Summing the counts in a data frame using sliding window

问题

I am new to R. I have a data frame in R like following

df <- data.frame(ID=c(rep("A1",10),rep("A2",13),rep("A3",12)),
                 Values=c(10,2,4,23,10,5,20,15,13,21,15,9,19,5,14,25,18,19,31,26,4,21,4,6,7,12,15,18,25,20,16,29,21,19,10))

For every ID I would like to sum the counts in column "Values" in a sliding windows for every 3 positions. Following data frame is an excerpt from df which includes only the records corresponding to A1:

ID    Values
A1     10
A1      2
A1      4
A1     23
A1     10
A1      5
A1     20
A1     15
A1     13
A1     21

I would like to take 3 entries at time and sum and move to next 3 entries. When the sliding windows can't accommodate 3 entries then I skip those values.

For an example, Window_1 starts from first value (10) while window_2 starts from second value (2) and window_3 starts from third value (4).

 window_1 = [10+2+4] + [23+10+5] + [20+15+13] = 102 
 window_2 = [2+4+23] + [10+5+20] + [15+13+21] = 113
 window_3 = [4+23+10] + [5+20+15] = 77

and report it in a data frame like following:

ID  Window_1 Window_2 Window_3
A1   102       113      77

Likewise I would like sum the counts in column Values for everyid in the data frame "df" and report in a data.frmae like following:

ID    window_1   window_2   window_3
A1      102       113         77
A2      206       195         161
A3      198       163         175

I tried the following code

sum_win_3=0
sum_win_2=0
sum_win_1=0
win_1_counts=0
win_2_counts=0
win_3_counts=0

for (i in seq(1,length(df$Values),3))
{

  if((i+i+1+i+2) %% 3 == 0)
  {
    win_1_counts=df$Values[i]+df$Values[i+1]+df$Values[i+2]
    win_1_counts[is.na(win_1_counts)]=0
    #print(win_1_counts)
  }
  sum_win_1=sum_win_1+win_1_counts
}
#print(sum_win_1)

for (j in seq(2,length(df$Values),3))
{
  if((j+j+1+j+2) %% 3 == 0)
  {
    win_2_counts=df$Values[j]+df$Values[j+1]+df$Values[j+2]
    win_2_counts[is.na(win_2_counts)]=0
    #print(win_2_counts)
  }
  sum_win_2=sum_win_2+win_2_counts
}
#print(sum_win_2)

for (k in seq(3,length(df$Values),3))
{
  if((k+k+1+k+2) %% 3 == 0)
  {
    win_3_counts=df$Values[k]+df$Values[k+1]+df$Values[k+2]
    win_3_counts[is.na(win_3_counts)]=0
    #print(win_3_counts)
  }
  #sum_win_3=sum_win_3+win_3_counts
}
print(sum_win_3)
output=data.frame(ID=df[1],Window_1=sum_win_1,Window_2=sum_win_2,Window_3=sum_win_3)

The above code sums the counts for window_1, windows_2 and window_3 by taking all the IDs together rather working on every ID separately.
Kindly guide me in getting the the output in the desired format stated above. Thanks in advance

回答1:

Using the data.table package, I would approach it as follows:

library(data.table)
setDT(df)[, .(w1 = sum(Values[1:(3*(.N%/%3))]),
              w2 = sum(Values[2:(3*((.N-1)%/%3)+1)]),
              w3 = sum(Values[3:(3*((.N-2)%/%3)+2)]))
          , by = ID]

which gives:

   ID  w1  w2  w3
1: A1 102 113  77
2: A2 206 195 161
3: A3 198 163 175

Or to avoid the repetition (thanx to @Cath):

setDT(df)[, lapply(1:3, function(i) {sum(Values[i:(3*((.N-i+1)%/%3)+(i-1))])})
          , by = ID]

If you want to rename the V1, V2 & V3 variables, you can do that afterwards, but you can also do:

cols <- c("w1","w2","w3")

setDT(df)[, (cols) := lapply(1:3, function(i) {sum(Values[i:(3*((.N-i+1)%/%3)+(i-1))])})
          , by = ID]

回答2:

This could be done using tapplyand aggregate

sumf <- function(x1){
             sum(tapply(x1, 
                        (seq_along(x1) -1) %/%3, 
                        function(x) ifelse(length(x) == 3, sum(x), 0)))
        }

aggregate(Values ~ ID, data = df, 
          FUN = function(y){
              cbind(sumf(y), sumf(y[-1]), sumf(y[-c(1,2)]))
          })

#  Group.1 x.1 x.2 x.3
#1      A1 102 113  77
#2      A2 206 195 161
#3      A3 198 163 175

This can also be done using filter

sum.filter <- function(z) tapply(head(tail(as.numeric(
    filter(z, c(1,1,1))),-1), -1), 
    0:(length(z)-3) %% 3 +1, sum)

aggregate(Values ~ ID, data = df, FUN = function(y){ cbind(sum.filter(y) )})

回答3:

This seems to work:

library(zoo)
wins = function(x, w) 
  rollapply(x, width = w*((length(x)-seq(w)+1) %/% w), align = "left", sum)

aggregate(Values ~ ID, df, wins, 3)
#   ID Values.1 Values.2 Values.3
# 1 A1      102      113       77
# 2 A2      206      195      161
# 3 A3      198      163      175

This is the only answer so far to perform the calculation on a rolling basis, which is usually more efficient.

来源：https://stackoverflow.com/questions/35039361/summing-the-counts-in-a-data-frame-using-sliding-window

标签

dataframe

sliding-window