# Using R to create a table comparing 5yr mean

Hello,

I have linelisted surveillance data of disease notifications between 2015 and 2021. I am comparing notification trends for 2020 and 2021 to the five-year mean before the pandemic. I have calculated the following table in Excel but would like to know whether it is possible to streamline the process in R for future reproducibility. The table looks like this:

I donβt have any code to share as I have not been successful in finding a solution/I used excel to perform the calculations.

Thanks very much!

Thanks for posting this question! I will respond by suggesting approaches and building blocks for this table rather than providing a full chunk of code to create it.

The first step is to create an aggregate dataset, with columns for disease, sex, age group, year, and corresponding counts. From there you can write the following to generate most of the content for your table.

table β df_aggregate %>%
group_by(disease, sex, agegroup) %>%
summarise(count_mean = sum(count)[year %in% c(2014:2019)/5,
count_2020 = sum(count)[year==2020],
count_2021 = sum(count)[year==2021] %>%
arrange(disease, sex, agegroup)

You will need to make sure agegroup is a factor and ordered from youngest to oldest group, or arranging them will not give the right order.

You can create additional columns based on the comparison of βcount_2021β to βcount_meanβ, to create those βtrendβ columns, and can use unicode characters for arrows. Arrow types are on this website (Arrow symbols β β β β β² βΌ β βΆ Arrow and direction symbols in Unicode - Unicode Character Table), but note how the values are typed in the very simple example below (i.e. β\U2191β not βU+2191β).

mutate(arrow_2021 = if_else(trend_2020-trend_mean>0, β\U2191β, β\U2193β)

Once your table is created, you can pipe this object into flextable(). This function will create a table ready for presentation that can be outputted to word/powerpoint etc, and the unicode values will actually look like arrows when viewing this table, rather than strings.

The flextable will need further formatting though. E.g.: you can use the merge_v function to merge cells vertically so that the disease and sex values are not repeated.

df_table %>%
flextable() %>%
merge_v(j = c(βdiseaseβ, βsexβ))

Other ways of editing the table (e.g. changing column names, adding in additional header rows with horizontally merged sections) are described in the epiRhandbook: 29 Tables for presentation | The Epidemiologist R Handbook

2 Likes

Hello @mdal,

What you are looking to do is definitely feasible within R. Below is an example of how I would tackle the problem. However, there are two options to generalize: you could write a function that takes the years you want to include in the 5-year average calculation and then another to take the years you want to include as counts. Alternatively, you could try using the slider package to calculate 5-year rolling averages for each year.

``````# loading packages
library(tidyverse)
library(gt)

# creating linelist data
linelist_data <- tibble(
year = sample(
x = 2015:2021,
size = 500,
replace = TRUE
),
disease = sample(
x = c("Disease A", "Disease B"),
size = 500,
replace = TRUE
),
sex = sample(
x = c("Female", "Male"),
size = 500,
replace = TRUE
),
age_group = sample(
x = c("0-14", "15-29", "30-39", "40-49", ">=50"),
size = 500,
replace = TRUE
),
n = rpois(n = 500, lambda = 65)
) |>
mutate(
disease = factor(x = disease, levels = c("Disease A", "Disease B")),
sex = factor(x = sex, levels = c("Female", "Male")),
age_group = factor(age_group, levels = c("0-14", "15-29", "30-39", "40-49", ">=50"))
)

# aggregating linelist data
aggregated_data <- linelist_data |>
count(disease,
year,
sex,
age_group) |>
complete(disease,
year = 2015:2021,
sex,
age_group,
fill = list(n = 0))

# creating average data for 2015 to 2019
averages <- aggregated_data |>
filter(between(year, 2015, 2019)) |>
group_by(disease,
sex,
age_group) |>
summarise(average = mean(n),
.groups = "drop")

# creating counts data for 2020 and 2021 in a wide format
counts <- aggregated_data |>
filter(between(year, 2020, 2021)) |>
pivot_wider(
id_cols = c(disease, sex, age_group),
names_prefix = "count_",
names_from = year,
values_from = n
)

# joining the counts and average data, you can adjust how trend is defined
joined_data <- counts |>
left_join(averages,
by = c("disease", "sex", "age_group")) |>
mutate(
trend_2020 = sign(count_2020 - average),
trend_2021 = sign(count_2021 - average)
) |>
relocate(disease,
sex,
age_group,
average,
count_2020,
trend_2020,
count_2021,
trend_2021)

# using gt to create a nice table - you could look into creating group rows
joined_data |>
gt() |>
cols_label(
disease = "Disease",
sex = "Sex",
age_group = "Age Group",
average = "5 year mean",
count_2020 = "Count",
trend_2020 = "Trend",
count_2021 = "Count",
trend_2021 = "Trend"
) |>
tab_spanner(label = "2020",
columns = c("count_2020", "trend_2020")) |>
tab_spanner(label = "2021",
columns = c("count_2021", "trend_2021"))
``````

Created on 2022-05-19 by the reprex package (v2.0.1)

Session info
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#> ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ
``````

All the best,

Tim

1 Like

Sorry for my slow reply, @machupovirus - I have been on leave.
Thank you very much for providing this code. I applied it to my existing linelist and it worked perfectly.
Unfortunately, I had a problem installing the gt package so used a combined approach of your code and the flextable mentioned by @paulablomquist.
Thanks again!
Meg

1 Like

Thanks very much for your response @paulablomquist.
Itβs helpful to learn of the Unicode characters and their use within R. My if_else statement only returns one symbol i.e. all up or down arrows depending on whether I state the trend to be > or < 0. Do you have any advice on this?
Thanks,
Meg

1 Like

Please disregard previous Q @paulablomquist - I was able to identify a solution.