Convert tibble to time series - r

Question

I tried to download data on covid provided by the Economist's Github repository.
library(readr)
library(knitr)
myfile <- "https://raw.githubusercontent.com/TheEconomist/covid-19-excess-deaths-tracker/master/output-data/excess-deaths/all_weekly_excess_deaths.csv"
test <- read_csv(myfile)
What I get is a tibble data frame and I am unable to easily access the data stored in that tibble. I would like to look at one column, say test$covid_deaths_per_100k and re-shape that into a matrix or ts object with rows referring to time and columns referring to countries.
I tried it manually, but I failed. Then I tried with the tsibble package and failed again:
tsibble(test[c("covid_deaths_per_100k","country")],index=test$start_date)
Error: Must extract column with a single valid subscript.
x Subscript `var` has the wrong type `date`.
ℹ It must be numeric or character.
So, I guess the problem is that the data are stacked by countries and hence the time index is duplicated. I would need some of these magic pipe functions to make this work? Is there an easy way to do that, perhaps without piping?

Answer 1

A valid tsibble must have distinct rows identified by key and index:
as_tsibble(test,index = start_date,key=c(country,region))
# A tsibble: 11,715 x 17 [1D]
# Key: country, region [176]
country region region_code start_date end_date days year week population total_deaths
<chr> <chr> <chr> <date> <date> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Australia Australia 0 2020-01-01 2020-01-07 7 2020 1 25734100 2497
2 Australia Australia 0 2020-01-08 2020-01-14 7 2020 2 25734100 2510
3 Australia Australia 0 2020-01-15 2020-01-21 7 2020 3 25734100 2501
4 Australia Australia 0 2020-01-22 2020-01-28 7 2020 4 25734100 2597
5 Australia Australia 0 2020-01-29 2020-02-04 7 2020 5 25734100 2510
6 Australia Australia 0 2020-02-05 2020-02-11 7 2020 6 25734100 2530
7 Australia Australia 0 2020-02-12 2020-02-18 7 2020 7 25734100 2613
8 Australia Australia 0 2020-02-19 2020-02-25 7 2020 8 25734100 2608
9 Australia Australia 0 2020-02-26 2020-03-03 7 2020 9 25734100 2678
10 Australia Australia 0 2020-03-04 2020-03-10 7 2020 10 25734100 2602
# ... with 11,705 more rows, and 7 more variables: covid_deaths <dbl>, expected_deaths <dbl>,
# excess_deaths <dbl>, non_covid_deaths <dbl>, covid_deaths_per_100k <dbl>,
# excess_deaths_per_100k <dbl>, excess_deaths_pct_change <dbl>

Answer 2

ts works best with monthly, quarterly or annual series. Here we show a few approaches.
1) monthly This creates a monthly zoo object z from the indicated test columns splitting by country and aggregating to produce a monthly time series. It then creates a ts object from that.
library(zoo)
z <- read.zoo(test[c("start_date", "country", "covid_deaths")],
split = "country", FUN = as.yearmon, aggregate = sum)
as.ts(z)
2) weekly To create a weekly ts object with frequency 53
to_weekly <- function(x) {
yr <- as.integer(as.yearmon(x))
wk <- as.integer(format(as.Date(x), "%U"))
yr + wk/53
}
z <- read.zoo(test[c("start_date", "country", "covid_deaths")],
split = "country", FUN = to_weekly, aggregate = sum)
as.ts(z)
3) daily If you want a series where the times are dates then omit the FUN argument and use zoo directly.
z <- read.zoo(test[c("end_date", "country", "covid_deaths")],
split = "country", aggregate = sum)