zoo FAQ

1. I know that duplicate times are not allowed but my data has them. What do I do?

zoo objects should not normally contain duplicate times. If you try to create such an object using zoo or read.zoo then warnings will be issued but the objects will be created. The user then has the opportunity to fix them up – typically by using aggregate.zoo or duplicated.

Merging is not well defined for duplicate series with duplicate times and rather than give an undesired or unexpected result, merge.zoo issues an error message if it encounters such illegal objects. Since merge.zoo is the workhorse behind many zoo functions, a significant portion of zoo will not accept duplicates among the times.

Typically duplicates are eliminated by (1) averaging over them, (2) taking the last among each run of duplicates or (3) interpolating the duplicates and deleting ones on the end that cannot be interpolated. These three approaches are shown here using the aggregate.zoo function. Another way to do this is to use the aggregate argument of read.zoo which will aggregate the zoo object read in by read.zoo all in one step.

Note that in the example code below that identity is the identity function (i.e. it just returns its argument). It is an R core function:

A "zoo" series with duplicated indexes

z <- suppressWarnings(zoo(1:8, c(1, 2, 2, 2, 3, 4, 5, 5)))
z
## 1 2 2 2 3 4 5 5 
## 1 2 3 4 5 6 7 8

Fix it up by averaging duplicates:

aggregate(z, identity, mean)
##   1   2   3   4   5 
## 1.0 3.0 5.0 6.0 7.5

Or, fix it up by taking last in each set of duplicates:

aggregate(z, identity, tail, 1)
## 1 2 3 4 5 
## 1 4 5 6 8

Fix it up via interpolation of duplicate times

time(z) <- na.approx(ifelse(duplicated(time(z)), NA, time(z)), na.rm = FALSE)

If there is a run of equal times at end they wind up as NAs and we cannot have NA times.

z[!is.na(time(z))]
##      1      2 2.3333 2.6667      3      4      5 
##      1      2      3      4      5      6      7

The read.zoo command has an aggregate argument that supports arbitrary summarization. For example, in the following we take the last value among any duplicate times and sum the volumes among all duplicate times. We do this by reading the data twice, once for each aggregate function. In this example, the first three columns are junk that we wish to suppress which is why we specified colClasses; however, in most cases that argument would not be necessary.

Lines <- "1|BHARTIARTL|EQ|18:15:05|600|1
2|BHARTIARTL|EQ|18:15:05|600|99
3|GLENMARK|EQ|18:15:05|238.1|5
4|HINDALCO|EQ|18:15:05|43.75|100
5|BHARTIARTL|EQ|18:15:05|600|1
6|BHEL|EQ|18:15:05|1100|11
7|HINDALCO|EQ|18:15:06|43.2|1
8|CHAMBLFERT|EQ|18:15:06|46|10
9|CHAMBLFERT|EQ|18:15:06|46|90
10|BAJAUTOFIN|EQ|18:15:06|80|100"

library("zoo")
library("chron")

tail1 <- function(x) tail(x, 1)
cls <- c("NULL", "NULL", "NULL", "character", "numeric", "numeric")
nms <- c("", "", "", "time", "value", "volume")

z <- read.zoo(text = Lines, aggregate = tail1,
  FUN = times, sep = "|", colClasses = cls, col.names = nms)

z2 <- read.zoo(text = Lines, aggregate = sum,
  FUN = times, sep = "|", colClasses = cls, col.names = nms)

z$volume <- z2$volume
z
##          value volume
## 18:15:05  1100    217
## 18:15:06    80    201

If the reason for the duplicate times is that the data is stored in long format then use read.zoo (particlarly the split argument) to convert it to wide format. Wide format is typically a time series whereas long format is not so wide format is the suitable one for zoo.

Lines <- "Date Stock Price
2000-01-01 IBM 10
2000-01-02 IBM 11
2000-01-01 ORCL 12
2000-01-02 ORCL 13"

stocks <- read.zoo(text = Lines, header = TRUE, split = "Stock")
stocks
##            IBM ORCL
## 2000-01-01  10   12
## 2000-01-02  11   13

2. When I try to specify a log axis to plot.zoo a warning is issued. What is wrong?

Arguments that are part of ... are passed to the panel function and the default panel function, lines, does not accept log.
Either ignore the warning, use suppressWarnings (see ?suppressWarnings) or create your own panel function which excludes the log:

z <- zoo(1:100)
plot(z, log = "y", panel = function(..., log) lines(...))

