计算基于一年的减少百分比

Question

我有这个数据框：

dat = read.delim(text = "LM, year
  85.20000, 2020
  56.70000, 2021
  49.00000, 2022
  71.00000, 2023
  33.00000, 2024
  96.50000, 2025
  26.30000, 2026
  21.30000, 2027", sep = ",", header = T)

我想计算特定年份的减少百分比变化，我为此编写了这个函数：

pChange = function( dat , startYear){

  lastYear = max(dat$year)
  
  res = dat %>% arrange(year) %>%  
    mutate(change = (LM[year == lastYear] - LM[year == startYear]) / LM[year == startYear])
  res

}

pChange( dat, startYear = 2022)

但是，我每年都得到相同的值，我认为存在问题但我不知道如何解决它。有什么想法吗？

    LM year     change
1 85.2 2020 -0.5653061
2 56.7 2021 -0.5653061
3 49.0 2022 -0.5653061
4 71.0 2023 -0.5653061
5 33.0 2024 -0.5653061
6 96.5 2025 -0.5653061
7 26.3 2026 -0.5653061
8 21.3 2027 -0.5653061

Answer 1

如果你想计算startYear和每年之间的百分比变化，我相信你不想包括lastYear。

pChange = function(dat, startYear){
  
  res = dat %>% 
    arrange(year) %>%  
    mutate(change = (LM - LM[year == startYear]) / LM[year == startYear])
  res
}

pChange(dat, startYear = 2022)

    LM year     change
1 85.2 2020  0.7387755
2 56.7 2021  0.1571429
3 49.0 2022  0.0000000
4 71.0 2023  0.4489796
5 33.0 2024 -0.3265306
6 96.5 2025  0.9693878
7 26.3 2026 -0.4632653
8 21.3 2027 -0.5653061

Answer 2

我建议您先计算与上一年相比的所有百分比变化，然后再对您要查找的那个进行子集化：

library(tidyverse)

dat <- tribble(
  ~ LM, ~ year,
  85.20000, 2020,
  56.70000, 2021,
  49.00000, 2022,
  71.00000, 2023,
  33.00000, 2024,
  96.50000, 2025,
  26.30000, 2026,
  21.30000, 2027,
)

pchange <- function(dat, start_year) {
  output <- dat %>%
    mutate(change = (LM - lag(LM, order_by = year)) / LM) %>%
    filter(year == start_year)
  output$change
}

pchange(dat, 2025)

计算基于一年的减少百分比

compute percentage reduction based on a year

r

percentage

dplyr