我给出三个参数，输入 df，我要清理的列，我要添加的新列，名称已清理。我哪里错了？

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

"Error: Column new_col_name must be a 1d atomic #vector or a list"

Answer 1

如果我们传递不带引号的列名，则使用

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

您要做的是确保您使用的函数的输出是一个向量或只有一维的列表，以便您可以将其作为新列添加到所需的数据框中。您可以使用基础包中的 Class 函数验证对象的 class。

mutate 函数本身应该做你想做的，它 returns 相同的数据框但具有新列：

     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项更改，但我认为这样更容易阅读。

我给出三个参数，输入 df，我要清理的列，我要添加的新列，名称已清理。我哪里错了？

I give three arguments, the input df, the column I want to clean,the new column I want to be added with cleansed names. Where am I going wrong?

regex

r

gsub

数据