I have two tables. One has info from 2012 till 2014 with the period of 3 hours. It looks like this:
B C
1 01.06.2012 00:00 10 0
2 01.06.2012 03:00 10 0
3 01.06.2012 06:00 10 6
4 01.06.2012 09:00 7,5 0
5 01.06.2012 12:00 6 2,5
6 01.06.2012 15:00 6 0
7 01.06.2012 18:00 4 0
8 01.06.2012 21:00 4 0
9 02.06.2012 00:00 0 0
10 02.06.2012 03:00 0 0
The other table is the same time but sampled by 1 minute:
1 01.06.2012 00:00
2 01.06.2012 00:01
3 01.06.2012 00:01
4 01.06.2012 00:03
5 01.06.2012 00:03
6 01.06.2012 00:05
7 01.06.2012 00:05
8 01.06.2012 00:07
9 01.06.2012 00:08
10 01.06.2012 00:09
11 01.06.2012 00:10
Now, I need the values of 2nd and 3rd rows of the second table to correlate to the first, so that if a timestamp from the second table is between timestamp(i) and timestamp(i+1) of the first table it will take the B(i) and C(i) and copy them. I have this code and I know it works, but it takes more than 12 hours to run it and I have many of such files that I need to work with in the same fashion.
clouds <- read.csv('~/2012-2014 clouds info.csv', sep=";", header = FALSE)
cloudFull <- read.csv('~/2012-2014 clouds.csv', sep=";", header = FALSE)
for (i in 1:nrow(cloudFull)){
dateOne <- strptime(cloudFull[i,1], '%d.%m.%Y %H:%M')
for (j in 1:nrow(clouds)){
bottomDate = strptime(clouds[j,1], '%d.%m.%Y %H:%M')
upperDate = strptime(clouds[j+1,1], '%d.%m.%Y %H:%M')
if ((dateOne >= bottomDate) && (dateOne < upperDate)) {
cloudFull[i,2] <- clouds[j,2]
cloudFull[i,3] <- clouds[j,3]
break
}
}
}
write.csv(cloudFull, file = 'cc.csv')
Now how do I make it run faster? The object.size(cloudFull) gives me 39580744 bytes, it has 470000 rows but other files will have even more data. I'm just beginning with R (have worked in it for 2 days only so far) and I'd be grateful for any advice in a very simple language :D
Its hard to know what your real data looks like, but along the lines of
full <- strptime(cloudFull[,1], '%d.%m.%Y %H:%M')
ref <- strptime(clouds[,1], '%d.%m.%Y %H:%M')
## ref <- sort(ref)
cloudsFull[, 2:3] <- clouds[findInterval(full, ref), 2:3]
Use of findInterval() changes the problem into one that scales linearly rather than quadratic.
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