Question

D <- "06.12.1948"                 # which is dd.mm.yyyy
as.Date(D, "%d.%m.%y")            # convert to date
[1] "2019-12-06"                  # ????

我错过了什么？

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”

Answer 1

格式区分大小写（“％y”不明确且系统依赖，我相信）：

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’.

Answer 2

为避免记住日期格式，我们可以使用打包的解决方案。

1）使用lubridate

lubridate::dmy(D)
#[1] "1948-12-06"

2）使用anytime

anytime::anydate(D)
#[1] "1948-06-12"

Answer 3

可能会对某人有所帮助。我在Cole Beck的教程“在R中处理日期时间”中找到了此功能。该功能可以识别您的数据格式。

# FUNCTION guessDateFormat @x vector of character dates/datetimes @returnDates return
# actual dates rather than format convert character datetime to POSIXlt datetime, or 
# at least guess the format such that you could convert to datetime
guessDateFormat <- function(x, returnDates = FALSE, tzone = "") {
x1 <- x
# replace blanks with NA and remove
x1[x1 == ""] <- NA
x1 <- x1[!is.na(x1)]
if (length(x1) == 0)
  return(NA)
# if it's already a time variable, set it to character
if ("POSIXt" %in% class(x1[1])) {
x1 <- as.character(x1)
}
dateTimes <- do.call(rbind, strsplit(x1, " "))
for (i in ncol(dateTimes)) {
dateTimes[dateTimes[, i] == "NA"] <- NA
}
# assume the time part can be found with a colon
timePart <- which(apply(dateTimes, MARGIN = 2, FUN = function(i) {
                        any(grepl(":", i))
                        }))
# everything not in the timePart should be in the datePart
datePart <- setdiff(seq(ncol(dateTimes)), timePart)
# should have 0 or 1 timeParts and exactly one dateParts
if (length(timePart) > 1 || length(datePart) != 1)
  stop("cannot parse your time variable")
timeFormat <- NA
if (length(timePart)) {
# find maximum number of colons in the timePart column
ncolons <- max(nchar(gsub("[^:]", "", na.omit(dateTimes[, timePart]))))
if (ncolons == 1) {
timeFormat <- "%H:%M"
} else if (ncolons == 2) {
timeFormat <- "%H:%M:%S"
} else stop("timePart should have 1 or 2 colons")
}
# remove all non-numeric values
dates <- gsub("[^0-9]", "", na.omit(dateTimes[, datePart]))
# sep is any non-numeric value found, hopefully / or -
sep <- unique(na.omit(substr(gsub("[0-9]", "", dateTimes[, datePart]), 1, 1)))
if (length(sep) > 1)
  stop("too many seperators in datePart")
# maximum number of characters found in the date part
dlen <- max(nchar(dates))
dateFormat <- NA
# when six, expect the century to be omitted
if (dlen == 6) {
if (sum(is.na(as.Date(dates, format = "%y%m%d"))) == 0) {
dateFormat <- paste("%y", "%m", "%d", sep = sep)
} else if (sum(is.na(as.Date(dates, format = "%m%d%y"))) == 0) {
dateFormat <- paste("%m", "%d", "%y", sep = sep)
} else stop("datePart format [six characters] is inconsistent")
}else if (dlen == 8) {
if (sum(is.na(as.Date(dates, format = "%Y%m%d"))) == 0) {
dateFormat <- paste("%Y", "%m", "%d", sep = sep)
} else if (sum(is.na(as.Date(dates, format = "%m%d%Y"))) == 0) {
dateFormat <- paste("%m", "%d", "%Y", sep = sep)
} else stop("datePart format [eight characters] is inconsistent")
} else {
stop(sprintf("datePart has unusual length: %s", dlen))
}
if (is.na(timeFormat)) {
format <- dateFormat
} else if (timePart == 1) {
format <- paste(timeFormat, dateFormat)
} else if (timePart == 2) {
format <- paste(dateFormat, timeFormat)
} else stop("cannot parse your time variable")
if (returnDates)
  return(as.POSIXlt(x, format = format, tz = tzone))
format
}

# generate some dates
mydates <- format(as.POSIXct(sample(31536000, 20), origin = "2011-01-01", tz = "UTC"), "%m.%d.%Y %H:%M")

mydates
## [1] "02/07/2011 06:51" "11/21/2011 17:03" "09/17/2011 22:42" "02/16/2011 13:45"
## [5] "12/14/2011 19:11" "09/08/2011 09:22" "12/06/2011 14:06" "02/02/2011 11:00"
## [9] "03/27/2011 06:12" "01/05/2011 15:09" "04/15/2011 04:17" "10/20/2011 14:20"
## [13] "11/13/2011 21:46" "02/26/2011 03:24" "12/29/2011 11:02" "03/17/2011 02:24"
## [17] "02/27/2011 13:51" "06/27/2011 08:36" "03/14/2011 10:54" "01/28/2011 14:14"
guessDateFormat(mydates)

[1]“％m。％d。％Y％H：％M”

将字符串转换为日期，格式为：“dd.mm.yyyy”

3 个答案: