R: Matching rows by two columns - r

Question

I am currently trying to figure out a vectorized way to match by two values in the same row. I have the following two simplified data frames:
# Dataframe 1: Displaying all my observations
df1 <- data.frame(c(1, 2, 3, 4, 5, 6, 7, 8),
c("A", "B", "C", "D", "A", "B", "A", "C"),
c("B", "E", "D", "A", "C", "A", "D", "A"))
colnames(df1) <- c("ID", "Number1", "Number2")
> df1
ID Number1 Number2
1 1 A B
2 2 B E
3 3 C D
4 4 D A
5 5 A C
6 6 B A
7 7 A D
8 8 C A
# Dataframe 2: Matrix of observations I am interested in
df2 <- matrix(c("A", "B",
"D", "A",
"C", "B",
"E", "D"),
ncol = 2,
byrow = TRUE)
> df2
[,1] [,2]
[1,] "A" "B"
[2,] "D" "A"
[3,] "C" "B"
[4,] "E" "D"
What I am trying to accomplish is to create a new column in df1 that states TRUE only if the exact combination is present in df2 (for example ID = 1 is equivalent to the first row in df2 because both of them consist of A and B). Additionally, if there is a shortcut, I would also like the status to be TRUE if the numbers are reversed, i.e. df1$Number1 matches df2[i,2] and df1$Number2 matches df2[i,1] (for example for ID = 7, the combination in df1 is A,D and in df2, the combination is D,A --> TRUE).
My desired output looks like this:
> df1
ID Number1 Number2 Status
1 1 A B TRUE
2 2 B E FALSE
3 3 C D FALSE
4 4 D A TRUE
5 5 A C FALSE
6 6 B A TRUE
7 7 A D TRUE
8 8 C A FALSE
All I have gotten so far is this:
for (i in 1:nrow(df1)) {
for (j in 1:nrow(df2)) {
Status <- ifelse(df1$Number1[i] %in% df2[j,1] &&
df1$Number2[i] %in% df2[j,2], TRUE, FALSE)
StatusComb[i,j] <- Status
}
df1$Status[i] <- ifelse(any(StatusComb[i,]) == TRUE, TRUE, FALSE)
}
It is really inefficient (you can clearly tell I am new to R) and does not look very nice either. I would appreciate any help!

Answer 1

One method would be to merge things together.
Adapting your data, to account for reversed labels, I'll reverse df2 on itself and rbind it:
df2 <- rbind.data.frame(df2, df2[,c(2,1)])
colnames(df2) <- c("Number1", "Number2")
df2$a <- TRUE
df2
# Number1 Number2 a
# 1 A B TRUE
# 2 D A TRUE
# 3 C B TRUE
# 4 E D TRUE
# 5 B A TRUE
# 6 A D TRUE
# 7 B C TRUE
# 8 D E TRUE
I added a so that it'll be merged in. From there:
df3 <- merge(df1, df2, all.x = TRUE)
df3$a <- !is.na(df3$a)
df3[ order(df3$ID), ]
# Number1 Number2 ID a
# 1 A B 1 TRUE
# 5 B E 2 FALSE
# 7 C D 3 FALSE
# 8 D A 4 TRUE
# 2 A C 5 FALSE
# 4 B A 6 TRUE
# 3 A D 7 TRUE
# 6 C A 8 FALSE
If you look at it before !is.na(df3$a), you'll see that the column is wholly TRUE and NA (the NA were not present in df2); if that is enough for you, then you can omit the middle step. The order step is just because row-order with merge is not assured (in fact I find it always inconveniently different). Since it was previously ordered by ID, I reverted to that, but it was entirely for aesthetics here to match your desired output.

Answer 2

You could define the combination variable that you want to search for in alphabetical order as below:
combination <- apply(df2, 1, function(x) {
paste(sort(x), collapse = '')
})
combination
[1] "AB" "AD" "BC" "DE"
And then mutate the Status field based on the concatenation of the Number field
library(dplyr)
df1 %>%
rowwise() %>%
mutate(S = paste(sort(c(Number1, Number2)), collapse = "")) %>%
mutate(Status = ifelse(S %in% combination, TRUE, FALSE))
Source: local data frame [8 x 5]
Groups: <by row>
# A tibble: 8 x 5
ID Number1 Number2 S Status
<dbl> <chr> <chr> <chr> <lgl>
1 1 A B AB TRUE
2 2 B E BE FALSE
3 3 C D CD FALSE
4 4 D A AD TRUE
5 5 A C AC FALSE
6 6 B A AB TRUE
7 7 A D AD TRUE
8 8 C A AC FALSE
Data:
I set stringsAsFactors = F in the dataframe
df1 <- data.frame(c(1, 2, 3, 4, 5, 6, 7, 8),
c("A", "B", "C", "D", "A", "B", "A", "C"),
c("B", "E", "D", "A", "C", "A", "D", "A"), stringsAsFactors = F)
colnames(df1) <- c("ID", "Number1", "Number2")