3. How do I create right and a left vertical axes in plot.zoo?

The following shows an example of creating a plot containing a single panel and both left and right axes.

set.seed(1)
z.Date <- as.Date(paste(2003, 02, c(1, 3, 7, 9, 14), sep = "-"))
z <- zoo(cbind(left = rnorm(5), right = rnorm(5, sd = 0.2)), z.Date)

plot(z[,1], xlab = "Time", ylab = "")
opar <- par(usr = c(par("usr")[1:2], range(z[,2])))
lines(z[,2], lty = 2)

axis(side = 4)
legend("bottomright", lty = 1:2, legend = colnames(z), bty="n")

par(opar)

4. I have data frame with both numeric and factor columns. How do I convert that to a "zoo" object?

A "zoo" object may be (1) a numeric vector, (2) a numeric matrix or (3) a factor but may not contain both a numeric vector and factor.
The underlying reason for this constraint is that "zoo" was intended to generalize R’s "ts" class, which is also based on matrices, to irregularly spaced series with an arbitrary index class. The main reason to stick to matrices is that operations on matrices in R are much faster than on data frames.

If you have a data frame with both numeric and factor variables that you want to convert to "zoo", you can do one of the following.

Use two "zoo" variables instead:

DF <- data.frame(time = 1:4, x = 1:4, f = factor(letters[c(1, 1, 2, 2)]))
zx <- zoo(DF$x, DF$time)
zf <- zoo(DF$f, DF$time)

These could also be held in a "data.frame" again:

DF2 <- data.frame(x = zx, f = zf)

Or convert the factor to numeric and create a single "zoo" series:

z <- zoo(data.matrix(DF[-1]), DF$time)

5. Why does lag give slightly different results on a "zoo" and a "zooreg" series which are otherwise the same?

To be definite let us consider the following examples, noting how both lag and diff give a different answer with the same input except its class is "zoo" in one case and "zooreg" in another:

z <- zoo(11:15, as.Date("2008-01-01") + c(-4, 1, 2, 3, 6))
zr <- as.zooreg(z)

lag(z)
## 2007-12-28 2008-01-02 2008-01-03 2008-01-04 
##         12         13         14         15
lag(zr)
## 2007-12-27 2008-01-01 2008-01-02 2008-01-03 2008-01-06 
##         11         12         13         14         15
diff(log(z))
## 2008-01-02 2008-01-03 2008-01-04 2008-01-07 
## 0.08701138 0.08004271 0.07410797 0.06899287
diff(log(zr))
## 2008-01-03 2008-01-04 
## 0.08004271 0.07410797

lag.zoo and lag.zooreg work differently. For "zoo" objects the lagged version is obtained by moving values to the adjacent time point that exists in the series but for "zooreg" objects the time is lagged by deltat, the time between adjacent regular times.

A key implication is that "zooreg" can lag a point to a time point that did not previously exist in the series and, in particular, can lag a series outside of the original time range whereas that is not possible in a "zoo" series.

Note that lag.zoo has an na.pad= argument which in some cases may be what is being sought here.

The difference between diff.zoo and diff.zooreg stems from the fact that diff(x) is defined in terms of lag like this: x-lag(x,-1).

6. How do I subtract the mean of each month from a "zoo" series?

Suppose we have a daily series. To subtract the mean of Jan 2007 from each day in that month, subtract the mean of Feb 2007 from each day in that month, etc. try this:

set.seed(123)
z <- zoo(rnorm(100), as.Date("2007-01-01") + seq(0, by = 10, length = 100))
z.demean1 <- z - ave(z, as.yearmon(time(z)))

This first generates some artificial data and then employs ave to compute monthly means.

