R从数据框中的字符串中删除单词

Question

假设我有以下数据集：

Date_Received = c("Addition 1/2/2018", "Swimming Pool 1/8/2018", "Abandonment 1/9/2018", "Existing Approval 3/14/2018", "Holding Tank 5/11/2018")

Date_Approved = c("1/2/2018", "1/8/2018", "1/9/2018", "SB 3/21/2018", "JW 5/11/2018")

我想删除Date_Received列中date之前的characters ，以便以后可以使用lubridate将其转换为date类型数据格式。

我尝试使用以下代码，但它只删除了first大写字母。

我怎样才能解决这个问题？

期望的输出：

Date_Received Date_Approved 
1/2/2018      1/2/2018
1/8/2018      1/8/2018
1/9/2018      1/9/2018
3/14/2018     SB 3/21/2018
5/11/2018     JW 5/11/2018

代码

library(tidyverse)

df = data.frame(Date_Received, Date_Approved)

df= df%>% mutate(Date.Received = trimws(Date_Received, whitespace = "[A-Z]*\\s*")) %>% filter(nzchar(Date.Received)) 

Answer 1

我们可以使用trimws ，它有一个空格参数（如您在代码中使用的那样），可用于指定空格。

library(dplyr)

df %>% 
  mutate(Date_Received = trimws(Date_Received, "left", "\\D"))

或使用str_replace_all ：

library(stringr)

df %>% 
  mutate(Date_Received = str_replace_all(Date_Received, "^\\D+", ""))

输出

  Date_Received Date_Approved
1      1/2/2018      1/2/2018
2      1/8/2018      1/8/2018
3      1/9/2018      1/9/2018
4     3/14/2018  SB 3/21/2018
5     5/11/2018  JW 5/11/2018

使用sub的另一个选项：

df$Date_Received <- sub("^\\D+", "", df$Date_Received)

Answer 2

让生活简单：

Date_Received = c("Addition 1/2/2018", "Swimming Pool 1/8/2018", "Abandonment 1/9/2018", "Existing Approval 3/14/2018", "Holding Tank 5/11/2018")
stringr::word(Date_Received, -1)
[1] "1/2/2018"  "1/8/2018"  "1/9/2018"  "3/14/2018" "5/11/2018"