在日期中添加一个月

Question

我想把一个月添加到我拥有的日期.但到目前为止它不可能以直接的方式进行.以下是我试过的.

d <- as.Date("2004-01-31")
d + 60
# [1] "2004-03-31"

添加不会有帮助,因为月份不会重叠.

seq(as.Date("2004-01-31"), by = "month", length = 2) 
# [1] "2004-01-31" "2004-03-02"

以上可能有用,但同样不是直截了当.此外它还增加了30天或某些东西到日期,如下所示的问题

seq(as.Date("2004-01-31"), by = "month", length = 10) 
#  [1] "2004-01-31" "2004-03-02" "2004-03-31" "2004-05-01" "2004-05-31" "2004-07-01" "2004-07-31" "2004-08-31" "2004-10-01" "2004-10-31"

在上面,对于前2个日期,月份没有变化.

以下方法也在一个月内失败,但是一年的成功

d <- as.POSIXlt(as.Date("2010-01-01"))
d$year <- d$year +1
d
# [1] "2011-01-01 UTC"
d <- as.POSIXlt(as.Date("2010-01-01"))
d$month <- d$month +1
d

错误format.POSIXlt(x, usetz = TRUE):无效'x'参数

这样做的正确方法是什么？

Answer 1

%m+%来自lubridate的功能增加一个月而不超过新月的最后一天.

library(lubridate)
(d <- ymd("2012-01-31"))
 1 parsed with %Y-%m-%d
[1] "2012-01-31 UTC"
d %m+% months(1)
[1] "2012-02-29 UTC"

Answer 2

当你说"为一个日期添加一个月"时,这是不明确的.

你的意思是

1. 加30天？
2. 将月份的一部分增加1？

在这两种情况下,简单添加的整个包装似乎有点夸张.

当然,对于第一点,简单的+运算符将执行:

d=as.Date('2010-01-01') 
d + 30 
#[1] "2010-01-31"

至于第二个,我只会创建一个简单的单行函数(并且具有更通用的范围):

add.months= function(date,n) seq(date, by = paste (n, "months"), length = 2)[2]

您可以使用任意月份,包括否定:

add.months(d, 3)
#[1] "2010-04-01"
add.months(d, -3)
#[1] "2009-10-01"

当然,如果您只想添加,通常只需一个月:

add.month=function(date) add.months(date,1)
add.month(d)
#[1] "2010-02-01"

如果你在1月31日增加一个月,那么从2月31日开始就没有意义了,最好完成工作就是将缺少的3天添加到下个月,即3月.如此正确:

add.month(as.Date("2010-01-31"))
#[1] "2010-03-03"

如果由于一些非常特殊的原因,你需要在一个月的最后一天提供一个上限,它会更长一些:

add.months.ceil=function (date, n){

  #no ceiling
  nC=add.months(date, n)

  #ceiling
  day(date)=01
  C=add.months(date, n+1)-1

  #use ceiling in case of overlapping
  if(nC>C) return(C)
  return(nC)
}

像往常一样,你可以添加一个月的版本:

add.month.ceil=function(date) add.months.ceil(date,1)    

所以:

  d=as.Date('2010-01-31')
  add.month.ceil(d)
  #[1] "2010-02-28"
  d=as.Date('2010-01-21')
  add.month.ceil(d)
  #[1] "2010-02-21"

随着减少:

  d=as.Date('2010-03-31')
  add.months.ceil(d, -1)
  #[1] "2010-02-28"
  d=as.Date('2010-03-21')
  add.months.ceil(d, -1)
  #[1] "2010-02-21"

此外,您没有告诉您是否对标量或矢量解决方案感兴趣.至于后者:

add.months.v= function(date,n) as.Date(sapply(date, add.months, n), origin="1970-01-01")

注意:*applyfamily会破坏类数据,这就是为什么必须重建它.矢量版本带来:

d=c(as.Date('2010/01/01'), as.Date('2010/01/31'))
add.months.v(d,1)
[1] "2010-02-01" "2010-03-03"

希望你喜欢它))

Answer 3

Vanilla R有一个天真的difftime类,但Lubridate CRAN包让你做你想要的:

require(lubridate)
d <- as.Date('2004-01-01')
month(d) <- month(d) + 1
day(d) <- days_in_month(d)
d
[1] "2004-02-29"

希望有所帮助.