To subtract the mean of all Januaries from each January, etc. try this:

z.demean2 <- z - ave(z, format(time(z), "%m"))

7. How do I create a monthly series but still keep track of the dates?

Create a S3 subclass of "yearmon" called "yearmon2" that stores the dates as names on the time vector. It will be sufficient to create an as.yearmon2 generic together with an as.yearmon2.Date methods as well as the inverse: as.Date.yearmon2.

as.yearmon2 <- function(x, ...) UseMethod("as.yearmon2")
as.yearmon2.Date <- function(x, ...) {
  y <- as.yearmon(with(as.POSIXlt(x, tz = "GMT"), 1900 + year + mon/12))
  names(y) <- x
  structure(y, class = c("yearmon2", class(y)))
}

as.Date.yearmon2 is inverse of as.yearmon2.Date

as.Date.yearmon2 <- function(x, frac = 0, ...) {
  if (!is.null(names(x))) return(as.Date(names(x)))
  x <- unclass(x)
  year <- floor(x + .001)
  month <- floor(12 * (x - year) + 1 + .5 + .001)
  dd.start <- as.Date(paste(year, month, 1, sep = "-")) 
  dd.end <- dd.start + 32 - as.numeric(format(dd.start + 32, "%d"))
  as.Date((1-frac) * as.numeric(dd.start) + frac * as.numeric(dd.end),
    origin = "1970-01-01")
}

This new class will act the same as "yearmon" stores and allows recovery of the dates using as.Date and aggregate.zoo.

dd <- seq(as.Date("2000-01-01"), length = 5, by = 32)
z <- zoo(1:5, as.yearmon2(dd))
z
## Jan 2000 Feb 2000 Mar 2000 Apr 2000 May 2000 
##        1        2        3        4        5
aggregate(z, as.Date, identity) 
## 2000-01-01 2000-02-02 2000-03-05 2000-04-06 2000-05-08 
##          1          2          3          4          5

8. How are axes added to a plot created using plot.zoo?

On single panel plots axis or Axis can be used just as with any classic graphics plot in R.

The following example adds custom axis for single panel plot. It labels months but uses the larger year for January. Months, quarters and years should have successively larger ticks.

z <- zoo(0:500, as.Date(0:500))
plot(z, xaxt = "n")
tt <- time(z)
m <- unique(as.Date(as.yearmon(tt)))
jan <- format(m, "%m") == "01"
mlab <- substr(months(m[!jan]), 1, 1)
axis(side = 1, at = m[!jan], labels = mlab, tcl = -0.3, cex.axis = 0.7)
axis(side = 1, at = m[jan], labels = format(m[jan], "%y"), tcl = -0.7)
axis(side = 1, at = unique(as.Date(as.yearqtr(tt))), labels = FALSE)

abline(v = m, col = grey(0.8), lty = 2)

A multivariate series can either be generated as (1) multiple single panel plots:

z3 <- cbind(z1 = z, z2 = 2*z, z3 = 3*z)
opar <- par(mfrow = c(2, 2))
tt <- time(z)
m <- unique(as.Date(as.yearmon(tt)))
jan <- format(m, "%m") == "01"
mlab <- substr(months(m[!jan]), 1, 1)
for(i in 1:ncol(z3)) {
  plot(z3[,i], xaxt = "n", ylab = colnames(z3)[i], ylim = range(z3))
  axis(side = 1, at = m[!jan], labels = mlab, tcl = -0.3, cex.axis = 0.7)
  axis(side = 1, at = m[jan], labels = format(m[jan], "%y"), tcl = -0.7)
  axis(side = 1, at = unique(as.Date(as.yearqtr(tt))), labels = FALSE)
}
par(opar)

or (2) as a multipanel plot. In this case any custom axis must be placed in a panel function.

plot(z3, screen = 1:3, xaxt = "n", nc = 2, ylim = range(z3),
  panel = function(...) {
    lines(...)
    panel.number <- parent.frame()$panel.number
    nser <- parent.frame()$nser
    # place axis on bottom panel of each column only
    if (panel.number %% 2 == 0 || panel.number == nser) { 
      tt <- list(...)[[1]]
      m <- unique(as.Date(as.yearmon(tt)))
      jan <- format(m, "%m") == "01"
      mlab <- substr(months(m[!jan]), 1, 1)
      axis(side = 1, at = m[!jan], labels = mlab, tcl = -0.3, cex.axis = 0.7)
      axis(side = 1, at = m[jan], labels = format(m[jan], "%y"), tcl = -0.7)
      axis(side = 1, at = unique(as.Date(as.yearqtr(tt))), labels = FALSE)
    }
})

