如何从R中的日期中减去月份?

问题描述：

我正在尝试从日期中减去 n 个月，如下所示:

I'm trying to subtract n months from a date as follows:

maturity <- as.Date("2012/12/31")

m <- as.POSIXlt(maturity)

m$mon <- m$mon - 6

但结果日期是 01-Jul-2012，而不是 30-Jun-2012，正如我所期望的.有没有什么捷径可以得到这样的结果?

but the resulting date is 01-Jul-2012, and not 30-Jun-2012, as I should expect. Is there any short way to get such result?

提前致谢

答

1) seq.Date.请注意，6 月只有 30 天，因此它不能给出 6 月 31 日，而是给出了 7 月 1 日.

1) seq.Date. Note that June has only 30 days so it cannot give June 31st thus instead it gives July 1st.

seq(as.Date("2012/12/31"), length = 2, by = "-6 months")[2]
## [1] "2012-07-01"

如果我们知道是在月底，我们可以这样做:

If we knew it was at month end we could do this:

seq(as.Date(cut(as.Date("2012/12/31"), "month")), length=2, by="-5 month")[2]-1
## "2012-06-30"

2) 年月.此外，如果我们知道现在是月底，那么我们可以像这样使用动物园包的 "yearmon" 类:

2) yearmon. Also if we knew it was month end then we could use the "yearmon" class of the zoo package like this:

library(zoo)
as.Date(as.yearmon(as.Date("2012/12/31")) -.5, frac = 1)
## [1] "2012-06-30"

这会将日期转换为 "yearmon" 减去 6 个月(一年的 0.5)，然后使用 frac= 将其转换回 "Date"1 表示月底(frac=0 表示月初).与之前的解决方案相比，这还具有自动矢量化的优点，即 as.Date(...) 可以是日期矢量.

This converts the date to "yearmon" subtracts 6 months (.5 of a year) and then converts it back to "Date" using frac=1 which means the end of the month (frac=0 would mean the beginning of the month). This also has the advantage over the previous solution that it is vectorized automatically, i.e. as.Date(...) could have been a vector of dates.

请注意，如果 "Date" 类仅用作表示月份的一种方式，那么我们可以完全摆脱它并直接使用 "yearmon" 因为那建模我们首先想要的:

Note that if "Date" class is only being used as a way of representing months then we can get rid of it altogether and directly use "yearmon" since that models what we want in the first place:

as.yearmon("2012-12") - .5
## [1] "Jun 2012"

3) 星期一.第三种解决方案是 mondate 包，它的优点是它返回 6 个月前的月底，而不必知道我们是月底:

3) mondate. A third solution is the mondate package which has the advantage here that it returns the end of the month 6 months ago without having to know that we are month end:

library(mondate)
mondate("2011/12/31") - 6
## mondate: timeunits="months"
## [1] 2011/06/30

这也是矢量化的.

4) 润滑.此 lubridate 答案已根据包中的更改进行了更改:

4) lubridate. This lubridate answer has been changed in line with changes in the package:

library(lubridate)
as.Date("2012/12/31") %m-% months(6)
## [1] "2012-06-30"

lubridate 也是矢量化的.

lubridate is also vectorized.

5) sqldf/SQLite

library(sqldf)
sqldf("select date('2012-12-31', '-6 months') as date")
##         date
## 1 2012-07-01

或者如果我们知道我们在月底:

or if we knew we were at month end:

sqldf("select date('2012-12-31', '+1 day', '-6 months', '-1 day') as date")
##         date
## 1 2012-06-30

如何从R中的日期中减去月份?

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