问题描述
我有一个分组的小标题,其中假设一个函数从上一行获取其值,则必须从其他参数中计算出几个参数。我试图找到涉及lag
,mutate
,case_when
和aggregate
的答案,但在以下玩具数据集中实现这些答案没有运气:
library(tidyverse)
set.seed(42)
df <- tibble(
gr = c(1,1,2,2),t = rep((seq(1:3)),v1 = c(1,NA,1.6,NA),v2 = rnorm(6),v3 = c(-0.2,0.3,-0.6,-0.2,0.2)
)
# 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]
if ia.na(df$v1[i]) is TRUE
,然后计算:
mutate(R1=abs(v1-5*v2)) %>% mutate(R2=(R1^(1/2))) %>% mutate(RI3=R1/R2)
逐行以便fill
排序和分组的数据集中的空白;
Rdf <- df
Rdf$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$RI3[5]
Rdf <- Rdf %>% mutate(R1=abs(v1-5*v2)) %>%
mutate(R2=abs(R1*v2)^(1/2)) %>% mutate(RI3=R1/R2)
Rdf
并导致:
# A tibble: 6 x 8
# Groups: gr [2]
gr t v1 v2 v3 R1 R2 RI3
<dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 -1.39 -0.2 7.94 3.32 2.39
2 1 2 2.39 -0.279 0.3 3.79 1.03 3.68
3 1 3 3.68 -0.133 -0.6 4.35 0.762 5.71
4 2 1 1.6 0.636 -0.2 1.58 1.00 1.58
5 2 2 1.58 -0.284 1 3.00 0.923 3.25
6 2 3 3.25 -2.66 0.2 16.5 6.63 2.49
我想将for-loop
中的if-condition
应用于nested df
即可。
任何建议实现此效果都很好!
解决方法
我实现了一个for循环。但是我不确定我从给定种子的相同df开始。希望它能满足您的需求。
当我需要编写看起来很复杂的for循环时,可以使用browser()进行构建。
library(tibble)
library(dplyr)
set.seed(42)
df <- tibble(
gr = c(1,1,2,2),t = rep((seq(1:3)),v1 = c(1,NA,1.6,NA),v2 = rnorm(6),v3 = c(-0.2,0.3,-0.6,-0.2,0.2)
)
# Data prep
df <- df %>%
group_by(gr) %>%
arrange(t,.by_group = TRUE) %>%
mutate(R1=abs(v1-5*v2)) %>%
mutate(R2=abs(R1*v2)^(1/2)) %>% #
mutate(RI3=R1/R2) %>%
ungroup()
#going through df row by row
for (i in 1:nrow(df)) {
#browser()
# run into problems with i == 1 for the lagged operation,hence made two cases
if (i == 1) {
df$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]
}