如何根据R中的组计算两行之间的百分比

Question

我有一个 datafrema，其中包含 2005 年到 2018 年之间 6 个点的土地使用数据。我想计算 2005 年到 2018 年之间的百分比变化。

df<-structure(list(place = c("F01", "F01", "F02", "F02", "F03", "F03", "F04", "F04", "F05", "F05", "F06", "F06"), year = c(2005, 2018, 2005, 2018, 2005, 2018, 2005, 2018, 2005, 2018, 2005, 2018), Veg = c(12281.5824712026, 12292.2267477317, 7254.98919713131, 7488.9138055415, 864.182200710528, 941.602680778032, 549.510775817472, 584.104674537216, 5577.10195081334, 5688.28474549675, 1244.96456185886, 1306.41862713264), Agri = c(113.178596532624, 1376.68748390712, 85.2373706436, 1048.71071335262, 0, 46.236076173504, 0, 46.236076173504, 85.2373706436, 1002.47463717912, 1.413692976528, 228.851945376768 ), Past = c(9190.16856517738, 7855.55923692456, 5029.33750161394, 3776.9718412309, 983.015569149264, 800.981808818688, 710.255983089744, 572.213021852304, 3726.66100294858, 2700.40306039963, 879.982298683488, 597.410020198656), Urb = c(146.026168634304, 200.910719487744, 146.026168634304, 200.910719487744, 141.119822421648, 194.840155529712, 141.119822421648, 194.840155529712, 4.906346212656, 6.070563958032, NA, NA), SoloExp = c(61.12143163224, 67.940421283728, 61.12143163224, 62.451966198384, 50.144521461552, 54.801392443056, 49.146620536944, 52.639273773072, 9.895850835696, 7.650573755328, 6.320039189184, 1.164217745376), Hidro = c(9.230583552624, 7.983207396864, 9.230583552624, 7.983207396864, NA, NA, NA, NA, 7.401098524176, 6.320039189184, 5.654771906112, 4.490554160736), total = c(691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971, 691953.981181971)), row.names = c(NA, -12L), class = "data.frame")

我尝试使用lead命令计算2005年和2018年之间的差异，但我没有成功：

df2<-df%>%
  select(-c(total))%>%
  replace(is.na(.), 0)%>%
  pivot_longer(cols = c(3:8),
               names_to = 'classe',
               values_to = 'area')%>%
  group_by(place, classe)%>%
  mutate(percent=(((((area)-lead(area))/area)*100)*-1))%>%
  pivot_wider(names_from = 'classe',
              values_from = 'percent')%>%
  select(-c(area, year))

例如对于 Veg class 我希望得到：

place   Veg
F01 0.09
F02 3.22
F03 8.96
F04 6.30
F05 1.99
F06 4.94

Answer 1

如果您需要所有参数的百分比，这是一种解决方案

library(dplyr)
library(tidyr)

df_new<-df %>%
select(-(total))%>%
  replace(is.na(.), 0)%>%
  pivot_longer(cols = c(3:8),
               names_to = 'classe',
               values_to = 'area') %>%
  pivot_wider(names_from=year, values_from=area) %>%
  mutate(percent=(`2018`-`2005`)/`2005`) %>%
  select(-`2018`,-`2005`) %>%
  pivot_wider( names_from="classe", values_from="percent")
  

df_new
#> # A tibble: 6 × 7
#>   place      Veg  Agri   Past     Urb SoloExp   Hidro
#>   <chr>    <dbl> <dbl>  <dbl>   <dbl>   <dbl>   <dbl>
#> 1 F01   0.000867  11.2 -0.145   0.376  0.112   -0.135
#> 2 F02   0.0322    11.3 -0.249   0.376  0.0218  -0.135
#> 3 F03   0.0896   Inf   -0.185   0.381  0.0929 NaN    
#> 4 F04   0.0630   Inf   -0.194   0.381  0.0711 NaN    
#> 5 F05   0.0199    10.8 -0.275   0.237 -0.227   -0.146
#> 6 F06   0.0494   161.  -0.321 NaN     -0.816   -0.206

^{由 reprex package (v2.0.1)}

创建于 2022-01-09

Answer 2

另一个可能的解决方案：

library(tidyverse)

df %>% 
  group_by(place) %>% 
  summarise(across(-year, ~ 100*(last(.x) / first(.x) - 1)))

#> # A tibble: 6 × 8
#>   place    Veg   Agri  Past   Urb SoloExp Hidro total
#>   <chr>  <dbl>  <dbl> <dbl> <dbl>   <dbl> <dbl> <dbl>
#> 1 F01   0.0867  1116. -14.5  37.6   11.2  -13.5     0
#> 2 F02   3.22    1130. -24.9  37.6    2.18 -13.5     0
#> 3 F03   8.96     Inf  -18.5  38.1    9.29  NA       0
#> 4 F04   6.30     Inf  -19.4  38.1    7.11  NA       0
#> 5 F05   1.99    1076. -27.5  23.7  -22.7  -14.6     0
#> 6 F06   4.94   16088. -32.1  NA    -81.6  -20.6     0

如何根据R中的组计算两行之间的百分比

how to calculate the percentage between two rows based on a group in R

r

percentage