r-将tibble转换为时间序列

Question

我试图下载《经济学人》的Github存储库提供的新冠肺炎数据。

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)

我得到的是一个tibble数据帧，我无法轻松访问存储在该tibble中的数据。我想查看一列，比如test$covid_deaths_per_100k，并将其重新成形为矩阵或ts对象，其中行表示时间，列表示国家。

我手动尝试了一下，但失败了。然后我尝试了tsibble包，但再次失败：

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.

所以，我想问题是数据是按国家堆叠的，因此时间指数是重复的。我需要一些神奇的管道函数来实现这一点？有没有一种简单的方法可以做到这一点，也许不用管道？

Answer 1

有效的tsibble必须具有由关键字和索引标识的不同行：

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与月度、季度或年度系列配合使用效果最佳。这里我们展示一些方法。

1(monthly这将根据所示的test列创建一个月度动物园对象z，按国家划分并聚合以生成月度时间序列。然后创建一个ts对象。

library(zoo)
z <- read.zoo(test[c("start_date", "country", "covid_deaths")], 
split = "country", FUN = as.yearmon, aggregate = sum)
as.ts(z)

2(每周创建频率为53 的每周ts对象

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如果您想要一个时间是日期的系列，那么省略FUN参数，直接使用zoo。

z <- read.zoo(test[c("end_date", "country", "covid_deaths")], 
split = "country", aggregate = sum)