计算r中数据帧中连续值的比率

Question

I have a dataframe with 5 second intraday data of a stock. The dataframe exists of a column for the date, one for the time and one for the price at that moment. I want to make a new column in which it calculates the ratio of two consecutive price values. I tried it with a for loop, which works but is really slow.

data["ratio"]<- 0
i<-2
for(i in 2:nrow(data))
{
  if(is.na(data$price[i])== TRUE){
    data$ratio[i] <- 0
  } else {
    data$ratio[i] <- ((data$price[i] / data$price[i-1]) - 1) 
  }
}

I was wondering if there is a faster option, since my dataset contains more than 500.000 rows. I was already trying something with ddply:

data["ratio"]<- 0
fun <- function(x){
  data$ratio <- ((data$price/lag(data$price, -1))-1)
}
ddply(data, .(data), fun)

and mutate:

data<- mutate(data, (ratio =((price/lag(price))-1)))

but both don't work and I don't know how to solve it... Hopefully somebody can help me with this!

Answer 1

You can use the lag function to shift the your data by one row and then take the ratio of the original data to the shifted data. This is vectorized, so you don't need a for loop, and it should be much faster. Also, the number of lag units in the lag function has to be positive, which may be causing an error when you run your code.

# Create some fake data
set.seed(5)  # For reproducibility
dat = data.frame(x=rnorm(10))

dat$ratio = dat$x/lag(dat$x,1)

dat
             x       ratio
1  -0.84085548          NA
2   1.38435934 -1.64637013
3  -1.25549186 -0.90691183
4   0.07014277 -0.05586875
5   1.71144087 24.39939227
6  -0.60290798 -0.35228093
7  -0.47216639  0.78314834
8  -0.63537131  1.34565131
9  -0.28577363  0.44977422
10  0.13810822 -0.48327840

Answer 2

for loop in R can be extremely slow. Try to avoid it if you can.

datalen=length(data$price)

data$ratio[2:datalen]=data$price[1:datalen-1]/data$price[2:datalen]

You don't need to do the is.NA check, you will get NA in the result either the numerator or the denominator is NA.