I see you've already accepted an answer, but here is another possible solution.
This function was just hacked together, but could be worked on some more to be made more generalized.
myfun = function(DATA1, DATA2, MATCH1, MIN, MAX) {
temp = sapply(1:nrow(DATA2),
function(x) DATA1[[MATCH1]] >= DATA2[[MIN]][x] &
DATA1[[MATCH1]] <= DATA2[[MAX]][x])
if (isTRUE(any(rowSums(temp) == 0))) {
temp1 = DATA1[-(which(rowSums(temp) == 0)), ]
}
OUT = cbind(temp1[order(temp1[[MATCH1]]), ],
DATA2[order(DATA2[[MIN]]), ], row.names=NULL)
condition = ((OUT[4] <= OUT[2] & OUT[2] <= OUT[5]) == 0)
if (isTRUE(any(condition))) {
OUT[-which(condition), ]
} else {
OUT
}
}
Here's what the function does:
- It first compares, row by row, the value in the second column of the first
data.frame
with the values in the second and third columns of the second data.frame
.
- It then checks to find if any of those has
FALSE
for both conditions, and removes them from the first data.frame
.
- Then, it sorts the first
data.frame
by the second column, and the second data.frame
by the "min" match column.
- Finally, it does one more check to ensure that the value from the first dataset is between the provided values; if not, that row is removed.
Now, here is some sample data. A
and B
are the same as your provided data. X
and Y
have been changed for further testing purposes. In the merge between X
and Y
, there should be only one row.
A = read.table(header=TRUE, text="A B
rs10 23353
rs100 10000
rs234 54440")
B = read.table(header=TRUE, text="A B C
E235 20000 30000
E255 50000 60000")
X = A[c(3, 1, 2), ]
X[1, 2] = 57000
Y = B
Y[2, 3] = 55000
Here's how you would use the function and the output you would get.
myfun(A, B, 2, 2, 3)
# A B A B C
# 1 rs10 23353 E235 20000 30000
# 2 rs234 54440 E255 50000 60000
myfun(X, Y, 2, 2, 3)
# A B A B C
# 1 rs10 23353 E235 20000 30000
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