9. Why is nothing plotted except axes when I plot an object with many NAs?

Isolated points surrounded by NA values do not form lines:

z <- zoo(c(1, NA, 2, NA, 3))
plot(z)

So try one of the following:

Plot points rather than lines.

plot(z, type = "p") 

Omit NAs and plot that.

plot(na.omit(z))

Fill in the NAs with interpolated values.

plot(na.approx(z))

Plot points with lines superimposed.

plot(z, type = "p")
lines(na.omit(z))

Note that this is not specific to zoo. If we plot in R without zoo we get the same behavior.

10. Does zoo work with Rmetrics?

Yes. timeDate class objects from the timeDate package can be used directly as the index of a zoo series and as.timeSeries.zoo and as.zoo.timeSeries can convert back and forth between objects of class zoo and class timeSeries from the timeSeries package.

library("timeDate")
dts <- c("1989-09-28", "2001-01-15", "2004-08-30", "1990-02-09")
tms <- c(  "23:12:55",   "10:34:02",   "08:30:00",   "11:18:23")
td <- timeDate(paste(dts, tms), format = "%Y-%m-%d %H:%M:%S")

library("zoo")
z <- zoo(1:4, td)
zz <- merge(z, lag(z))
plot(zz)

library("timeSeries")
## 
## Attaching package: 'timeSeries'
## The following object is masked from 'package:zoo':
## 
##     time<-
## The following objects are masked from 'package:graphics':
## 
##     lines, points
zz
##                     z lag(z)
## 1989-09-28 23:12:55 1      4
## 1990-02-09 11:18:23 4      2
## 2001-01-15 10:34:02 2      3
## 2004-08-30 08:30:00 3     NA
as.timeSeries(zz)
## GMT
##                     z lag(z)
## 1989-09-28 23:12:55 1      4
## 1990-02-09 11:18:23 4      2
## 2001-01-15 10:34:02 2      3
## 2004-08-30 08:30:00 3     NA
as.zoo(as.timeSeries(zz))
##                     z lag(z)
## 1989-09-28 23:12:55 1      4
## 1990-02-09 11:18:23 4      2
## 2001-01-15 10:34:02 2      3
## 2004-08-30 08:30:00 3     NA

11. What other packages use zoo?

The CRAN page of the package at https://CRAN.R-project.org/package=zoo lists all reverse dependencies on CRAN, stratified by Depends, Imports, Suggests, Linking to, and Enhances.

These can also be queried from within R using the tools package:

db <- tools::CRAN_package_db()
db[db$Package == "zoo", "Reverse depends"]

12. Why does ifelse not work as I expect?

The ordinary R ifelse function only works with zoo objects if all three arguments are zoo objects with the same time index. zoo provides an ifelse.zoo function that should be used instead. The .zoo part must be written out since ifelse is not generic.

z <- zoo(c(1, 5, 10, 15))
# wrong !!!
ifelse(diff(z) > 4, -z, z)
##   2   3   4 
##   1  -5 -10
# ok
ifelse.zoo(diff(z) > 4, -z, z)
##   1   2   3   4 
##  NA   5 -10 -15
# or if we merge first we can use ordinary ifelse
xm <- merge(z, dif = diff(z))
with(xm, ifelse(dif > 4, -z, z))
##   1   2   3   4 
##  NA   5 -10 -15
# or in this case we could also use orindary ifelse if we 
# use fill = NA to ensure all three have same index
ifelse(diff(z, fill = NA) > 4, -z, z)
##   2   3   4 
##   1  -5 -10

13. In a series which is regular except for a few missing times or for which we wish to align to a grid how is it filled or aligned?

