通过R根据规则在同一组的两行之间执行减法

Perform subtraction between two rows of the same group based on a rule via R

我有一个数据框,其中包含 2005 年和 2018 年的土地使用数据。我想生成一个新的 data.frame,显示每一列的 2005 年和 2018 年之间的差异,以便如果减少,减号(-)。例如,如果变量 Veg 在 2005 年有 1000 公顷,在 2018 年有 8000 公顷,那么 data.frame 应该表示 -200。

data.frame 示例:

df

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")

我想要这样的输出

这是一个对行顺序具有鲁棒性的 dplyr 管道(例如,如果 2005 曾经出现 after 2018 无论出于何种原因)。

library(dplyr)
library(tidyr) # pivot_*
df %>%
  pivot_longer(-c("place", "year")) %>%
  pivot_wider(c("place", "name"), names_from = "year", values_from = "value") %>%
  mutate(results = coalesce(`2005` - `2018`, 0)) %>%
  transmute(place, name, results = dplyr::coalesce(results, 0)) %>%
  pivot_wider(place, names_from = "name", values_from = "results")
# # A tibble: 6 x 8
#   place    Veg    Agri  Past    Urb SoloExp Hidro total
#   <chr>  <dbl>   <dbl> <dbl>  <dbl>   <dbl> <dbl> <dbl>
# 1 F01    -10.6 -1264.  1335. -54.9    -6.82  1.25     0
# 2 F02   -234.   -963.  1252. -54.9    -1.33  1.25     0
# 3 F03    -77.4   -46.2  182. -53.7    -4.66  0        0
# 4 F04    -34.6   -46.2  138. -53.7    -3.49  0        0
# 5 F05   -111.   -917.  1026.  -1.16    2.25  1.08     0
# 6 F06    -61.5  -227.   283.   0       5.16  1.16     0

我们可以使用 across 函数来做到这一点:

library(dplyr)

df %>% 
  mutate(across(-c(place, year), ~ lag(., default = .[1]) - .)) %>% 
  filter(year==2018) %>% 
  select(-year)
  place        Veg        Agri      Past        Urb   SoloExp    Hidro total
1   F01  -10.64428 -1263.50889 1334.6093 -54.884551 -6.818990 1.247376     0
2   F02 -233.92461  -963.47334 1252.3657 -54.884551 -1.330535 1.247376     0
3   F03  -77.42048   -46.23608  182.0338 -53.720333 -4.656871       NA     0
4   F04  -34.59390   -46.23608  138.0430 -53.720333 -3.492653       NA     0
5   F05 -111.18279  -917.23727 1026.2579  -1.164218  2.245277 1.081059     0
6   F06  -61.45407  -227.43825  282.5723         NA  5.155821 1.164218     0