Answer 4

"mondate"有点类似于"Date"添加n添加n月而不是n天:

> library(mondate)
> d <- as.Date("2004-01-31")
> as.mondate(d) + 1
mondate: timeunits="months"
[1] 2004-02-29

Answer 5

最简单的方法是将Date转换为POSIXlt格式.然后执行如下算术运算:

date_1m_fwd     <- as.POSIXlt("2010-01-01")
date_1m_fwd$mon <- date_1m_fwd$mon +1

此外,如果您想要处理data.table中的Date列,不幸的是,不支持POSIXlt格式.

您仍然可以使用基本R代码执行添加月份,如下所示:

library(data.table)  
dt <- as.data.table(seq(as.Date("2010-01-01"), length.out=5, by="month"))  
dt[,shifted_month:=tail(seq(V1[1], length.out=length(V1)+3, by="month"),length(V1))]

希望能帮助到你.

Answer 6

这是一个不需要安装任何软件包的函数。您为它提供一个Date对象（或character它可以转换为的对象Date），它会n向该日期添加月份，而不会更改该月的某天（除非您所登陆的月份没有足够的天数，在这种情况下，默认为返回月份的最后一天）。以防万一没有意义的阅读，下面有一些示例。

功能定义

addMonth <- function(date, n = 1){
  if (n == 0){return(date)}
  if (n %% 1 != 0){stop("Input Error: argument 'n' must be an integer.")}

  # Check to make sure we have a standard Date format
  if (class(date) == "character"){date = as.Date(date)}

  # Turn the year, month, and day into numbers so we can play with them
  y = as.numeric(substr(as.character(date),1,4))
  m = as.numeric(substr(as.character(date),6,7))
  d = as.numeric(substr(as.character(date),9,10))

  # Run through the computation
  i = 0
  # Adding months
  if (n > 0){
    while (i < n){
      m = m + 1
      if (m == 13){
        m = 1
        y = y + 1
      }
      i = i + 1
    }
  }
  # Subtracting months
  else if (n < 0){
    while (i > n){
      m = m - 1
      if (m == 0){
        m = 12
        y = y - 1
      }
      i = i - 1
    }
  }

  # If past 28th day in base month, make adjustments for February
  if (d > 28 & m == 2){
      # If it's a leap year, return the 29th day
      if ((y %% 4 == 0 & y %% 100 != 0) | y %% 400 == 0){d = 29}
      # Otherwise, return the 28th day
      else{d = 28}
    }
  # If 31st day in base month but only 30 days in end month, return 30th day
  else if (d == 31){if (m %in% c(1, 3, 5, 7, 8, 10, 12) == FALSE){d = 30}}

  # Turn year, month, and day into strings and put them together to make a Date
  y = as.character(y)

  # If month is single digit, add a leading 0, otherwise leave it alone
  if (m < 10){m = paste('0', as.character(m), sep = '')}
  else{m = as.character(m)}

  # If day is single digit, add a leading 0, otherwise leave it alone
  if (d < 10){d = paste('0', as.character(d), sep = '')}
  else{d = as.character(d)}

  # Put them together and convert return the result as a Date
  return(as.Date(paste(y,'-',m,'-',d, sep = '')))
}

一些例子

加数个月

> addMonth('2014-01-31', n = 1)
[1] "2014-02-28"  # February, non-leap year
> addMonth('2014-01-31', n = 5)
[1] "2014-06-30"  # June only has 30 days, so day of month dropped to 30
> addMonth('2014-01-31', n = 24)
[1] "2016-01-31"  # Increments years when n is a multiple of 12 
> addMonth('2014-01-31', n = 25)
[1] "2016-02-29"  # February, leap year

减去月份

> addMonth('2014-01-31', n = -1)
[1] "2013-12-31"
> addMonth('2014-01-31', n = -7)
[1] "2013-06-30"
> addMonth('2014-01-31', n = -12)
[1] "2013-01-31"
> addMonth('2014-01-31', n = -23)
[1] "2012-02-29"

Answer 7

addedMonth <- seq(as.Date('2004-01-01'), length=2, by='1 month')[2]
addedQuarter <- seq(as.Date('2004-01-01'), length=2, by='1 quarter')[2]