r - 有效地用累积频率替换数据帧

Question

我正在尝试编写一个程序，该程序采用一个大数据框并用这些值的累积频率(升序排序)替换每一列值。例如，如果一列值是:5, 8, 3, 5, 4, 3, 8, 5, 5, 1。那么相对频率和累积频率是:

1:rel_freq=0.1，cum_freq = 0.1

3:rel_freq=0.2，cum_freq=0.3

4:rel_freq=0.1，cum_freq = 0.4

5:rel_freq=0.4，cum_freq = 0.8

8:rel_freq=0.2，cum_freq = 1.0

那么原列变为:0.8, 1.0, 0.3, 0.8, 0.4, 0.3, 1.0, 0.8, 0.8, 0.1

以下代码正确执行此操作，但由于嵌套循环，它的扩展性很差。知道如何更有效地执行此任务吗？

mydata = read.table(.....)

totalcols = ncol(mydata)
totalrows = nrow(mydata)

for (i in 1:totalcols) {
    freqtable = data.frame(table(mydata[,i])/totalrows)  # create freq table
    freqtable$CumSum = cumsum(freqtable$Freq)   # calc cumulative freq

    hashtable = new.env(hash=TRUE)
    nrows = nrow(freqtable)

    # store cum freq in hash
    for (x in 1:nrows) {
        dummy = toString(freqtable$Var1[x])
        hashtable[[dummy]] = freqtable$CumSum[x]
    }

    # replace original data with cum freq
    for (j in 1:totalrows) {
        dummy = toString(mydata[j,i])
        mydata[j,i] = hashtable[[dummy]]
    }
}

Answer 1

这处理没有 for 的单列-环形:

R> x <- c(5, 8, 3, 5, 4, 3, 8, 5, 5, 1)
R> y <- cumsum(table(x)/length(x))
R> y[as.character(x)]
  5   8   3   5   4   3   8   5   5   1 
0.8 1.0 0.3 0.8 0.4 0.3 1.0 0.8 0.8 0.1