Question

请考虑以下我的数据字段子集：

  Pack  side row col     v1      v2
1   P1  Left   1   1 0.4094 -3.8700
2   P1 Right   1   1 0.4110 -3.5245
3   P1  Left   1   2 0.4118 -3.4876
4   P1 Right   1   2 0.4108 -3.7268
5   P1  Left   1   3 0.4119 -3.5322
6   P1 Right   1   3 0.4110 -3.6101

我对v1和v2左右之间的差异感兴趣，特别是v1的百分比差异和v2的直线差异。

我想要的输出是一个新的数据字段，如下所示：

 Pack   row col dv1              dv2
1 P1    1   1   0.389294404      0.3455
2 P1    1   2   -0.243427459    -0.2392
3 P1    1   3   -0.218978102    -0.0779

其中，对于v1，dv1的计算是(Right-Left)/Left*100，对于v2，dv2的计算是Right-Left。

这是df数据：

df <- structure(list(Pack = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("P1", 
"P2", "P3", "P4"), class = "factor"), side = structure(c(1L, 
2L, 1L, 2L, 1L, 2L), .Label = c("Left", "Right"), class = "factor"), 
    row = c(1L, 1L, 1L, 1L, 1L, 1L), col = c(1L, 1L, 2L, 2L, 
    3L, 3L), v1 = c(0.4094, 0.411, 0.4118, 0.4108, 0.4119, 0.411
    ), v2 = c(-3.87, -3.5245, -3.4876, -3.7268, -3.5322, -3.6101
    )), .Names = c("Pack", "side", "row", "col", "v1", "v2"), row.names = c(NA, 
6L), class = "data.frame")

谢谢！

Answer 1

我们可以首先按side对行进行排序，并确保首先有Left，然后是Right。这给出了

library(tidyverse)
df %>% arrange(side) %>% group_by(Pack, row, col) %>% 
  summarise(dv1 = (v1[2] - v1[1]) / v1[1] * 100, dv2 = v2[2] - v2[1])
# A tibble: 3 x 5
# Groups:   Pack, row [?]
#   Pack    row   col    dv1     dv2
#   <fct> <int> <int>  <dbl>   <dbl>
# 1 P1        1     1  0.391  0.345 
# 2 P1        1     2 -0.243 -0.239 
# 3 P1        1     3 -0.218 -0.0779

或者只是

df %>% arrange(side) %>% group_by(Pack, row, col) %>% 
  summarise(dv1 = diff(v1) / v1[1] * 100, dv2 = diff(v2))

Answer 2

另一种使用dplyr和lead的{{1}}方法

mutate

编辑-更改了过滤器以删除偶数行号

在R中，根据行之间的差异计算新列

2 个答案: