将R中的数据分组以执行功能

Question

Here is an example of my data:

           id   score
1          82   0.50000
2          82   0.39286
3          82   0.56250
4         328   0.50000
5         328   0.67647
6         328   0.93750
7         328   0.91667

I want to make a column of moving average's of scores for each id.

So I need to somehow group the data by id then apply a MA function to that grouped data and then have the output as another column "MA_score"

I would like my output to look like this:

           id   score    MA_score
1          82   0.50000   NULL
2          82   0.39286   0.xxxx
3          82   0.56250   NULL
4         328   0.50000   NULL
5         328   0.67647   0.yyyy
6         328   0.93750   0.qqqq
7         328   0.91667   NULL

Answer 1

You could use split and rollapply from the zoo package as one of many ways to approach this. Note that in the example below I set the width of the rollapply function to 1 so it just returns each value. For widths greater than one it will take the mean of that number of values.

require(zoo)
sapply( split( df , df$id) , function(x) rollapply( x , width = 1 , align = 'left' , mean) )
#Note that by setting width = 1 we just return the value
$`82`
     id   score
[1,] 82 0.50000
[2,] 82 0.39286
[3,] 82 0.56250

$`328`
      id   score
[1,] 328 0.50000
[2,] 328 0.67647
[3,] 328 0.93750
[4,] 328 0.91667

If we were to set width = 3 you would get:

$`82`
     id   score
[1,] 82 0.48512

$`328`
      id     score
[1,] 328 0.7046567
[2,] 328 0.8435467

Or you could use aggregate in base R:

aggregate(  score ~ id , data = df , function(x) rollapply( x , width = 1 , align = 'left' , mean)  )
   id                              score
1  82          0.50000, 0.39286, 0.56250
2 328 0.50000, 0.67647, 0.93750, 0.91667

There are quite a few ways to do this. I would precisely define your moving average function though, because there are many ways to calculate it (check out for example TTR:::SMA )

---

Or more straightforward using ave :

within(df, { MA_score <- ave(score, id, FUN=function(x) 
                rollmean(x, k=3, na.pad = TRUE))})

Answer 2

You could split your data by unique ID values, calculate the rolling mean (from 'zoo' package) for each of these unique IDs and append the results to your initial dataframe:

# Required packages
library(zoo)

# Data setup
df <- data.frame(id = c(82, 82, 82, 328, 328, 328, 328), 
                 score = c(0.5, 0.39286, 0.5625, 0.5, 0.67647, 0.9375, 0.91667))

# Split data by unique IDs
df.sp <- split(df, df$id)

# Calculate rolling mean for each unique ID
df.ma <- lapply(seq(df.sp), function(i) {
  rollmean(df.sp[[i]]$score, k = 3, na.pad = TRUE)
})

# Append column 'MA_score' to dataframe
for (i in seq(names(df.sp))) {
  df[which(df$id == names(df.sp)[i]), "MA_score"] <- df.ma[[i]]
}

df
   id   score  MA_score
1  82 0.50000        NA
2  82 0.39286 0.4851200
3  82 0.56250        NA
4 328 0.50000        NA
5 328 0.67647 0.7046567
6 328 0.93750 0.8435467
7 328 0.91667        NA