Matching a fixed point with an interval in a data frame - r

Question

I'm trying to match stock trades from one data frame with the mid-quote that was prevailing during that time. Thus, the time stamps don't match exactly but I have just a corresponding time interval of quotes for the time the trade happened.
I wrote a loop which works but since I know that loops should be avoided whenever possible, I looked out for an alternative.
First, this is my loop:
t=dim(x1)[1]
z=1
for (i in 1:t) {
flag=FALSE
while(flag==FALSE){
if(x1[z,1]>x2[i,1]){
x2[i,2]=x1[z-1,2]
flag=TRUE
}
else {
z=z+1
}
}
}
I've found the advice on Stack Overflow to merge the two arrays, so I added the upper bound of the interval as another column and matched the corresponding times with the subset-function.
Unfortunately, this method takes far more time than the loop. I assume it's due to the huge array that is created by merging. The data frames with the quotes have like 500.000 observations and the transaction data 100.000.
Is there a more elegant (and especially faster) way to solve this problem?
Furthermore, for some data I get the error message "missing value where TRUE/FALSE needed", even though the if-condition works when I do it manually.
edit:
My quote data would look like this:
Time midquote
[1,] 35551 50.85229
[2,] 35589 53.77627
[3,] 36347 54.27945
[4,] 37460 52.01283
[5,] 37739 53.65414
[6,] 38249 52.34947
[7,] 38426 50.59568
[8,] 39858 53.75646
[9,] 40219 51.38876
[10,] 40915 52.09319
and my transaction data:
Time midquote
[1,] 36429 0
[2,] 38966 0
[3,] 39334 0
[4,] 39998 0
[5,] 40831 0
So I want to know the midquotes from the time in the latter from the corresponding time of the former. The time in the example is in seconds from midnight.

Answer 1

For your example datasets, the following approach is faster:
x2[ , 2] <- x1[vapply(x2[, 1], function(x) which(x <= x1[, 1])[1] - 1L,
FUN.VALUE = integer(1)), 2]
# Time midquote
# [1,] 36429 54.27945
# [2,] 38966 50.59568
# [3,] 39334 50.59568
# [4,] 39998 53.75646
# [5,] 40831 51.38876
A second approach:
o <- order(c(x1[ , 1], x2[ , 1]))
tmp <- c(x1[ , 2], x2[ , 2])[o]
idx <- which(!tmp)
x2[ , 2] <- tmp[unlist(tapply(idx, c(0, cumsum(diff(idx) > 1)),
function(x) x - seq_along(x)), use.names = FALSE)]
# Time midquote
# [1,] 36429 54.27945
# [2,] 38966 50.59568
# [3,] 39334 50.59568
# [4,] 39998 53.75646
# [5,] 40831 51.38876