R 中的重新排序因子基于对应于原始因子的“子因子”的特定级别的值

Question

I have a dataset in R with the following structure:

sub ses value

sub1 0.1454

sub1 1.0857

sub2 0.1123

sub2 1.2345

sub3 0.2105

sub3 1.0665

df <- data.frame(sub=c("sub1","sub1","sub2","sub2","sub3","sub3"),ses=c(0,1,0,1,0,1),value=c(.1454,.0857,.1123,.2345,.2105,.0665))

For the purposes of plotting the change in value between session 0 and 1, I'm interested in reordering the dataset so that the order of Sub reflects the variable Value only for Ses == 0 in descending order, so you can easily visualize differences in "Baseline" (Ses == 0) values across subjects, while keeping both of each subject's values together in the data frame:

sub ses value

sub2 0.1123

sub2 1.2345

sub1 0.1454

sub1 1.0857

sub3 0.2105

sub3 1.0665

df <- data.frame(sub=c("sub2","sub2","sub1","sub1","sub3","sub3"),ses=c(0,1,0,1,0,1),value=c(.1123,.2345,.1454,.0857,.2105,.0665))

This has turned out to be a nontrivial solution for a relatively large dataset. I want something akin to

df.ordered <- df %>% group_by(sub) %>% arrange(desc(ses == 0, .by_group=TRUE))

but this syntax is incorrect, as you cannot specify a value of ses this way (it will just reorder by ses in this case). Much appreciation for anyone who can give me guidance on how to implement this. Thanks!

Answer 1

We can use arrange on a factor with levels specified according to the custom values we want it to order

library(dplyr)
df %>%
   dplyr::arrange(factor(sub, levels = 
               unique(sub[ses ==0][order(value[ses == 0])])))
#  sub ses  value
#1 sub2   0 0.1123
#2 sub2   1 0.2345
#3 sub1   0 0.1454
#4 sub1   1 0.0857
#5 sub3   0 0.2105
#6 sub3   1 0.0665

Answer 2

You can first extract the correct order in which you want to arrange the data by filtering only the 0 values and arranging them by value

library(dplyr)
correct_order <- df %>%
                  filter(ses == 0) %>%
                  arrange(value) %>%
                  pull(sub)
correct_order
#[1] "sub2" "sub1" "sub3"

You can then arrange them using match :

df %>% arrange(match(sub, correct_order), ses)
#  sub ses  value
#1 sub2   0 0.1123
#2 sub2   1 0.2345
#3 sub1   0 0.1454
#4 sub1   1 0.0857
#5 sub3   0 0.2105
#6 sub3   1 0.0665