假设我有一个这样的变量
c<-c("9/21/2011 0:00:00", "9/25/2011 0:00:00", "10/2/2011 0:00:00",
"9/28/2011 0:00:00", "9/27/2011 0:00:00")
有什么快速方法可以删除所有 0:00:00
以便
c
[1] "9/21/2011" "9/25/2011" "10/2/2011" "9/28/2011" "9/27/2011"
最佳答案
您可以将它们转换为日期,然后根据需要设置格式,例如:
v <- c("9/21/2011 0:00:00", "9/25/2011 0:00:00", "10/2/2011 0:00:00",
"9/28/2011 0:00:00", "9/27/2011 0:00:00")
v <- format(as.POSIXct(v,format='%m/%d/%Y %H:%M:%S'),format='%m/%d/%Y')
> v
[1] "09/21/2011" "09/25/2011" "10/02/2011" "09/28/2011" "09/27/2011"
或者,您可以简单地使用 gsub 删除"0:00:00"
子字符串:
v <- gsub(x=v,pattern=" 0:00:00",replacement="",fixed=T)
> v
[1] "9/21/2011" "9/25/2011" "10/2/2011" "9/28/2011" "9/27/2011"
关于string - 如何从日期字符变量中删除时间字段字符串?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23089895/