Multirow deletion: delete row depending on other row - r

Question

I'm stuck with a quite complex problem. I have a data frame with three rows: id, info and rownum. The data looks like this:
id info row
1 a 1
1 b 2
1 c 3
2 a 4
3 b 5
3 a 6
4 b 7
4 c 8
What I want to do now is to delete all other rows of one id if one of the rows contains the info a. This would mean for example that row 2 and 3 should be removed as row 1's coloumn info contains the value a. Please note that the info values are not ordered (id 3/row 5 & 6) and cannot be ordered due to other data limitations.
I solved the case using a for loop:
# select all id containing an "a"-value
a_val <- data$id[grep("a", data$info)]
# check for every id containing an "a"-value
for(i in a_val) {
temp_data <- data[which(data$id == i),]
# only go on if the given id contains more than one row
if (nrow(temp_data) > 1) {
for (ii in nrow(temp_data)) {
if (temp_data$info[ii] != "a") {
temp <- temp_data$row[ii]
if (!exists("delete_rows")) {
delete_rows <- temp
} else {
delete_rows <- c(delete_rows, temp)
}
}
}
}
}
My solution works quite well. Nevertheless, it is very, very, very slow as the original data contains more than 700k rows and more that 150k rows with an "a"-value.
I could use a foreach loop with 4 cores to speed it up, but maybe someone could give me a hint for a better solution.
Best regards,
Arne
[UPDATE]
The outcome should be:
id info row
1 a 1
2 a 4
3 a 6
4 b 7
4 c 8

Answer 1

Here is one possible solution.
First find ids where info contains "a":
ids <- with(data, unique(id[info == "a"]))
Subset the data:
subset(data, (id %in% ids & info == "a") | !id %in% ids)
Output:
id info row
1 1 a 1
4 2 a 4
6 3 a 6
7 4 b 7
8 4 c 8
An alternative solution (maybe harder to decipher):
subset(data, info == "a" | !rep.int(tapply(info, id, function(x) any(x == "a")),
table(id)))
Note. #BenBarnes found out that this solution only works if the data frame is ordered according to id.

Answer 2

You might want to investigate the data.table package:
EDIT: If the row variable is not a sequential numbering of each row in your data (as I assumed it was), you could create such a variable to obtain the original row order:
library(data.table)
# Create data.table of your data
dt <- as.data.table(data)
# Create index to maintain row order
dt[, idx := seq_len(nrow(dt))]
# Set a key on id and info
setkeyv(dt, c("id", "info"))
# Determine unique ids
uid <- dt[, unique(id)]
# subset your data to select rows with "a"
dt2 <- dt[J(uid, "a"), nomatch = 0]
# identify rows of dataset where the id doesn't have an "a"
dt3 <- dt[J(dt2[, setdiff(uid, id)])]
# rbind those two data.tables together
(dt4 <- rbind(dt2, dt3))
# id info row idx
# 1: 1 a 1 1
# 2: 2 a 4 4
# 3: 3 a 6 6
# 4: 4 b 7 7
# 5: 4 c 8 8
# And if you need the original ordering of rows,
dt5 <- dt4[order(idx)]
Note that setting a key for the data.table will order the rows according to the key columns. The last step (creating dt5) sets the row order back to the original.

Answer 3

Here is a way using ddply:
df <- read.table(text="id info row
1 a 1
1 b 2
1 c 3
2 a 4
3 b 5
3 a 6
4 b 7
4 c 8",header=TRUE)
library("plyr")
ddply(df,.(id),subset,rep(!'a'%in%info,length(info))|info=='a')
Returns:
id info row
1 1 a 1
2 2 a 4
3 3 a 6
4 4 b 7
5 4 c 8

Answer 4

if df is this (RE Sacha above) use match which just finds the index of the first occurrence:
df <- read.table(text="id info row
1 a 1
1 b 2
1 c 3
2 a 4
3 b 5
3 a 6
4 b 7
4 c 8",header=TRUE)
# the first info row matching 'a' and all other rows that are not 'a'
with(df, df[c(match('a',info), which(info != 'a')),])
id info row
1 1 a 1
2 1 b 2
3 1 c 3
5 3 b 5
7 4 b 7
8 4 c 8

Answer 5

try to take a look at subset, it's quite easy to use and it will solve your problem.
you just need to specify the value of the column that you want to subset based on, alternatively you can choose more columns.
http://stat.ethz.ch/R-manual/R-devel/library/base/html/subset.html
http://www.statmethods.net/management/subset.html