CODEWITHSUNDEEP
datetimertime-series
Janvb
Janvb

Reputation: 1290

Convert a irregular time series to a regular time series

I am having a problem when converting irregular time series to regular time series. Below a simplified example can be found:

require(zoo)
t <- as.character(c(1981,1984,1985))
d <- c(1,3,6)
dt <- data.frame(d,t)
t <- as.Date(t,"%Y")
z <- zoo(d,t)
plot(z)
ts.d <- as.ts(as.zooreg(z,freq=1)) # create a regular ts object
ts.d # regular time series

I would like to create a regular time series ts.d that looks like this c(1981,NA,NA,1984,1985).

The amazing thing is that the first time that I run this: it works! but when I want to run it again or repeat it (the as.ts()line) it stops workings and I obtain a very long time series:

ts.d # regular time series
Time Series:
Start = 4299 
End = 5760 
Frequency = 1 
  [1]  1 NA NA NA NA NA NA NA NA NA NA NA NA NA
 [15] NA NA NA NA NA NA NA NA

etc.

What is going wrong?

Upvotes: 9

Views: 7304

Answers (2)

G. Grothendieck
G. Grothendieck

Reputation: 531

As has been pointed out the as.Date(as.character(t), "%Y") is incorrect as it does not give the desired month and day. If we wanted to convert years to "Date" class we could do this as.Date(as.yearmon(t)) using zoo's as.yearmon; however, then we would have the further problem that different years have different numbers of days so there is no way to have a regular series using dates to represent years.

Really we don't want dates in the first place. We just want to work with years in which case it simplifies to just:

> z <- zoo(c(1, 3, 6), c(1981, 1984, 1985))
> 
> as.ts(z)
Time Series:
Start = 1981 
End = 1985 
Frequency = 1 
[1]  1 NA NA  3  6

or if we want to be safe we could do this which will force it to be annual even if the input has, by chance, a lower frequency: frequency(z) <- 1; as.ts(z) or just define the original zoo series to have a frequency of 1 right from the beginning:

> z <- zoo(c(1, 3, 6), c(1981, 1984, 1985), frequency = 1)
> as.ts(z)
Time Series:
Start = 1981 
End = 1985 
Frequency = 1 
[1]  1 NA NA  3  6

With this example it does not make a difference but in this case z <- zoo(c(1, 3, 6), c(1981, 1983, 1985), frequency = 1) the explicit frequency would be needed to prevent it from having a frequency of 0.5 .

Upvotes: 7

Joshua Ulrich
Joshua Ulrich

Reputation: 176648

It's not a bug. There are 1,461 days spanning the 4 years in your time series. And it doesn't work for me the first time I run it. as.Date(t,"%Y") doesn't know what month/day to use to make a date, so it uses today's month/day. That does not make for reproducible analysis. Try this instead:

t <- c(1981,1984,1985)
d <- c(1,3,6)
z <- zoo(d,t)
z <- merge(z,zoo(,c(1981,1982,1983,1984,1985)))
ts.d <- as.ts(z)

Which yields:

> ts.d
Time Series:
Start = 1981 
End = 1985 
Frequency = 1 
[1]  1 NA NA  3  6

Upvotes: 3

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