how to subset data frame if all the conditions are met? [duplicate] - r

Question

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Finding ALL duplicate rows, including "elements with smaller subscripts"
(9 answers)
Closed 5 years ago.
i have two message ids say 197, 198 . I want to subset the data frame for those users who have got the messages from these ids. I only want those rows which contains both these message ids.
The data frame is m
I have used the code
a = c(197,198)
n = subset(m$userid,m$mid %in% a)
I also tried
n = m[m$mid == 197 & m$mid == 198]
both of these codes are creating OR output whereas I want AND output.
here is the sample dataframe:
mid userid opened
197 1022 Y
197 1036 N
197 1100 Y
198 1000 Y
198 1022 N
198 1036 Y
I want output as records containing userid for both mid 197 &198
mid userid opened
197 1022 Y
197 1036 N
198 1022 N
198 1036 Y

Answer 1

Using sqldf one solution could be achieved as:
# data
m <- read.table(text = "mid userid opened
197 1022 Y
197 1036 N
197 1100 Y
198 1000 Y
198 1022 N
198 1036 Y", header = T, stringsAsFactors = F)
library(sqldf)
result <- sqldf("SELECT * FROM m
WHERE userid in (SELECT userid FROM m WHERE mid == 197) AND
userid in (SELECT userid FROM m WHERE mid == 198)")
result
# mid userid opened
# 1 197 1022 Y
# 2 197 1036 N
# 3 198 1022 N
# 4 198 1036 Y

Answer 2

Using duplicated :
m[duplicated(m$userid) | duplicated(m$userid,fromLast = T), ]
# mid userid opened
# 1 197 1022 Y
# 2 197 1036 N
# 5 198 1022 N
# 6 198 1036 Y
With your real data you may need first : m2 <- subset(m,mid %in% a) to make sure you have only mid from a in your table before applying my solution.

Answer 3

For the sake of completeness, here are two data.table approaches. Both are capable of handling a of arbitrary length, i.e., with more than just 2 selected mid.
Join
library(data.table)
setDT(m)[m[mid %in% a][, uniqueN(mid), by = .(userid)][V1 == uniqueN(a)],
on = "userid"]
mid userid opened V1
1: 197 1022 Y 2
2: 198 1022 N 2
3: 197 1036 N 2
4: 198 1036 Y 2
The expression
m[mid %in% a][, uniqueN(mid), by = .(userid)][V1 == uniqueN(a)]
userid V1
1: 1022 2
2: 1036 2
filters m, then counts the number of unique mid by userid and returns those userid which have matches with all entries in a. (Instead of uniqueN(a), length(a) can be used but the former is safer).
Subsetting by row indices
There is an alternative approach which returns the row id's of m which are then used for subsetting:
m[mid %in% a][, .I[uniqueN(mid) == uniqueN(a)], by = .(userid)]
userid V1
1: 1022 1
2: 1022 5
3: 1036 2
4: 1036 6
m[m[mid %in% a][, .I[uniqueN(mid) == uniqueN(a)], by = .(userid)]$V1]
mid userid opened
1: 197 1022 Y
2: 198 1022 N
3: 197 1036 N
4: 198 1036 Y