Subseting data frame based on multiple criteria for deletion of rows - r

Question

Consider the following data frame consisting of column names "id" and "x", where each id is repeated four times. Data is as follows:
df<-data.frame("id"=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4),
"x"=c(2,2,1,1,2,3,3,3,1,2,2,3,2,2,3,3))
The question is about how to subset the data frame by the following criteria:
(1) keep all entries of each id, if its corresponding values in column x does not contain 3 or it has 3 as the last number.
(2) for a given id with multiple 3s in column x, keep all the numbers up to the first 3 and delete the remaining 3s. The expected output would look like:
id x
1 1 2
2 1 2
3 1 1
4 1 1
5 2 2
6 2 3
7 3 1
8 3 2
9 3 2
10 3 3
11 4 2
12 4 2
13 4 3
I am familiar with the use of the 'filter' function in dplyr package to subset data, but this particular situation confuses me because of the complexity of the above criteria. Any help on this would be greatly appraciated.

Answer 1

Here's one solution that uses / creates some new columns to help you filter on:
library(dplyr)
df<-data.frame("id"=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4),
"x"=c(2,2,1,1,2,3,3,3,1,2,2,3,2,2,3,3))
df %>%
group_by(id) %>% # for each id
mutate(num_threes = sum(x == 3), # count number of 3s
flag = ifelse(unique(num_threes) > 0, # if there is a 3
min(row_number()[x == 3]), # keep the row of the first 3
0)) %>% # otherwise put a 0
filter(num_threes == 0 | row_number() <= flag) %>% # keep ids with no 3s or up to first 3
ungroup() %>%
select(-num_threes, -flag) # remove helpful columns
# # A tibble: 13 x 2
# id x
# <dbl> <dbl>
# 1 1 2
# 2 1 2
# 3 1 1
# 4 1 1
# 5 2 2
# 6 2 3
# 7 3 1
# 8 3 2
# 9 3 2
# 10 3 3
# 11 4 2
# 12 4 2
# 13 4 3

Answer 2

this works for me:
data
df<-data.frame("id"=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4),
"x"=c(2,2,1,1,2,3,3,3,1,2,2,3,2,2,3,3))
commands
library(dplyr)
df <- mutate(df, before = lag(x))
df$condition1 <- 1
df$condition1[df$x == 3 & df$before == 3] <- 0
final_df <- df[df$condition1 == 1, 1:2]
result
x id
1 2
1 2
1 1
1 1
2 2
2 3
3 1
3 2
3 2
3 3
4 2
4 2
4 3`

Answer 3

One idea is to pick out the rows with x==3 and use unique() over them. Then append the unique rows with just single 3 to the rest part of the data frame, and finally order the rows.
Here is a solution with base R for the idea above:
res <- (r <- with(df,rbind(df[x!=3,],unique(df[x==3,]))))[order(as.numeric(rownames(r))),]
rownames(res) <- seq(nrow(res))
which give
> res
id x
1 1 2
2 1 2
3 1 1
4 1 1
5 2 2
6 2 3
7 3 1
8 3 2
9 3 2
10 3 3
11 4 2
12 4 2
13 4 3
DATA
df<-data.frame("id"=c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4),
"x"=c(2,2,1,1,2,3,3,3,1,2,2,3,2,2,3,3))