How to Subtract Months from a Date in R

R subtracting 1 month from today's date gives NA

The calculation of months is indeed perfomed by base R but not the way your think. Months is used to get the month of a date object.

#Example
today <- Sys.Date()
months(today)
[1] "March"

To add or substract months, you should use %m+% from lubridate:

today <- Sys.Date()
today %m+% months(-1)
[1] "2017-02-28"

Subtract months from time in format year-month

As we want to subtract one month, we should subtract 1/12 which is 0.083 and not 0.1

library(zoo)
as.yearmon(global_date) - (1/12)
#[1] "Dec 2016"

If we need output in the mentioned format

format(as.yearmon(global_date) - (1/12), "%Y%m")
#[1] "201612"

Add (subtract) months without exceeding the last day of the new month

The lubridate function %m+% may be useful here:

Add and subtract months to a date without exceeding the last day of the new month

as.Date("2014-12-31") %m+% months(6)
# [1] "2015-06-30"

To also handle the second case, you will need to round up to nearest month using ceiling_date, and subtract one day using days.

ceiling_date(as.Date("2014-02-28") %m+% months(6), unit = "month") - days(1)
# [1] "2014-08-31"

How do I subtract date columns in R when date format is not recognized?

diff is the wrong function to calculate difference between dates. You can directly subtract the dates.

t1$date_admission - t1$date_symptoms
#Time differences in days
# [1]  -20   NA -172   NA   15   -1    3   11    1  -49    1   97   -1    9  101
#[16] -142   -1 -166 -116 -195   60  -18    9    0 -110   -1 -175  -80  -53 -227
#[31]  -14   41   71   13    8   16   16    1    4 -176   59   29   14  102    8
#[46] -143    0 -134   91  186    0   -1   40  -25 -168   60  -83    3   36   41
#[61]  -26   59 -168 -167  121    6 -111   99 -133   -1    6  -51   59 -214

You might be trying to use difftime :

difftime(t1$date_admission, t1$date_symptoms, units = "days")

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diff function subtracts consecutive values. See for example :

diff(c(5, 9, 4, 5))
#[1]  4 -5  1

where the calculation is (9 - 5 = 4), (4 - 9 = -5) and (5 - 4 = 1). In your case you are first subtracting the dates and then taking applying diff on them to get difference between consecutive numbers.

Add a month to a Date

Vanilla R has a naive difftime class, but the Lubridate CRAN package lets you do what you ask:

require(lubridate)
d <- ymd(as.Date('2004-01-01')) %m+% months(1)
d
[1] "2004-02-01"

Hope that helps.

How to subtract/add days from/to a date?

Just subtract a number:

> as.Date("2009-10-01")
[1] "2009-10-01"
> as.Date("2009-10-01")-5
[1] "2009-09-26"

Since the Date class only has days, you can just do basic arithmetic on it.

If you want to use POSIXlt for some reason, then you can use it's slots:

> a <- as.POSIXlt("2009-10-04")
> names(unclass(as.POSIXlt("2009-10-04")))
[1] "sec"   "min"   "hour"  "mday"  "mon"   "year"  "wday"  "yday"  "isdst"
> a$mday <- a$mday - 6
> a
[1] "2009-09-28 EDT"

Subtracting two dates of same dataset in R

X <- as.Date(df$date2) - as.Date(df$date1)

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