在R中将日期转换为as.POSIXct，并减去两个日期

Question

I'm trying to calculate a date - 1 (basically the day before the date) in R and when it converts it to a POSIXct it seems to subtract another date?

The column is of type POSIXct:

class(df$Date)
[1] "POSIXct" "POSIXt" 

Here's the initial value:

> df[12,"Date"]
[1] "2016-03-09 EST"

If I just do as.Date and subtract one it works fine:

as.Date(df[12,"Date"]-1, tz="EST")
[1] "2016-03-08"

But I'm saving it back to the same column so it converts is back to as.POSIXct automatically (I think). And then I end up with March 7 in that column. And 7 pm. If I type it out here I get this:

as.POSIXct(as.Date(df[12,"Date"]-1, tz="EST"))
[1] "2016-03-07 19:00:00 EST"

I've tried using America/New York for the tz. I've tried the as.Date around just the df[12,"Date"] or around the whole thing including the -1... I have no clue what to do!

Thanks!

Answer 1

Use as.difftime to take away amounts specified in units of time. It will work consistently regardless of your data being in Date or POSIXct formats. Eg:

This fails as you described:

x <- as.POSIXct(c("2016-03-09","2016-03-10"), tz="US/Eastern")
#[1] "2016-03-09 EST" "2016-03-10 EST"
x[1] <- as.Date(x[1]-1, tz="US/Eastern")
#[1] "2016-03-07 19:00:00 EST" "2016-03-10 00:00:00 EST"

This works:

x <- as.POSIXct(c("2016-03-09","2016-03-10"), tz="US/Eastern")
#[1] "2016-03-09 EST" "2016-03-10 EST"
x[1] <- x[1] - as.difftime(1, units="days")
#[1] "2016-03-08 EST" "2016-03-10 EST"

Answer 2

If you don't want to do what Frank mentioned in the comments,

You should consider using strptime instead of as.POSIXct

To ensure it returns in a POSIXct format use:

strptime(df[12,"Date"]-1,tz="EST",format="%Y-%m-%d")