Question

library(dplyr)
clean_name <- function(df,col_name,new_col_name){

 #remove whitespace and common titles.
 df$new_col_name <- mutate_all(df, 
                  trimws(gsub("MR.?|MRS.?|MS.?|MISS.?|MASTER.?","",df$col_name)))

 #remove any chunks of text where a number is present
 df$new_col_name<- transmute_all(df,
                  gsub("[^\\s]*[\\d]+[^\\s]*","",df$col_name,perl = TRUE))

}

我收到以下错误

“错误：列new_col_name必须是一维原子#vector或列表”

Answer 1

如果我们要传递未加引号的列名，请使用

library(tidyverse)
clean_name <- function(df,col_name, new_col_name){
   col_name <- enquo(col_name)
  new_col_name <- enquo(new_col_name)
   df %>% 
     mutate(!! new_col_name := 
     trimws(str_replace_all(!!col_name, "MR.?|MRS.?|MS.?|MISS.?|MASTER.?","")) ) %>%
     transmute(!! new_col_name := trimws(str_replace_all(!! new_col_name, 
              "[^\\s]*[\\d]+[^\\s]*","")))
     }


clean_name(dat1, col1, colN) 
#   colN
#1  one
#2  two

数据

dat1 <- data.frame(col1 = c("MR. one", "MS. two 24"), stringsAsFactors = FALSE)

Answer 2

您要做的是确保所使用函数的输出是矢量或仅具有一维的列表，以便可以将其作为新列添加到所需数据帧中。您可以使用基本包中的Class函数来验证对象的类。

mutate函数本身应该执行您想要的操作，它返回相同的数据帧但带有新列：

     library(dplyr)
     clean_name <- function(df, col_name, new_col_name) {

     # first_cleaning_to_colname = The first change you want to make to the col_name column. This should be a vector.
     # second_cleaning_to_colname = The change you're going to make to the col_name column after the first one. This should be a vector too.

     first_change <- mutate(df, col_name = first_cleaning_to_colname)

     second_change <- mutate(first_change, new_col_name = second_cleaning_to_colname)

     return(second_change)
     }

您可以同时进行这两个更改，但我认为这样更容易阅读。

我给了三个参数，输入df，我要清除的列，我要添加有清除名称的新列。我要去哪里错了？

2 个答案:

数据