Question

我有一个基本上我想要清理的电子邮件列表。我想声明如果'@'字符不在特定电子邮件中，我想删除该电子邮件 - 这样就会删除像'mywebsite.com'这样的输入。

我的代码如下：

  email_clean <- function(email, invalid = NA){
    email <- trimws(email)                                                          # Removes whitespace
    email[(nchar(email) %in% c(1,2)) ] <- invalid                                   # Removes emails with 1 or 2 character length
    bad_email <- c("\\@no.com", "\\@na.com","\\@none.com","\\@email.com",           # List of bad emails - modify to the 
                   "\\@noemail.com", "\\@test.com",                                 # specifications of the request

    pattern = paste0("(?i)\\b",paste0(bad_email,collapse="\\b|\\b"),"\\b")          # Deletes names matching bad email
    email <-gsub(pattern, invalid, sapply(email,as.character))
    unname(email)
  }

  ## Define vector of SSN from origianl csv column
  Cleaned_Email <- email_clean(my_data$Email)


  ## Binds cleaned phone to csv
  my_data<-cbind(my_data,Cleaned_Email)

谢谢！

Answer 1

  email_clean <- function(email, invalid = NA){
    email <- trimws(email)                                                          # Removes whitespace
    email[(nchar(email) %in% c(1,2)) ] <- invalid                                   # Removes emails with 1 or 2 character length
    email[!grepl("@", email)] <- invalid  # <------------------ New line added here ------------
    bad_email <- c("\\@no.com", "\\@na.com","\\@none.com","\\@email.com",           # List of bad emails - modify to the 
                   "\\@noemail.com", "\\@test.com",                                 # specifications of the request

    pattern = paste0("(?i)\\b",paste0(bad_email,collapse="\\b|\\b"),"\\b")          # Deletes names matching bad email
    email <-gsub(pattern, invalid, sapply(email,as.character))
    unname(email)
  }

Answer 2

尝试此选项可以排除my_data中没有＆＃39; @＆＃39;的所有行。登录电子邮件列：

my_data <- my_data[grep('@', my_data$Email), ]

R部分字符串匹配 - 排除

2 个答案: