从R中给定的字符串中提取日期 [英] Extract date from a given string in R

问题描述

这是我拥有的一个字符串

Here is a string that I have

"7MA_S_VE_MS_FB_MEASURE_P1_2013-08-21_17-42-19.BMP"

我正在尝试以这种方式提取日期:

I am trying to extract dates this way:

library(stringr)
as.Date(str_extract(test,"[0-9]{4}/[0-9]{2}/[0-9]{2}"),"%Y-%m-%d")

我为此不适用.

所需的输出为

2013-08-21

有人可以指出我正确的方向吗?

Can someone point me in the right direction?

推荐答案

您已在正则表达式中将破折号-替换为斜线/.

You have replaced your dash - with a slash / in your regular expression.

as.Date(str_extract(string, "[0-9]{4}-[0-9]{2}-[0-9]{2}"), format="%Y-%m-%d")
# [1] "2013-08-21"

但是您也可以将[0-9]位替换为\d，它们代表同一件事.我不确定为什么，但是正则表达式专业人士似乎总是使用\d版本(请注意，您必须使用另一个反斜杠来转义反斜杠):

But you can also replace the [0-9] bits with \d, which represent the same thing. I'm not sure why, but regex pros seem to always use the \d version (note that you'll have to escape the backslash with another backslash):

as.Date(str_extract(string, "\\d{4}-\\d{2}-\\d{2}"), format="%Y-%m-%d")
# [1] "2013-08-21"

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