如何根据特定日期列排名正确地对行求和？

Question

想法是根据列名求和在 01/01/2021 和 01/08/2021 之间：

# define rank parameters {start-end}
first_date <- format(Sys.Date(), "01/01/%Y")

actual_date <- format(Sys.Date() %m-% months(1), "01/%m/%Y")


# get the sum of the rows between first_date and actual_date

df$ytd<- rowSums(df[as.character(seq(first_date,
                                       actual_date))])

但是，应用时出现下一个错误：

Error in seq.default(first_date, to_date) : 'from' must be a finite number

预期输出是一个新列，它采用指定排名之间的行总和。

数据

df <- structure(list(country = c("Mexico", "Mexico", "Mexico", "Mexico"
), `01/01/2021` = c(12, 23, 13, 12), `01/02/2021` = c(12, 23, 
13, 12), `01/03/2021` = c(12, 23, 13, 12), `01/04/2021` = c(12, 
23, 13, 12), `01/05/2021` = c(12, 23, 13, 12), `01/06/2021` = c(12, 
23, 13, 12), `01/07/2021` = c(12, 23, 13, 12), `01/08/2021` = c(12, 
23, 13, 12), `01/09/2021` = c(12, 23, 13, 12), `01/10/2021` = c(12, 
23, 13, 12), `01/11/2021` = c(12, 23, 13, 12), `01/12/2021` = c(12, 
23, 13, 12)), row.names = c(NA, -4L), class = c("tbl_df", "tbl", 
"data.frame"))

我怎样才能正确应用一个函数来获得这个输出？

Answer 1

format 和 seq 不起作用，即 seq 期望 Date class 而 format 是 character class。相反，在 across 或 select

中使用范围运算符

library(dplyr)
out <- df %>% 
    mutate(ytd = rowSums(across(all_of(first_date):all_of(actual_date))))

-输出

> out$ytd
[1]  96 184 104  96

Answer 2

使用 match -

的基础 R 方法

df$ytd <- rowSums(df[match(first_date, names(df)):match(actual_date, names(df))])
df$ytd
#[1]  96 184 104  96

如何根据特定日期列排名正确地对行求和？

how to properly sum rows based in an specific date column rank?

r

sum

date

dataframe

数据