D <- "06.12.1948" # which is dd.mm.yyyy
as.Date(D, "%d.%m.%y") # convert to date
[1] "2019-12-06" # ????
我想念的是什么?
Sys.getlocale(category =“ LC_ALL”)[1]“ LC_COLLATE = German_Austria.1252; LC_CTYPE = German_Austria.1252; LC_MONETARY = German_Austria.1252; LC_NUMERIC = C; LC_TIME = German_Austria.1252”
格式区分大小写(我相信“%y”是模棱两可的,并且取决于系统):
as.Date(D, "%d.%m.%Y")
[1] "1948-12-06"
帮助主题?strptime
包含详细信息:
‘%y’ Year without century (00-99). On input, values 00 to 68 are
prefixed by 20 and 69 to 99 by 19 - that is the behaviour
specified by the 2004 and 2008 POSIX standards, but they do
also say ‘it is expected that in a future version the default
century inferred from a 2-digit year will change’.
要对此进行扩展,
"%y"
是2位数字的年份,因此在1948年的“ 19”中进行了读取"%Y"
。