Add months of zero demand to zoo time series

Question

I have some intermittent demand data that only includes lines where demand is present. I bring it in via read.csv, and my 2 columns are Date (as date) and Quantity (as integ

Answer 1

You can use merge to add the missing rows and then set their values to zero.

First, let's create some fake data:

# Vector of dates from Jan 1, 2015, to Mar 31, 2015
dates = seq(as.Date("2015-01-01"), as.Date("2015-03-31"), by="1 day")

# Let's create data for few of these dates, leaving some out
set.seed(55)
dat = data.frame(dates=dates[sample(1:length(dates), 70)],
                 quantity=sample(1:10, 70, replace=TRUE))
dat = dat[order(dat$dates),]

Now let's make believe dat is what you imported from a csv file. We want to fill in quantity=0 for the missing dates. So first we need to add rows for the missing dates. You can do this by creating a date vector containing all dates from the first date to the last date in your csv file and using the merge function. In this case, we've already created that date vector above.

Now merge in rows for the missing dates. The new rows will have NA for quantity. We'll change those NAs to zero below.

dat = merge(data.frame(dates), dat, by="dates", all.x=TRUE)

# Set missing values to zero
dat$quantity[is.na(dat$quantity)] = 0

Now you can aggregate by month, convert to a zoo series, etc.

Answer 2

Because "df" is a zoo object, you may use merge.zoo and its fill argument. The current data set is merged with an empty zoo object which contains all the desired dates.

tt <- seq(min(Date), max(Date), "month")
merge(df, zoo(, tt), fill = 0)

# 2013-01-01 2013-02-01 2013-03-01 2013-04-01 2013-05-01 2013-06-01 2013-07-01 2013-08-01 2013-09-01 2013-10-01 2013-11-01 
#          2          0          0          0          0          0          0          0          0          3          6 

For further examples, see ?merge.zoo ("extend an irregular series to a regular one").