将四位数的年份值转换为日期类 [英] Convert four digit year values to class Date

问题描述

我的数据集中有一个整数列，该列具有四位数的年份值，例如:

I've an integer column in my dataset which has four digit year values, like:

 c(2001, 2002, 2002, 2002, 2003, 2005) 

我尝试使用 as.Date 将四位数的年份转换为 Date 类:

I try to convert the four digit year to class Date using as.Date:

year <- as.Date(as.character(data_file$evtYear), format = "%Y")

但是输出是:

"2001-05-15" "2002-05-15" "2002-05-15" "2002-05-15" "2003-05-15" "2005-05-15"

这给出了错误的输出.它在一个日期(2001年也是15年)中给出了两年的值.

This is giving the wrong output. It's giving two year values in one date (both 2001 and also 15).

我只想将我的四位数年份部分从原始数据转换为年"作为类 Date .预期的输出结果很简单:

I just want the convert my four digit year part from the original data to 'Year' as class Date. Expected out put is simply:

2001 2002 2002 2002 2003 2005 

但是他们的班级应该是 Date 类型.

But their class should be of Date type.

如何在R中实现这一目标?

How to achieve this in R?

推荐答案

基于评论，事实证明，提出问题的人不需要将数字年份更改为日期" 类;但是，问题被问到如何做，所以这里是一个答案.

Based on the comments it turned out that the person asking the question did not need to change a numeric year to "Date" class; nevertheless, the question asked how to do it so here is an answer.

以下是从4位数字年份创建日期" 类对象的几种方法.所有人都使用 as.Date :

Here are a few ways to create a "Date" class object from a 4 digit numeric year. All use as.Date:

yrs <- c(2001, 2002, 2002, 2002, 2003, 2005)

1)ISOdate

as.Date(ISOdate(yrs, 1, 1))  # beginning of year
as.Date(ISOdate(yrs, 12, 31))  # end of year

此ISOdate解决方案有点棘手，因为它创建了一个中间POSIXct对象，因此可能存在时区问题.您可能更喜欢以下其中一种.

This ISOdate solution is a bit tricky because it creates an intermediate POSIXct object so time zone problems could exist. You might prefer one of the following.

2)粘贴

as.Date(paste(yrs, 1, 1, sep = "-")) # beginning of year
as.Date(paste(yrs, 12, 31, sep = "-")) # end of year

3)动物园:: as.yearmon

library(zoo)

as.Date(as.yearmon(yrs)) # beginning of year
as.Date(as.yearmon(yrs) + 11/12, frac = 1) # end of year

注意:如果上述任何一项的结果都是 y ，则 format(y，％Y")给出字符年份和 as.numeric(format(y，％Y"))给出数字年份.

Note: If y is the result for any of the above then format(y, "%Y") gives the character year and as.numeric(format(y, "%Y")) gives the numeric year.

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