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    dput(x)
structure(list(Date = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 
3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L), .Label = c("1/1/2012", 
"2/1/2012", "3/1/2012", "4/1/2012", "5/1/2012", "6/1/2012"), class = "factor"), 
    Continent = structure(c(3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 
    3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L), .Label = c("Asia", "Europe", 
    "South America"), class = "factor"), Score = c(10L, 4L, 9L, 
    1L, 9L, 3L, 10L, 0L, 0L, 10L, 4L, 9L, 10L, 4L, 9L, 0L, 0L, 
    5L), Country = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 
    3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("Brasil", 
    "China", "Germany"), class = "factor"), mean = c(6.83333333333333, 
    3.5, 5.83333333333333, 6.83333333333333, 3.5, 5.83333333333333, 
    6.83333333333333, 3.5, 5.83333333333333, 6.83333333333333, 
    3.5, 5.83333333333333, 6.83333333333333, 3.5, 5.83333333333333, 
    6.83333333333333, 3.5, 5.83333333333333), sd = c(4.91596040125088, 
    3.33166624979154, 3.81663027639129, 4.91596040125088, 3.33166624979154, 
    3.81663027639129, 4.91596040125088, 3.33166624979154, 3.81663027639129, 
    4.91596040125088, 3.33166624979154, 3.81663027639129, 4.91596040125088, 
    3.33166624979154, 3.81663027639129, 4.91596040125088, 3.33166624979154, 
    3.81663027639129), outlier1 = c(FALSE, FALSE, FALSE, TRUE, 
    TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE, 
    FALSE, FALSE, TRUE, TRUE, FALSE)), .Names = c("Date", "Continent", 
"Score", "Country", "mean", "sd", "outlier1"), row.names = c(NA, 
-18L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000005e70788>)

各国の平均、標準偏差、外れ値1を計算しました。各国に outlier_score を適用してランク付けしたいと思います。このデータセットの外れ値スコアを計算する方法を誰かが教えてもらえますか?

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1 に答える 1

2 
# if the record is an outlier,
# take the absolute value of the difference
# between the score and the mean
# otherwise leave it blank
x$distance.to.mean <- ifelse( x$outlier1 , abs( x$Score - x$mean ) , NA )

# for all records with non-missing distances,
# add a `rank` variable based on its order in the data
x[ !is.na( x$distance.to.mean ) , 'rank' ] <- 
    rank( x[ !is.na( x$distance.to.mean ) , 'distance.to.mean' ] )

# see the result
x

# sum up the number of outliers in each country grouping
outliers.by.country <- tapply( x$outlier1 , x$Country , sum )

# take a look at those counts
outliers.by.country

# create a vector of all matches to the outliers.by.country table
y <- match( x$Country , names( outliers.by.country ) )

# and merge on the contents of the outliers.by.country table to x
x$sum.outliers <- 
    outliers.by.country[ y ]

# sort by the sum if you like
x <- x[ order( x$sum.outliers , decreasing = TRUE ) , ]
于 2013-01-23T20:06:28.390 に答える