按组划分的列之间的差异

Question

嗨，我需要实现以下目标：

grp value   diff
1   10       NA  # diff[1] = value[2]-value[0] of grp = 1
1   15       10  # diff[2] = value[3]-value[1] of grp = 1
1   20        5  # diff[3] = value[4]-value[3] of grp = 1
1   10       NA  # diff[4] = value[5]-value[3] of grp = 1
2   25       NA  # diff[5] = value[6]-value[4] of grp = 2
2   30       10  # diff[6] = value[7]-value[5] of grp = 2
2   35       NA  # diff[7] = value[8]-value[6] of grp = 2

我已经尝试过使用shift和lag之类的函数，但无法获得这种类型的解决方案，因为我需要将先前值的差值相减并减去它们，例如diff[i] = value[i+1] - value[i-1]

使用for loop遇到错误，所以有更好的方法吗？

Answer 1

我认为您在描述微分值时有错字。但是，如果您希望diff[i]成为value[i+1]-value[i-1]，则可以通过同时使用lead中的lag和dplyr来实现。

library(dplyr)
df %>% group_by(grp) %>% mutate(diff = lead(value) -lag(value))

# A tibble: 7 x 3
# Groups:   grp [2]
    grp value  diff
  <dbl> <dbl> <dbl>
1     1    10    NA
2     1    15    10
3     1    20    -5
4     1    10    NA
5     2    25    NA
6     2    30    10
7     2    35    NA
它看起来像您要找的东西吗？

Answer 2

我们可以做

library(dplyr)
df1 %>%
  group_by(grp)
  mutate(diff = lag(abs(lead(value, 2) - value)))
# A tibble: 7 x 3
# Groups:   grp [2]
#    grp value  diff
#  <int> <int> <int>
#1     1    10    NA
#2     1    15    10
#3     1    20     5
#4     1    10    NA
#5     2    25    NA
#6     2    30    10
#7     2    35    NA