CODEWITHSUNDEEP
rlapply
thatsciencegal
thatsciencegal

Reputation: 113

Linear interpolation over a list of data frames

I am trying to perform a linear interpolation of mean water level for each data frame contained in a list of data frames (1038 in total) using the apply family of functions, but I am having trouble defining the variables in the interpolation function to point to columns within each data frame.

Here is the head of my list:

head(df_list)
$`1928-01-01`
      station date_measured daily_mean max min mean      lon      lat
8267 20250000    1928-01-01          0 453 193  272 -49.5519 -15.2753
8268 20250000    1928-01-01          0 453 191  276 -49.5519 -15.2753

$`1928-02-01`
      station date_measured daily_mean max min mean      lon      lat
8269 20250000    1928-02-01          0  NA  NA   NA -49.5519 -15.2753
8270 20250000    1928-02-01          0  NA  NA   NA -49.5519 -15.2753

$`1928-03-01`
      station date_measured daily_mean max min mean      lon      lat
8271 20250000    1928-03-01          0 394 219  282 -49.5519 -15.2753
8272 20250000    1928-03-01          0 382 218  281 -49.5519 -15.2753

$`1928-04-01`
      station date_measured daily_mean max min mean      lon      lat
8273 20250000    1928-04-01          0 280 176  224 -49.5519 -15.2753
8274 20250000    1928-04-01          0 287 178  223 -49.5519 -15.2753

$`1928-05-01`
      station date_measured daily_mean max min mean      lon      lat
8275 20250000    1928-05-01          0 199 161  172 -49.5519 -15.2753
8276 20250000    1928-05-01          0 197 162  173 -49.5519 -15.2753

$`1928-06-01`
      station date_measured daily_mean max min mean      lon      lat
8277 20250000    1928-06-01          0 174 132  149 -49.5519 -15.2753
8278 20250000    1928-06-01          0 173 132  149 -49.5519 -15.2753

This is what I tried originally:

daily_int <- lapply(df_list, function(x) interp(x=lon,y=lat,z=mean, method="linear"))

Which resulted in the following error:

 Error in interp(x = lon, y = lat, z = mean, method = "linear") : 
  object 'lat' not found

I realize the interp function is not finding the column that I want it to look at, and being new to the apply family, I'm not sure how to do that (or if it's even possible). Essentially, I need to interpolate daily water levels across a river, and I'd like to do it in the most efficient way possible while keeping the days separate.

Edit based on comment: I'm trying to predict the mean, min, and max for each data frame.

Upvotes: 0

Views: 143

Answers (2)

Jason Johnson
Jason Johnson

Reputation: 451

I am unfamiliar with the interp function you are using but in general if you want to access the column names of the data frames in an lapply loop the following should work:

daily_int <- lapply(df_list, function(x) interp(x=x$lon,y=x$lat,z=x$mean, method="linear"))

Does that work? If not maybe this will work:

daily_int <- lapply(df_list, function(x) interp(x=x[["lon"]],y=x[["lat"]],z=x[["mean"]], method="linear"))

Upvotes: 1

JMilner
JMilner

Reputation: 531

Im not exactly sure which values you are trying to predict for using the interp function

As far as the lapply goes: each value in the list becomes the x in the function. So using with() you can assign the data frame you are working with in the interp function.

daily_int <- lapply(df_list, function(x) {
    out <- with(x,interp(x=lon,y=lat,z=mean, method="linear"))
    return(out)
})

i dont have R open to test this and am not familiar with interp off the top of my head, so will update later!

Upvotes: 2

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