# [SOLVED] Taking column means as part of a function

## Issue

I have the following function:

``````estimate = function(df, y_true) {

R = nrow(df)

y_estimated = apply(df, 2, mean)

((sqrt( (y_estimated - y_true)^2 / R)) / y_true) * 100
}

df = iris[1:10,2:4]
y_true = c(3, 1, 0.4)
estimate(df = df, y_true = y_true)
``````

user:bird provided this and works great, however, I also need to find the means by group. So if we change the df to `df= iris[,2:5]`, how to do I find the means of each column by Species to use in the function. I figured something like this would work- but not luck:

``````estimate = function(df, y_true, group) {

R = nrow(df)

y_estimated = df %>% group_by(group) %>% apply(df, 2, mean)

((sqrt( (y_estimated - y_true)^2 / R)) / y_true) * 100
}

df = iris[2:5]
y_true = c(3, 1, 0.4)
group=df\$Species

estimate(df = df, y_true = y_true, group=group)
``````

Using `colMeans` also did not work.

This is an extension of this post which explains the purpose of each variable.

## Solution

Rather than modifying your function, you can keep the function as-is and apply it group-wise to your data. If you use `group_by` and then `group_modify`, the input to the function you pass to `group_modify` is the data frame, subset to the rows in that specific group.

``````estimate = function(df, y_true) {

R = nrow(df)

y_estimated = apply(df, 2, mean)

((sqrt( (y_estimated - y_true)^2 / R)) / y_true) * 100
}

df = iris[2:5]
y_true = c(3, 1, 0.4)

library(dplyr, warn.conflicts = FALSE)

df %>%
group_by(Species) %>%
group_modify(~ as.data.frame.list(estimate(., y_true)))
#> # A tibble: 3 × 4
#> # Groups:   Species 
#>   Species    Sepal.Width Petal.Length Petal.Width
#>   <fct>            <dbl>        <dbl>       <dbl>
#> 1 setosa           2.02          6.53        5.44
#> 2 versicolor       1.08         46.1        32.7
#> 3 virginica        0.123        64.4        57.5
``````

Created on 2022-02-24 by the reprex package (v2.0.1)

Answered By – IceCreamToucan

Answer Checked By – Mary Flores (BugsFixing Volunteer)