na.aggregate

0th

Percentile

Replace NA by Aggregation

Generic function for replacing each NA with aggregated values. This allows imputing by the overall mean, by monthly means, etc.

Keywords
ts
Usage
na.aggregate(object, ...)
## S3 method for class 'default':
na.aggregate(object, by = 1, \dots, FUN = mean,
             na.rm = FALSE, maxgap = Inf)
Arguments
object
an object.
by
a grouping variable corresponding to object, or a function to be applied to time(object) to generate the groups.
...
further arguments passed to by if by is a function.
FUN
function to apply to the non-missing values in each group defined by by.
na.rm
logical. Should any remaining NAs be removed?
maxgap
maximum number of consecutive NAs to fill. Any longer gaps will be left unchanged.
Value

  • An object in which each NA in the input object is replaced by the mean (or other function) of its group, defined by by. This is done for each series in a multi-column object. Common choices for the aggregation group are a year, a month, all calendar months, etc. If a group has no non-missing values, the default aggregation function mean will return NaN. Specify na.rm = TRUE to omit such remaining missing values.

See Also

zoo

Aliases
  • na.aggregate
  • na.aggregate.default
Examples
z <- zoo(c(1, NA, 3:9),
         c(as.Date("2010-01-01") + 0:2,
           as.Date("2010-02-01") + 0:2,
           as.Date("2011-01-01") + 0:2))
## overall mean
na.aggregate(z)
## group by months
na.aggregate(z, as.yearmon)
## group by calendar months
na.aggregate(z, months)
## group by years
na.aggregate(z, format, "%Y")
Documentation reproduced from package zoo, version 1.7-10, License: GPL-2

Community examples

Looks like there are no examples yet.