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How to use multiple factors in na.aggregate() from zoo package

Stack Overflow Asked by Shadrack Kibet on December 11, 2021

How can I use multiple factor variables say cut,color,clarityas arguments of by parameter in zoo::na.aggregate() function and be able to append output back into the data frame. Below is a reproducible example using single factor i.e carat from diamonds dataset. If not possible any workaround using tidyverse?

library(ggplot2)
library(zoo)

diamonds$avg <-  na.aggregate(object = diamonds$carat, by = diamonds$cut, mean)
View(diamonds)

One Answer

If you're trying to get the mean of carat based on cut, color and clarity, then this is what you're looking for:

library(dplyr)
library(zoo)

diamonds %>%
  group_by(cut,color,clarity) %>% 
  mutate(mean=na.aggregate(carat))

# A tibble: 53,940 x 11
# Groups:   cut, color, clarity [276]
   carat cut       color clarity depth table price     x     y     z  mean
   <dbl> <ord>     <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
 1 0.23  Ideal     E     SI2      61.5    55   326  3.95  3.98  2.43 0.23 
 2 0.21  Premium   E     SI1      59.8    61   326  3.89  3.84  2.31 0.21 
 3 0.23  Good      E     VS1      56.9    65   327  4.05  4.07  2.31 0.23 
 4 0.290 Premium   I     VS2      62.4    58   334  4.2   4.23  2.63 0.290
 5 0.31  Good      J     SI2      63.3    58   335  4.34  4.35  2.75 0.31 
 6 0.24  Very Good J     VVS2     62.8    57   336  3.94  3.96  2.48 0.24 
 7 0.24  Very Good I     VVS1     62.3    57   336  3.95  3.98  2.47 0.24 
 8 0.26  Very Good H     SI1      61.9    55   337  4.07  4.11  2.53 0.26 
 9 0.22  Fair      E     VS2      65.1    61   337  3.87  3.78  2.49 0.22 
10 0.23  Very Good H     VS1      59.4    61   338  4     4.05  2.39 0.23

Answered by Maël on December 11, 2021

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