
#   answer     location clean_loc
#    <int>        <chr>     <chr>
# 1      3      at home   At Home
# 2      3         home   At Home
# 3      3    in a home   At Home
# 4      3       school At School
# 5      2    my school At School
# 6      4       School At School
# 7      5         Work   At Work
# 8      1         work   At Work
# 9      2      working   At Work
#10      1 work usually   At Work

# Store the find-replace pairs in a data frame

word_map <- data.frame(pattern = c("home", "school", "work"),
                       replacement = c("At Home", "At School", "At Work"), 
                       stringsAsFactors = FALSE)

word_map
pattern replacement
1    home     At Home
2  school   At School
3    work     At Work

# Iterate through the pairs

for ( i in 1:nrow(word_map) ) {

    pattern     <- word_map$pattern[i]
    replacement <- word_map$replacement[i]

    mydf$location <- ifelse(grepl(pattern, mydf$location, ignore.case = TRUE), replacement, mydf$location)
}

mydf
   answer  location
1       4   At Home
2       4   At Home
3       1   At Home
4       5 At School
5       1 At School
6       2 At School
7       5   At Work
8       2   At Work
9       1   At Work
10      3   At Work