R 2D binning of data frame with secondary complex calculations

Question

I have a data frame that looks generally like this

df.data <- data.frame(x=sample(1:9, 10, replace = T), y=sample(1:9, 10, replace=T), vx=sample(-1:1, 10, replace=T), vy=sample(-1:1, 10, replace=T))

x and y are positions. vx and vy are x, y values for a 2d vector. I want to take this data frame and "bin" based on the x and y values, but performing a calculation on the vx and vy. This function does this except it uses a loop which is going to be too slow for my data set.

slowWay <- function(df)
{
    df.bin <- data.frame(expand.grid(x=0:3, y=0:3, vx=0, vy=0, count=0))

    for(i in 1:nrow(df))
    {
        x.bin <- floor(df[i, ]$x / 3)
        y.bin <- floor(df[i, ]$y / 3)
        print(c(x.bin, y.bin))

        df.bin[df.bin$x == x.bin & df.bin$y == y.bin, ]$vx = df.bin[df.bin$x == x.bin & df.bin$y == y.bin, ]$vx + df[i, ]$vx
        df.bin[df.bin$x == x.bin & df.bin$y == y.bin, ]$vy = df.bin[df.bin$x == x.bin & df.bin$y == y.bin, ]$vy + df[i, ]$vy
        df.bin[df.bin$x == x.bin & df.bin$y == y.bin, ]$count = df.bin[df.bin$x == x.bin & df.bin$y == y.bin, ]$count + 1
    }

    return(df.bin)
}

Is this type of 2D binning possible in a non looping way?

Answer 1

Here's another faster way to do it, one that includes unpopulated bin combinations:

fasterWay <- function(df.data) {
  a1 <- aggregate(df.data[,3:4], list(x=floor(df.data$x/3), y=floor(df.data$y/3)), sum)
  a2 <- aggregate(list(count=rep(NA,nrow(df.data))), list(x=floor(df.data$x/3), y=floor(df.data$y/3)), length)
  result <- merge(expand.grid(y=0:3,x=0:3), merge(a1,a2), by=c("x","y"), all=TRUE)
  result[is.na(result)] <- 0
  result <- result[order(result$y, result$x),]
  rownames(result) <- NULL
  result
}

It gives me:

   x y vx vy count
1  0 0  0  0     1
2  0 1  0  0     0
3  0 2 -1 -1     1
4  0 3  0  0     0
5  1 0 -1 -1     1
6  1 1  0  0     0
7  1 2  0  0     0
8  1 3 -1  0     2
9  2 0 -1 -1     1
10 2 1  0  0     0
11 2 2 -1  1     2
12 2 3  0  0     1
13 3 0  0  0     0
14 3 1  0  0     0
15 3 2 -1  0     1
16 3 3  0  0     0

Answer 2

This is one way, but will probably need to do it in a couple of steps if you want the full record with unpopulated bin combinations:

> by(df.data[, c("vx", "vy")],        # input data
     list(x.bin=floor(df.data$x / 3), y.bin=floor(df.data$y / 3)), # grouping
     function(df) sapply(df, function(x) c(Sum=sum(x), Count=length(x) ) ) )  #calcs
x.bin: 0
y.bin: 1
      vx vy
Sum    0  1
Count  1  1
--------------------------------------------------------------------- 
x.bin: 1
y.bin: 1
      vx vy
Sum    0  1
Count  2  2
--------------------------------------------------------------------- 
x.bin: 2
y.bin: 1
      vx vy
Sum   -1 -2
Count  2  2
--------------------------------------------------------------------- 
x.bin: 0
y.bin: 2
      vx vy
Sum    1  0
Count  1  1
--------------------------------------------------------------------- 
x.bin: 1
y.bin: 2
NULL
--------------------------------------------------------------------- 
x.bin: 2
y.bin: 2
      vx vy
Sum    2  1
Count  4  4

Answer 3

Here is a data.table version:

library(data.table)
dt.data<-as.data.table(df.data) # Convert to data.table
dt.data[,c("x.bin","y.bin"):=list(floor(x/3),floor(y/3))] # Add bin columns
setkey(dt.data,x.bin,y.bin)

dt.bin<-CJ(x=0:3, y=0:3) # Cross join to create bin combinations
dt.data.2<-dt.data[dt.bin,list(vx=sum(vx),vy=sum(vy),count=.N)] # Join the bins and data; sum vx/vy and count matching rows
dt.data.2[is.na(vx),vx:=0L] # Replace NA with 0
dt.data.2[is.na(vy),vy:=0L] # Replace NA with 0
dt.data.2[order(y.bin,x.bin)] # Display the final data.table output

##     x.bin y.bin vx vy count
##  1:     0     0  0  0     0
##  2:     1     0  0  0     0
##  3:     2     0  1  1     1
##  4:     3     0  0  0     0
##  5:     0     1  0  0     0
##  6:     1     1  0 -2     3
##  7:     2     1  0  0     0
##  8:     3     1  0  0     0
##  9:     0     2  0  0     1
## 10:     1     2  0  0     0
## 11:     2     2  0  2     3
## 12:     3     2 -1  1     1
## 13:     0     3  0  0     0
## 14:     1     3  0  0     0
## 15:     2     3  0  0     0
## 16:     3     3  1 -1     1