得到与前一个值和下一个值的差异

Question

我有这样的数据。一个小样本实际上df更大：

LOW   1  4  NA 
MID   3  4  4
HIG   2  5  4

并且希望得到LOW和HIG与MID的区别，这样结束df将如下所示：

LOW   2  0  NA 
MID   3  4  4
HIG   1  1  0

所以你得到的是: LOW =3- 1 =2和HIG =3-2=1。我可以通过VBA宏来实现，但是想用R进行缩放。

Answer 1

这可以用mutate_if/mutate_at来完成

library(dplyr)
df1 %>%
   mutate_if(is.numeric, ~ case_when(grp != 'MID' ~ 
         abs(. - .[grp == 'MID']), TRUE ~ .))
#   grp v1 v2 v3
#1 LOW  2  0 NA
#2 MID  3  4  4
#3 HIG  1  1  0

或在base R中

i1 <- df1$grp == 'MID'
df1[!i1, -1] <- abs(df1[!i1, -1] - rep(unlist(df1[i1, -1]), each = sum(!i1)))

数据

df1 <- structure(list(grp = c("LOW", "MID", "HIG"), v1 = c(1L, 3L, 2L
), v2 = c(4L, 4L, 5L), v3 = c(NA, 4L, 4L)), class = "data.frame", row.names = c(NA, 
-3L))

Answer 2

您可以在减去'LOW'、'HIG'行之后更改'MID'：

df1[df1$grp == 'LOW', -1] <- abs(df1[df1$grp == 'MID',-1]- df1[df1$grp == 'LOW',-1])
df1[df1$grp == 'HIG', -1] <- abs(df1[df1$grp == 'MID',-1]- df1[df1$grp == 'HIG',-1])

df1
#  grp v1 v2 v3
#1 LOW  2  0 NA
#2 MID  3  4  4
#3 HIG  1  1  0