string 如何从日期字符变量中删除时间字段字符串?
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How to remove time-field string from a date-as-character variable?
提问by David Z
Suppose I have a variable like this
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")
what's a quick way to remove all 0:00:00
s so that
什么是删除所有0:00:00
s的快速方法,以便
c
[1] "9/21/2011" "9/25/2011" "10/2/2011" "9/28/2011" "9/27/2011"
回答by digEmAll
You can turn them into dates and then format as desired, e.g.:
您可以将它们转换为日期,然后根据需要进行格式化,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"
Or, you can simply remove the " 0:00:00"
substring using gsub:
或者,您可以简单地" 0:00:00"
使用 gsub删除子字符串:
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"
回答by Kayle Sawyer
From the lubridate package: Use mdy_hms()
to read in the characters as Month, Day, Year and Hours, Minutes, Seconds, then wrap with as.Date()
to strip the time.
从 lubridate 包中:mdy_hms()
用于读取字符为月、日、年和小时、分钟、秒,然后用 包裹as.Date()
以剥离时间。
library(lubridate)
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 <- as.Date(mdy_hms(v))
v
# [1] "2011-09-21" "2011-09-25" "2011-10-02" "2011-09-28" "2011-09-27"
If you want to maintain the vector as character type, not date type:
如果要将向量保持为字符类型,而不是日期类型:
v <- as.character(as.Date(mdy_hms(v)))