How can I write a function to apply an analysis on several datasets and merge the results in one dataframe?

Question

I have five data frames (home.dat2013, home.dat2015, home.dat2016, home.dat2017, home.dat2018)

a sample data frame looks like this

Note: all five data frames have the same these four columns

home.dat2013 <- structure(list(Yield = c(
  43.5773588, 47.4013604, 46.3847655,
  49.1999453
), Latitude = c(
  399.412927, 397.4667224, 395.5014969,
  393.5341704
), Longitude = c(3.27465, 3.257958, 3.216063, 3.257626), TimeStamp = c(
  "2013-09-30 18:47:00", "2013-09-30 18:47:01",
  "2013-09-30 18:47:02", "2013-09-30 18:47:03"
)), class = "data.frame", row.names = c(
  NA,
  -4L
))

I wrote a code to divide a farming field into 120 grid cells (6 col, 20 rows)

This is the code that divides each year's data into 120 grid cells and then calculates Yield means per each grid cell

Here is my code for one year home2013

#dividing the field into grid cells for home2013
# range of latitude

minlatitude120 <- 0
maxlatitude120 <- max(home2013.dat$Latitude)
rangelat120 <- maxlatitude120-minlatitude120

#range of longitude

minlong120 <- 0
maxlong120 <- max(home2013.dat$Longitude)
rangelong120 <- maxlong120 - minlong120

min120 <- min(home2013.dat$Latitude/rangelat120)
max120 <- max(home2013.dat$Latitude/rangelat120)

min120 <- min(2*home2013.dat$Latitude/rangelat120)
max120 <- max(2*home2013.dat$Latitude/rangelat120)

#creating unique identifiers for our data 

unique_i <- unique(2*ceiling(home2013.dat$Latitude/rangelat120))
unique_i <- unique(2*ceiling(home2013.dat$Longitude/rangelong120))

#creating col and row

home2013.dat$row <- ceiling(20*home2013.dat$Latitude/rangelat120)
home2013.dat$col <- ceiling(6*home2013.dat$Longitude/rangelong120)
home2013.dat$cell <- 1000*(home2013.dat$row) + home2013.dat$col
uniquecombo120 <- unique(home2013.dat$cell)
length(uniquecombo120)


#calculating mean estimates for yield 

means2013 <- tapply(home2013.dat$Yield, home2013.dat$cell, mean)
yield13 <- data.frame (cell.number, means2013)
yield13 

I want to avoid using this code five times since it is long. My question is how can I can write a function that divides the field into 120 grid cells across all years(2013,2015,2016,2017,2018)

sample output

grid cell   means2013  means2015  means2016  means2017  means2018 
1001        50         80         100        117        20
1002        55         88         102        120        17

Answer 1

This could be achieved like so.

1. Put your code for the analysis in a function
2. Put your df's in a list
3. loop over the list using e.g. lapply which gives a list of the results
4. Merge the list with the results using e.g. Reduce and merge

BTW: Running your code resulted in an error as cell.number was not defined. Therefore I had to guess what cell.number is and set it to the names(means).

home.dat2013 <- structure(list(Yield = c(
  43.5773588, 47.4013604, 46.3847655,
  49.1999453
), Latitude = c(
  399.412927, 397.4667224, 395.5014969,
  393.5341704
), Longitude = c(3.27465, 3.257958, 3.216063, 3.257626), TimeStamp = c(
  "2013-09-30 18:47:00", "2013-09-30 18:47:01",
  "2013-09-30 18:47:02", "2013-09-30 18:47:03"
)), class = "data.frame", row.names = c(
  NA,
  -4L
))

home.dat2015 <- home.dat2016 <- home.dat2017 <- home.dat2018 <- home.dat2013

home.dat <- list(home.dat2013, home.dat2015, home.dat2016, home.dat2017, home.dat2018)
names(home.dat) <- c(2013, 2015:2018)

my_analysis <- function(x) {
  # dividing the field into grid cells for home2013
  # range of latitude

  minlatitude120 <- 0
  maxlatitude120 <- max(x$Latitude)
  rangelat120 <- maxlatitude120 - minlatitude120

  # range of longitude

  minlong120 <- 0
  maxlong120 <- max(x$Longitude)
  rangelong120 <- maxlong120 - minlong120

  min120 <- min(x$Latitude / rangelat120)
  max120 <- max(x$Latitude / rangelat120)

  min120 <- min(2 * x$Latitude / rangelat120)
  max120 <- max(2 * x$Latitude / rangelat120)

  # creating unique identifiers for our data

  unique_i <- unique(2 * ceiling(x$Latitude / rangelat120))
  unique_i <- unique(2 * ceiling(x$Longitude / rangelong120))

  # creating col and row

  x$row <- ceiling(20 * x$Latitude / rangelat120)
  x$col <- ceiling(6 * x$Longitude / rangelong120)
  x$cell <- 1000 * (x$row) + x$col
  uniquecombo120 <- unique(x$cell)

  # calculating mean estimates for yield

  means <- tapply(x$Yield, x$cell, mean)
  
  yield <- data.frame(cell.number = names(means), means)
  yield
}

# Apply the function to each df
results <- lapply(home.dat, my_analysis)
# Rename the columns of the df so that the means col includes the year
results <- lapply(names(home.dat), function(x) setNames(results[[x]], c("cell.number", paste0("means", x))))

# Merge the five df's
Reduce(function(x, y) merge(x, y, by = "cell.number"), results)
#>   cell.number means2013 means2015 means2016 means2017 means2018
#> 1       20006  46.64086  46.64086  46.64086  46.64086  46.64086