R - Create (Mutate) a new column as a function of past observations - r

Question

Ok so i have a pretty large data set of around 500 observations and 3 variables. The first column refers to time.
For a test data set I am using:
dat=as.data.frame(matrix(c(1,2,3,4,5,6,7,8,9,10,
1,1.8,3.5,3.8,5.6,6.2,7.8,8.2,9.8,10.1,
2,4.8,6.5,8.8,10.6,12.2,14.8,16.2,18.8,20.1),10,3))
colnames(dat)=c("Time","Var1","Var2")
Time Var1 Var2
1 1 1.0 2.0
2 2 1.8 4.8
3 3 3.5 6.5
4 4 3.8 8.8
5 5 5.6 10.6
6 6 6.2 12.2
7 7 7.8 14.8
8 8 8.2 16.2
9 9 9.8 18.8
10 10 10.1 20.1
So what I need to do is create a new column that each observations is the slope respect to time of some past points. For example taking 3 past points it would be something like:
slopeVar1[i]=slope(Var1[i-2:i],Time[i-2:i]) #Not real code
slopeVar[i]=slope(Var2[i-2:i],Time[i-2:i]) #Not real code
Time Var1 Var2 slopeVar1 slopeVar2
1 1 1 2 NA NA
2 2 1.8 4.8 NA NA
3 3 3.5 6.5 1.25 2.25
4 4 3.8 8.8 1.00 2.00
5 5 5.6 10.6 1.05 2.05
6 6 6.2 12.2 1.20 1.70
7 7 7.8 14.8 1.10 2.10
8 8 8.2 16.2 1.00 2.00
9 9 9.8 18.8 1.00 2.00
10 10 10.1 20.1 0.95 1.95
I actually got as far as using a for() function, but for really large data sets (>100,000) it starts taking too long.
The for() argument that I used is shown bellow:
#CREATE DATA FRAME
rm(dat)
dat=as.data.frame(matrix(c(1,2,3,4,5,6,7,8,9,10,
1,1.8,3.333,3.8,5.6,6.2,7.8,8.2,9.8,10.1,
2,4.8,6.5,8.8,10.6,12.2,14.8,16.2,18.8,20.1),10,3))
colnames(dat)=c("Time","Var1","Var2")
dat
plot(dat)
#CALCULATE SLOPE OF n POINTS FROM i TO i-n.
#In this case I am taking just 3 points, but it should
#be possible to change the number of points taken.
attach(dat)
n=3 #number for points to take slope
l=dim(dat[1])[1] #number of iterations
y=0
x=0
slopeVar1=NA
slopeVar2=NA
for (i in 1:l) {
if (i<n) {slopeVar1[i]=NA} #For the rows where there are not enough previous observations, it outputs NA
if (i>=n) {
y1=Var1[(i-n+1):i] #y data sets for calculating slope of Var1
y2=Var2[(i-n+1):i]#y data sets for calculating slope of Var2
x=Time[(i-n+1):i] #x data sets for calculating slope of Var1&Var2
z1=lm(y1~x) #Temporal value of slope of Var1
z2=lm(y2~x) #Temporal value of slope of Var2
slope1=as.data.frame(z1[1]) #Temporal value of slope of Var1
slopeVar1[i]=slope1[2,1] #Populating string of slopeVar1
slope2=as.data.frame(z2[1])#Temporal value of slope of Var2
slopeVar2[i]=slope2[2,1] #Populating string of slopeVar2
}
}
slopeVar1 #Checking results.
slopeVar2
(result=cbind(dat,slopeVar1,slopeVar2)) #Binds original data with new calculated slopes.
This code actually outputs what I want; but again, for really large data sets is quite inefficient.

Answer 1

This quick rollapply implemenation seems to be speeding it up somewhat -
library("zoo")
slope_func = function(period) {
y1=period[,2] #y data sets for calculating slope of Var1
y2=period[,3] #y data sets for calculating slope of Var2
x=period[,1] #x data sets for calculating slope of Var1&Var2
z1=lm(y1~x) #Temporal value of slope of Var1
z2=lm(y2~x) #Temporal value of slope of Var2
slope1=as.data.frame(z1[1]) #Temporal value of slope of Var1
slopeVar1[i]=slope1[2,1] #Populating string of slopeVar1
slope2=as.data.frame(z1[1])#Temporal value of slope of Var2
slopeVar2[i]=slope2[2,1] #Populating string of slopeVar2
}
}
start = Sys.time()
rollapply(dat[1:3], FUN=slope_func, width=3, by.column=FALSE)
end=Sys.time()
print(end-start)
Time difference of 0.04980111 secs
OP's previous implementation was taking Time difference of 0.2666121 secs for the same