有没有办法沿着排序和分组的tibble进行条件操作和多行操作？

Question

我有一个分组的tibble，其中几个参数必须根据其他参数进行计算，假设一个函数从前一行获取其值。我试图找到涉及lag、mutate、case_when和aggregate的答案，但在下面的玩具数据集中没有实现它们：

library(tidyverse)

set.seed(42)

df <- tibble(
  gr = c(1,1,1,2,2,2),
  t = rep((seq(1:3)),2),
  v1 = c(1,NA,NA,1.6,NA,NA),
  v2 = rnorm(6),
  v3 = c(-0.2,0.3,-0.6,-0.2,1,0.2)
  ) 

# These operations 
(df <- df %>% group_by(gr) %>%  arrange(t, .by_group = TRUE) %>% 
  mutate(R1=abs(v1-5*v2)) %>% 
  mutate(R2=abs(R1*v2)^(1/2)) %>% mutate(RI3=R1/R2)) 

# A tibble: 6 x 8
# Groups:   gr [2]
     gr     t    v1     v2    v3    R1    R2   RI3
  <dbl> <int> <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     1   1   -1.39   -0.2  7.94  3.32  2.39
2     1     2  NA   -0.279   0.3 NA    NA    NA   
3     1     3  NA   -0.133  -0.6 NA    NA    NA   
4     2     1   1.6  0.636  -0.2  1.58  1.00  1.58
5     2     2  NA   -0.284   1   NA    NA    NA   
6     2     3  NA   -2.66    0.2 NA    NA    NA   

现在，我需要做的是使用df$RI3[i-1]作为df$v1[i]的输入

if ia.na(df$v1[i]) is TRUE，然后计算：

mutate(R1=abs(v1-5*v2))  %>%  mutate(R2=(R1^(1/2))) %>% mutate(RI3=R1/R2)  

逐行排列( fill )已排序和分组的数据集内的间隙；

一个接一个地执行此操作将如下所示：

Rdf <-   df
Rdf$v1[2] <- df$RI3[1]
Rdf$v1[5] <- df$RI3[4]
Rdf <- Rdf %>%  mutate(R1=abs(v1-5*v2)) %>% 
  mutate(R2=abs(R1*v2)^(1/2)) %>% mutate(RI3=R1/R2) 
Rdf
Rdf$v1[3] <- Rdf$RI3[2]
Rdf$v1[6] <- Rdf$RI3[5]
Rdf <- Rdf %>%  mutate(R1=abs(v1-5*v2)) %>% 
  mutate(R2=abs(R1*v2)^(1/2)) %>% mutate(RI3=R1/R2) 
Rdf

并将导致：

# A tibble: 6 x 8
# Groups:   gr [2]
     gr     t    v1     v2    v3    R1    R2   RI3
  <dbl> <int> <dbl>  <dbl> <dbl> <dbl> <dbl> <dbl>
1     1     1  1    -1.39   -0.2  7.94 3.32   2.39
2     1     2  2.39 -0.279   0.3  3.79 1.03   3.68
3     1     3  3.68 -0.133  -0.6  4.35 0.762  5.71
4     2     1  1.6   0.636  -0.2  1.58 1.00   1.58
5     2     2  1.58 -0.284   1    3.00 0.923  3.25
6     2     3  3.25 -2.66    0.2 16.5  6.63   2.49

我猜应用于nested df的if-condition中的for-loop应该可以工作。

任何实现这一点的建议都是很棒的！

Answer 1

我实现了一个for循环。但我不确定我从给定种子的相同df开始。希望它能满足您的需求。

当我需要编写看起来很复杂的for循环时，我会使用browser()来构建它。

library(tibble)
library(dplyr)

set.seed(42)
df <- tibble(
  gr = c(1,1,1,2,2,2),
  t = rep((seq(1:3)),2),
  v1 = c(1,NA,NA,1.6,NA,NA),
  v2 = rnorm(6),
  v3 = c(-0.2,0.3,-0.6,-0.2,1,0.2)
) 

# Data prep
df <- df %>%
  group_by(gr) %>%  
  arrange(t, .by_group = TRUE) %>% 
  mutate(R1=abs(v1-5*v2)) %>% 
  mutate(R2=abs(R1*v2)^(1/2)) %>% #
  mutate(RI3=R1/R2) %>%
  ungroup()

#going through df row by row
for (i in 1:nrow(df)) {
    #browser()
    # run into problems with i == 1 for the lagged operation, hence made two cases
    if (i == 1) {
      df$v1[i] <- if_else(is.na(df$v1[i]), df$RI3[i], df$v1[i])
    } else {
      df$v1[i] <- if_else(is.na(df$v1[i]), df$RI3[i-1], df$v1[i])
    }
  
  # rowwise calculation
  df$R1[i] <- abs(df$v1[i]-5*df$v2[i])
  df$R2[i] <- abs(df$R1[i]*df$v2[i])^(1/2)
  df$RI3[i]=df$R1[i]/df$R2[i]
  
}