是否有更优雅的方式将两位数的年份转化为四位数的年份？ [英] Is there a more elegant way to convert two-digit years to four-digit years with lubridate?

问题描述

如果一个日期向量有两位数的年份，那么 mdy（）将年龄介于00和68之间，成为二十一世纪和二十世纪六十九年九十九岁之间。例如：

If a date vector has two-digit years, mdy() turns years between 00 and 68 into 21st Century years and years between 69 and 99 into 20th Century years. For example:

library(lubridate)    
mdy(c("1/2/54","1/2/68","1/2/69","1/2/99","1/2/04"))

提供以下输出：

Multiple format matches with 5 successes: %m/%d/%y, %m/%d/%Y.
Using date format %m/%d/%y.
[1] "2054-01-02 UTC" "2068-01-02 UTC" "1969-01-02 UTC" "1999-01-02 UTC" "2004-01-02 UTC"

我可以解决这个事实后，从错误的日期减去100，将2054年和2068年变成1954年和1968年。但是，有一个更优雅，更容易出错的解析两位数日期的方法，以便在解析过程中得到正确处理？

I can fix this after the fact by subtracting 100 from the incorrect dates to turn 2054 and 2068 into 1954 and 1968. But is there a more elegant and less error-prone method of parsing two-digit dates so that they get handled correctly in the parsing process itself?

更新： 之后@JoshuaUlrich指向我 strptime 我发现这个问题，它处理类似于我的问题，但是使用基数R.

Update: After @JoshuaUlrich pointed me to strptime I found this question, which deals with an issue similar to mine, but using base R.

这似乎是一个很好的除了日期处理R将在解析函数的日期中处理世纪选择限制两位数的日期。

It seems like a nice addition to date handling in R would be some way to handle century selection cutoffs for two-digit dates within the date parsing functions.

推荐答案

这是一个允许您执行此操作的功能：

Here is a function that allows you to do this:

library(lubridate)
x <- mdy(c("1/2/54","1/2/68","1/2/69","1/2/99","1/2/04"))


foo <- function(x, year=1968){
  m <- year(x) %% 100
  year(x) <- ifelse(m > year %% 100, 1900+m, 2000+m)
  x
}

尝试一下：

x
[1] "2054-01-02 UTC" "2068-01-02 UTC" "1969-01-02 UTC" "1999-01-02 UTC"
[5] "2004-01-02 UTC"

foo(x)
[1] "2054-01-02 UTC" "2068-01-02 UTC" "1969-01-02 UTC" "1999-01-02 UTC"
[5] "2004-01-02 UTC"

foo(x, 1950)
[1] "1954-01-02 UTC" "1968-01-02 UTC" "1969-01-02 UTC" "1999-01-02 UTC"
[5] "2004-01-02 UTC"

---

这里的魔法是使用模数运算符 %% 返回部门的分数部分。所以 1968 %% 100 产生68。

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