从R中的strptime()字符串中提取时间,返回NA [英] Extracting time from character string with strptime() in R, returning NA
问题描述
我正在尝试从R中的字符串中提取出时间,因此不能停止获取NA。我已经尝试了许多变体的正则表达式标签,但似乎无法解决这个简单的问题。任何帮助/澄清表示赞赏。
以下是我的代码示例:
> x
[1]2/7/2013 7:43
> class(x)
[1]character
> z > z
[1] NA
不知道你的字符串是一个datetime。所以做一个第一:
y< - strptime(x,format ='%m /%d /%Y% H:%M')
如果你试图只得到日期,你可以这样做: / p>
strptime(x,'%m /%d /%Y')
/ pre>
因为
strptime
丢弃任何超出您的格式字符串的额外字符,但您不能抓住尾随%H:%M
因为该函数不知道从哪里开始。
一旦它是一个正确的datetime类,你可以做的事情:
strftime(y,'%H:%M')
我更喜欢使用
as.POSIXlt
而不是strptime
和格式
而不是strftime
...但它们大致相当。 >I am trying to extract out the time from a character string in R and can't stop getting NA as a result. I have tried numerous variations of the regular expression tags, but can't seem to get around this simple problem. Any help/clarifications are appreciated.
Here is my code for an example:
> x [1] "2/7/2013 7:43" > class(x) [1] "character" > z <- strptime(x, "%H:%M") > z [1] NA
解决方案R doesn't know that your string is a datetime. So make it one first:
y <- strptime(x, format='%m/%d/%Y %H:%M')
If you were trying to get just the date, you could do:
strptime(x, '%m/%d/%Y')
Because
strptime
discards any extra characters past your format string, but you cannot grab the trailing%H:%M
because the function doesn't know where to start.Once it is a proper datetime class, you can do things to it:
strftime(y, '%H:%M')
I prefer to use
as.POSIXlt
rather thanstrptime
andformat
instead ofstrftime
... but they are roughly equivalent.这篇关于从R中的strptime()字符串中提取时间,返回NA的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!