CODEWITHSUNDEEP
rinterpolationapply
user29609
user29609

Reputation: 2101

Interpolate over each column in a data frame to new data and return a data frame r

I have a data frame of time series data for many locations (rows as dates and columns as locations).

Dates <- c(1950, 1960, 1970, 1980)
Well1 <- c(25, 30, 40, 28)
Well2 <- c(26, 29, 38, 25)
Well3 <- c(20, 25, 35, 19)
Inputs <- cbind.data.frame(Dates, Well1, Well2, Well3)

I have a data frame of new dates for each location.

Well1new <- c(1955, 1965, 1975, 1985)
Well2new <- Well1new + 1
Well3new <- Well2new + 1
NewDates <- cbind.data.frame(Well1new, Well2new, Well3new)

I need to interpolate to each of the new dates for each location, based on an interpolation of each of the Input dates and locations and return a data frame. I can calculate it easily for one location at a time:

approx(Inputs$Dates, Inputs$Well1, NewDates$Well1new, rule = 2)$y
[1] 27.5 35.0 34.0 28.0
approx(Inputs$Dates, Inputs$Well2, NewDates$Well2new, rule = 2)$y
[1] 27.8 34.4 30.2 25.0
approx(Inputs$Dates, Inputs$Well3, NewDates$Well3new, rule = 2)$y
[1] 23.5 32.0 23.8 19.0

But, in reality I will have thousands of locations. I tried to use apply to loop over the columns in NewDates, but I did not understand how to index the Inputs columns accordingly. I would also like to avoid for loops as speed is a concern (or is apply no faster than for loops?).

Upvotes: 0

Views: 786

Answers (2)

thelatemail
thelatemail

Reputation: 93938

Take advantage of Map to loop over both objects.

Map(approx, xout=NewDates, x=Inputs["Dates"], y=Inputs[-1], rule=2)

Output:

#$Well1new
#$Well1new$x
#[1] 1955 1965 1975 1985
# 
#$Well1new$y
#[1] 27.5 35.0 34.0 28.0
#...

Upvotes: 1

whalea
whalea

Reputation: 301

you can use lapply to perform the calculation for all your wells as follows. The performance of lapply vs for loops is a long debate with mixed opinions. I personally still like to use lapply where feasible

all.wells <- names(Inputs)[-1]
lapply(all.wells, function(x) {
    approx(Inputs$Dates, Inputs[,x], NewDates[,paste0(x, "new")], rule = 2)$y
  }
)

Upvotes: 0

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