Delete rows with value if not only value in group - r

Question

Somewhat new to R and I find myself needing to delete rows based on multiple criteria. The data frame has 3 columns and I need to delete rows where bid=99 and there are values less than 99 grouping by rid and qid. The desired output at an rid and qid level are bid has multiple values less than 99 or bid=99.
rid qid bid
1 1 5
1 1 6
1 1 99
1 2 6
2 1 7
2 1 99
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
3 2 99
4 1 2
4 1 6
4 2 1
4 2 2
4 2 99
5 1 99
5 2 99
The expected output...
rid qid bid
1 1 5
1 1 6
1 2 6
2 1 7
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
4 1 2
4 1 6
4 2 1
4 2 2
5 1 99
5 2 99
Any assistance would be appreciated.

Answer 1

You can use the base R function ave to generate a dropping variable like this:
df$dropper <- with(df, ave(bid, rid, qid, FUN= function(i) i == 99 & length(i) > 1))
ave calculates a function on bid, grouping by rid and qid. The function tests if each element of the grouped bid values i is 99 and if i has a length greater than 1. Also, with is used to reduce typing.
which returns
df
rid qid bid dropper
1 1 1 5 0
2 1 1 6 0
3 1 1 99 1
4 1 2 6 0
5 2 1 7 0
6 2 1 99 1
7 2 2 2 0
8 2 2 3 0
9 3 1 7 0
10 3 1 8 0
11 3 2 1 0
12 3 2 99 1
13 4 1 2 0
14 4 1 6 0
15 4 2 1 0
16 4 2 2 0
17 4 2 99 1
18 5 1 99 0
19 5 2 99 0
then drop the undesired observations with df[dropper == 0, 1:3] which will simultaneously drop the new variable.

Answer 2

If you want to just delete rows where bid = 99 then use dplyr.
library(dplyr)
df <- df %>%
filter(bid != 99)
Where df is your data frame. and != means not equal to
Updated solution using dplyr
df %>%
group_by(rid, qid) %>%
mutate(tempcount = n())%>%
ungroup() %>%
mutate(DropValue =ifelse(bid == 99 & tempcount > 1, 1,0) ) %>%
filter(DropValue == 0) %>%
select(rid,qid,bid)

Answer 3

Here is another option with all and if condition in data.table to subset the rows after grouping by 'rid' and 'qid'
library(data.table)
setDT(df1)[, if(all(bid==99)) .SD else .SD[bid!= 99], .(rid, qid)]
# rid qid bid
# 1: 1 1 5
# 2: 1 1 6
# 3: 1 2 6
# 4: 2 1 7
# 5: 2 2 2
# 6: 2 2 3
# 7: 3 1 7
# 8: 3 1 8
# 9: 3 2 1
#10: 4 1 2
#11: 4 1 6
#12: 4 2 1
#13: 4 2 2
#14: 5 1 99
#15: 5 2 99
Or without using the if
setDT(df1)[df1[, .I[all(bid==99) | bid != 99], .(rid, qid)]$V1]

Answer 4

Here is a solution using dplyr, which is a very expressive framework for this kind of problems.
df <- read.table(text =
" rid qid bid
1 1 5
1 1 6
1 1 99
1 2 6
2 1 7
2 1 99
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
3 2 99
4 1 2
4 1 6
4 2 1
4 2 2
4 2 99
5 1 99
5 2 99",
header = TRUE, stringsAsFactors = FALSE)
Dplyr verbs allow to express the program in a way that is close to the very terms of your questions:
library(dplyr)
res <-
df %>%
group_by(rid, qid) %>%
filter(!(any(bid < 99) & bid == 99)) %>%
ungroup()
# # A tibble: 15 × 3
# rid qid bid
# <int> <int> <int>
# 1 1 1 5
# 2 1 1 6
# 3 1 2 6
# 4 2 1 7
# 5 2 2 2
# 6 2 2 3
# 7 3 1 7
# 8 3 1 8
# 9 3 2 1
# 10 4 1 2
# 11 4 1 6
# 12 4 2 1
# 13 4 2 2
# 14 5 1 99
# 15 5 2 99
Let's check we get the desired output:
desired_output <- read.table(text =
" rid qid bid
1 1 5
1 1 6
1 2 6
2 1 7
2 2 2
2 2 3
3 1 7
3 1 8
3 2 1
4 1 2
4 1 6
4 2 1
4 2 2
5 1 99
5 2 99",
header = TRUE, stringsAsFactors = FALSE)
identical(as.data.frame(res), desired_output)
# [1] TRUE