# April is missing
zym <- zoo(1:5, as.yearmon("2000-01-01") + c(0, 1, 2, 4, 5)/12)
g <- seq(start(zym), end(zym), by = 1/12)
na.locf(zym, xout = g)
## Jan 2000 Feb 2000 Mar 2000 Apr 2000 May 2000 Jun 2000 
##        1        2        3        3        4        5

A variation of this is where the grid is of a different date/time class than the original series. In that case use the x argument. In the example that follows the series z is of "Date" class whereas the grid is of "yearmon" class:

z <- zoo(1:3, as.Date(c("2000-01-15", "2000-03-3", "2000-04-29")))
g <- seq(as.yearmon(start(z)), as.yearmon(end(z)), by = 1/12)
na.locf(z, x = as.yearmon, xout = g)
## Jan 2000 Feb 2000 Mar 2000 Apr 2000 
##        1        1        2        3

Here is a chron example where we wish to create a 10 minute grid:

Lines <- "Time,Value
2009-10-09 5:00:00,210
2009-10-09 5:05:00,207
2009-10-09 5:17:00,250
2009-10-09 5:30:00,193
2009-10-09 5:41:00,205
2009-10-09 6:00:00,185"

library("chron")
z <- read.zoo(text = Lines, FUN = as.chron, sep = ",", header = TRUE)
g <- seq(start(z), end(z), by = times("00:10:00"))
na.locf(z, xout = g)
## (10/09/09 05:00:00) (10/09/09 05:10:00) (10/09/09 05:20:00) (10/09/09 05:30:00) 
##                 210                 207                 250                 193 
## (10/09/09 05:40:00) (10/09/09 05:50:00) (10/09/09 06:00:00) 
##                 193                 205                 185

What is the difference between as.Date in zoo and as.Date in the core of R?

zoo has extended the origin argument of as.Date.numeric so that it has a default of origin="1970-01-01" (whereas in the core of R it has no default and must always be specified).
Note that this is a strictly upwardly compatible extensions to R and any usage of as.Date in R will also work in zoo.

This makes it more convenient to use as.Date as a function input. For example, one can shorten this:

z <- zoo(1:2, c("2000-01-01", "2000-01-02"))
aggregate(z, function(x) as.Date(x, origin = "1970-01-01"))
## 2000-01-01 2000-01-02 
##          1          2

to just this:

aggregate(z, as.Date) 
## 2000-01-01 2000-01-02 
##          1          2

As another example, one can shorten

Lines <- "2000-01-01 12:00:00,12
2000-01-02 12:00:00,13"
read.zoo(text = Lines, sep = ",", FUN = function(x) as.Date(x, origin = "1970-01-01"))
## 2000-01-01 2000-01-02 
##         12         13

to this:

read.zoo(text = Lines, sep = ",", FUN = as.Date)
## 2000-01-01 2000-01-02 
##         12         13

Note to package developers of packages that use zoo: Other packages that work with zoo and define as.Date methods should either import zoo or else should fully export their as.Date methods in their NAMESPACE file, e.g. export(as.Date.X), in order that those methods be registered with zoo’s as.Date generic and not just the as.Date generic in base.

15. How can I speed up zoo?

The main area where you might notice slowness is if you do indexing of zoo objects in an inner loop. In that case extract the data and time components prior to the loop. Since most calculations in R use the whole object approach there are relatively few instances of this.

For example, the following shows two ways of performing a rolling sum using only times nearer than 3 before the current time. The second one eliminates the zoo indexing to get a speedup:

n <- 50
z <- zoo(1:n, c(1:3, seq(4, by = 2, length = n-3)))

system.time({
    zz <- sapply(seq_along(z), 
        function(i) sum(z[time(z) <= time(z)[i] & time(z) > time(z)[i] - 3]))
    z1 <- zoo(zz, time(z))
})
##    user  system elapsed 
##   0.014   0.000   0.014
system.time({
    zc <- coredata(z)
    tt <- time(z)
    zr <- sapply(seq_along(zc), 
        function(i) sum(zc[tt <= tt[i] & tt > tt[i] - 3]))
    z2 <- zoo(zr, tt)
})
##    user  system elapsed 
##   0.005   0.000   0.005
identical(z1, z2) 
## [1] TRUE