顺序日期的格式（带后缀-st，-nd，-rd，-th的月份中的日期） [英] Format for ordinal dates (day of month with suffixes -st, -nd, -rd, -th)

问题描述

我错过了什么吗？我不知道如何将以下内容转换为 Date s，即每月的哪一天（％d ）的后缀为 -st ， -nd ， -rd ， -th ：

Am I missing something? I can't figure out how to convert the following to Dates, where day of the month (%d) has the ordinal suffixes -st, -nd, -rd, -th:

ord_dates <- c("September 1st, 2016", "September 2nd, 2016",
               "September 3rd, 2016", "September 4th, 2016")

？strptime 似乎没有列出序数后缀的简写，并且不能自动处理：

?strptime doesn't appear to list a shorthand for the ordinal suffix, and it isn't handled automagically:

as.Date(ord_dates, format = c("%B %d, %Y"))
#[1] NA NA NA NA

是否存在用于处理 format 参数中被忽略字符的令牌？我想念的令牌是什么？

Is there a token for handling ignored characters in the format argument? A token I'm missing?

我能想到的最好的方法是（可能有一个较短的正则表达式，但思路相同）：

Best I can come up with is (there may a shorter regex, but same idea):

as.Date(gsub("([0-9]+)(st|nd|rd|th)", "\\1", ord_dates), format = "%B %d, %Y")
# [1] "2016-09-01" "2016-09-02" "2016-09-03" "2016-09-04"

这类数据似乎相对常见；我是否缺少某些东西？

Seems like this sort of data should be relatively common; am I missing something?

推荐答案

享受 lubridate 的力量：

library(lubridate)    
mdy(ord_dates)

[1] "2016-09-01" "2016-09-02" "2016-09-03" "2016-09-04"

在内部， lubridate 没有任何启用此功能的特殊转换规范。相反， lubridate 首先（通过智能猜测）使用格式％B％dst，％Y 。这将得到 ord_dates 的第一个元素。

Internally, lubridate doesn't have any special conversion specifications which enable this. Rather, lubridate first uses (by smart guessing) the format "%B %dst, %Y". This gets the first element of ord_dates.

然后检查 NA s并在其余元素上重复其智能猜测，对％B％dnd，％Y 进行设置以获得第二个元素。它以这种方式继续下去，直到没有 NA 剩下了（在这种情况下，在4次迭代之后就会发生），或者直到它的聪明的猜测未能找到可能的格式候选者为止。

It then checks for NAs and repeats its smart guessing on the remaining elements, settling on "%B %dnd, %Y" to get the second element. It continues in this way until there are no NAs left (which happens in this case after 4 iterations), or until its smart guessing fails to turn up a likely format candidate.

您可以想像，这会使 lubridate 变慢，而且的确-仅使用智能手表的速度的一半上面@alistaire建议的正则表达式：

You can imagine this makes lubridate slower, and it does -- about half the speed of just using the smart regex suggested by @alistaire above:

set.seed(109123)
ord_dates <- sample(
  c("September 1st, 2016", "September 2nd, 2016",
    "September 3rd, 2016", "September 4th, 2016"),
  1e6, TRUE
  )

library(microbenchmark)

microbenchmark(times = 10L,
               lubridate = mdy(ord_dates),
               base = as.Date(sub("\\D+,", "", ord_dates),
                              format = "%B %e %Y"))
# Unit: seconds
#       expr      min       lq     mean   median       uq      max neval cld
#  lubridate 2.167957 2.219463 2.290950 2.252565 2.301725 2.587724    10   b
#       base 1.183970 1.224824 1.218642 1.227034 1.228324 1.229095    10  a 

<$ c中的明显优势$ c> lubridate 的优点是其简洁性和灵活性。

The obvious advantage in lubridate's favor being its conciseness and flexibility.

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