将posix时间加一年

本文介绍了将posix时间加一年的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个类似"2016-01-01" (YYYY-MM-DD)的日期，并且我正在使用 as.numeric(as.POSIXct(...))用作整数.

I have a date like "2016-01-01" (YYYY-MM-DD) and I'm using as.numeric(as.POSIXct(...)) to use it as an Integer.

我的问题是，有没有办法将这个日期添加一年，一个月或一天?我的意思是，如果我在2016年增加一年，就不会与2015年(双性恋的东西)增加一年相同.

My question is, is there a way to add to this date a year, a month or a day ? I mean, if I add one year to 2016 it wont be the same as adding a year to 2015 (bissextile stuff).

与1月1日增加32天相同，与2月1日增加32天相同(因为可能会更改的天数)

Same as adding 32 days to January 01 wont be the same as adding 32 days to February 01 (because of number of days that may change)

我设法使它工作了好几年甚至几个月，但我也想实施天

I managed to have something that works for years and months but I'd like to implement days as well

how_long_is_simul <- function(lst){ # Format DATE_START greg_date = intDate_to_gregorianDate(DATE_START) reg = "^([0-9]{4})\\-([0-9]{2})\\-([0-9]{2})$" # black-magic splited_date = str_match(greg_date, reg) # Manage months limit lst$years = lst$years + floor(lst$months/12) lst$months = lst$months%%12 # Build new date my_vector = c(lst$years, lst$months, 0) end_date = paste(as.numeric(splited_date[2:4]) + my_vector, collapse = "-") return(round((gregorianDate_to_intDate(end_date)-DATE_START)/86400)) } # AND the vars used by the function DATE_START <- 1451606400 # 2016-01-01 GMT lst = list( # no days, because of bissextile years years = 1, months = 0 )

基本上我在做什么是从 DATE_START 整数将其转换为公历，然后从 lst 添加月/年，然后重建一个干净的字符串并将其转换为整数

Basically what I'm doing is from a DATE_START integer transform it into gregorian then add the months / years from lst then rebuild a clean string and reconvert it to integer.

转换int< ---> gregorian是使用POSIXct

N.B. Conversion int <---> gregorian are done with POSIXct

我不确定我的解释是否很好，但是还是要谢谢你:)

I'm not sure if I explained it well, but thank you anyway :)

推荐答案

使用lubridate软件包中的％m +％添加日期，月份或年份非常简单:

Adding a day, month or year is quite straightforward with %m+% from the lubridate package:

library(lubridate) x <- as.Date("2016-01-01") x %m+% days(1)

给出:

[1] "2016-01-02"

要添加月份或年份，可以使用 months 或 years 代替 days :

For adding months or years, you can use months or years instead of days:

> x %m+% months(1) [1] "2016-02-01" > x %m+% years(1) [1] "2017-01-01"