一次用多个列中的值替换字符串答案

【问题标题】：Replace strings with values across multiple columns at once一次用多个列中的值替换字符串
【发布时间】：2019-09-02 04:33:12
【问题描述】：

我需要用多列中的数字替换字符串。下面是一个样本数据集：

x <- c("Low Outlier", "High Outlier", "Novice", "Novice", "Emerging", NA, "Proficient", "Approaching")
y <- c("Novice", "Approaching", "Proficient", "Approaching", "High Outlier", "Proficient",NA, "Emerging")
z <- c("High Outlier", "Proficient", "Approaching", "Emerging", "Low Outlier", "Approaching", "Approaching", "Emerging")

sam <- cbind(x,y,z)

我需要将“高/低异常值”转换为 0，将 NA 保留为 NA，将“新手”转换为 1，将“新兴”转换为 2，将“接近”转换为 3，将“精通”转换为 4。

我尝试用

转换单个变量

sam$x.r <- recode(sam$x.r,'Low Outlier'=0,'High Outlier'=0,'Novice'=1,'Emerging'=2,'Approaching'=3, 'Proficient'=4)

我收到一条错误消息“警告消息：在 recode.numeric(Dat17_18.1$I.E.ScoreStat, Low Outlier = 0, High Outlier = 0, : 强制引入的 NA"

我不确定如何一次重新编码所有变量。

【问题讨论】：

标签： r string replace numbers recode

【解决方案1】：

只需这样做-

sam[] <- recode(sam,'Low Outlier'=0,
                    'High Outlier'=0,
                    'Novice'=1,
                    'Emerging'=2,
                    'Approaching'=3, 
                    'Proficient'=4)

> sam
     x   y   z  
[1,] "0" "1" "0"
[2,] "0" "3" "4"
[3,] "1" "4" "3"
[4,] "1" "3" "2"
[5,] "2" "0" "0"
[6,] NA  "4" "3"
[7,] "4" NA  "3"
[8,] "3" "2" "2"

【讨论】：

【解决方案2】：

真的很快就重复了。这是一个简单的函数：

my_replacer<-function(df,y,z){    
df<-as.data.frame(apply(df,2,function(x) gsub(y,z,x)))
    #y is what you want to replace
    #z is the replacement
    #This uses regex
      df
    }
    my_replacer(sam,"Emerging.*","2")

这是我的使用方法：

library(dplyr)#can use ifelse. Still repetitive

    sam<-as.data.frame(sam)

    sam %>% 
      mutate_if(is.factor,as.character)->sam
    my_replacer(sam,"Emerging.*","2")

结果：

               x            y            z
    1  Low Outlier       Novice High Outlier
    2 High Outlier  Approaching   Proficient
    3       Novice   Proficient  Approaching
    4       Novice  Approaching            2
    5            2 High Outlier  Low Outlier
    6         <NA>   Proficient  Approaching
    7   Proficient         <NA>  Approaching
    8  Approaching            2            2

替换其他人：

my_replacer(sam,"Novi.*","1")
             x            y            z
1  Low Outlier            1 High Outlier
2 High Outlier  Approaching   Proficient
3            1   Proficient  Approaching
4            1  Approaching     Emerging
5     Emerging High Outlier  Low Outlier
6         <NA>   Proficient  Approaching
7   Proficient         <NA>  Approaching
8  Approaching     Emerging     Emerging

【讨论】：

【解决方案3】：

对于这种情况，我们可以使用dplyr 中的case_when

library(dplyr)

sam %>%
   mutate_all(~case_when(. %in% c("Low Outlier", "High Outlier") ~ '0', 
                   . == "Novice" ~ '1', 
                   . == "Emerging" ~ '2', 
                   . == "Approaching" ~ '3', 
                   . == "Proficient" ~ '4', 
                   TRUE ~ NA_character_))


#     x    y z
#1    0    1 0
#2    0    3 4
#3    1    4 3
#4    1    3 2
#5    2    0 0
#6 <NA>    4 3
#7    4 <NA> 3
#8    3    2 2

但是，最终输出包含字符列，因为我们的原始列也是字符。如果需要，我们可以添加mutate_all(as.numeric) 将它们转换为数字。

数据

x <- c("Low Outlier", "High Outlier", "Novice", "Novice", "Emerging", NA, 
      "Proficient", "Approaching")
y <- c("Novice", "Approaching", "Proficient", "Approaching", "High Outlier", 
      "Proficient",NA, "Emerging")
z <- c("High Outlier", "Proficient", "Approaching", "Emerging", "Low Outlier", 
      "Approaching", "Approaching", "Emerging")
sam <- data.frame(x,y,z, stringsAsFactors = FALSE)

【讨论】：

【解决方案4】：

我会使用命名向量作为映射

library(dplyr)
mapping = c("High Outlier" = 0, "Low Outlier" = 0, "Novice" = 1, "Emerging" = 2, "Approaching" = 3, "Proficient" = 4)

sam %>% 
  as.data.frame() %>% 
  mutate_all(function(i) mapping[i])

【讨论】：

【解决方案5】：

另一个使用factors重新编码，approxfun赋值的解决方案：

sam[] <- approxfun(1:5, c(0:3, 0))(
      as.numeric(factor(sam, 
                        c("Low Outlier", "Novice",  
                          "Emerging", "Approaching", 
                          "Proficient", "High Outlier"))))

#      x   y   z  
# [1,] "0" "1" NA 
# [2,] NA  "3" "0"
# [3,] "1" "0" "3"
# [4,] "1" "3" "2"
# [5,] "2" NA  "0"
# [6,] NA  "0" "3"
# [7,] "0" NA  "3"
# [8,] "3" "2" "2"

【讨论】：