我如何將所有因子列轉換為具有與字符串列表匹配的別名的數字？

Question

我還沒有在這里找到這個確切的問題。 我有很多列，對於所有與字符串列表中的任何一個匹配的列，我都想從factor-> character-> numeric轉換。

下面顯示了一個示例，其中包含字符串之一的列被轉換，並且在多個字符串失敗的情況下我嘗試了兩件事

#Making fake data where every column is a factor. At the end I'd like to convert all factors that contain either "alcium" or "zinc" in the column name.
library(reshape2)
fake <-data.frame(id=c(1,1,1,2,2,2,3,3,3,1,1,1,2,2,2,3,3,3),              
              time=c(rep("Time1",9), rep("Time2",9)), 
            test=c("calcium","magnesium","zinc","calcium","magnesium","zinc","calcium","magnesium","zinc","calcium","magnesium","zinc","calcium","magnesium","zinc","calcium","magnesium","zinc"), 
              score=floor(runif(18, min=1, max=5)))

fake <- dcast(fake, id ~ time + test)
fake <- fake %>% mutate_if(is.numeric,as.factor)

#This works, but only for columns containing one of the strings
fake <- fake %>% mutate_at(vars(contains('alcium')),  function(x) as.numeric(as.character(x))) 

#Now trying to convert all columns containing either "alcium" or "zinc"
fake <- fake %>% mutate_at(vars(contains('alcium'| 'zinc')),  function(x) as.numeric(as.character(x))) 
#gives an error

#2nd attempt:
strings <- c("alcium", "zinc")
fake <- fake %>% mutate_at(vars(contains(strings)),  function(x) as.numeric(as.character(x)))  
#gives an error

Answer 1

使用選擇助手matches()而不是contains()允許傳遞折疊成正則表達式友好格式的字符串。

library(dplyr)
strings <- c("alcium", "zinc")

fake %>%
  as_tibble %>%
  mutate_at(vars(matches(paste0(strings, collapse = "|"))),  as.numeric) 

# A tibble: 3 x 8
  id    Time1_calcium `Time1_ma    gnesium` Time1_magnesium Time1_zinc Time2_calcium Time2_magnesium Time2_zinc
  <fct>         <dbl> <fct>                 <fct>                <dbl>         <dbl> <fct>                <dbl>
1 1                 2 NA                    4                        1             3 4                        1
2 2                 2 NA                    3                        2             1 1                        3
3 3                 1 3                     NA                       1             2 3                        2

Answer 2

我已經更新了您的代碼。

如果您沒有太多，可以分開進行。 否則，我將無法使用多個字符串。

sofaWa <- fake %>% mutate_at(vars(contains('alcium')), list(as.numeric)) %>% 
  mutate_at(vars(contains('zinc')), list(as.numeric)) 

產生這個

# A tibble: 3 x 7
  id    Time1_calcium Time1_magnesium Time1_zinc Time2_calcium Time2_magnesium Time2_zinc
  <fct>         <dbl> <fct>                <dbl>         <dbl> <fct>                <dbl>
1 1                 3 2                        1             1 2                        1
2 2                 2 1                        2             1 1                        2
3 3                 1 4                        2             2 3                        1