使用data.table计算百分比和其他函数

时间:2017-06-13 03:36:48

标签: r data.table dplyr

我想将聚合函数和百分比函数应用于列。我发现讨论聚合的线程(Calculating multiple aggregations with lapply(.SD, ...) in data.table R package)和讨论百分比(How to obtain percentages per value for the keys in R using data.table?Use data.table to calculate the percentage of occurrence depending on the category in another column)的线程,但不是两者。

请注意,我正在寻找基于data.table的方法。 dplyr不适用于实际数据集。

以下是生成样本数据的代码:

set.seed(10)
  IData <- data.frame(let = sample( x = LETTERS, size = 10000, replace=TRUE), numbers1 = sample(x = c(1:20000),size = 10000), numbers2 = sample(x = c(1:20000),size = 10000))
  IData$let<-as.character(IData$let)

  data.table::setDT(IData)

以下是使用dplyr

生成输出的代码
Output <- IData %>%
    dplyr::group_by(let) %>%
    dplyr::summarise(numbers1.mean = as.double(mean(numbers1)),numbers1.median = as.double(median(numbers1)),numbers2.mean=as.double(mean(numbers2)),sum.numbers1.n = sum(numbers1)) %>%
    dplyr::ungroup() %>%
    dplyr::mutate(perc.numbers1 = sum.numbers1.n/sum(sum.numbers1.n)) %>%
    dplyr::select(numbers1.mean,numbers1.median,numbers2.mean,perc.numbers1)

示例输出(标题) 如果我运行head(output),我会得到:

  let numbers1.mean numbers1.median numbers2.mean perc.numbers1
  <chr>         <dbl>           <dbl>         <dbl>         <dbl>
    N     10320.951         10473.0      9374.435    0.03567927
    H      9683.590          9256.5      9328.035    0.03648391
    L     10223.322         10226.0      9806.210    0.04005400
    S      9922.486          9618.0     10233.849    0.03678742
    C      9592.620          9226.0      9791.221    0.03517997
    F     10323.867         10382.0     10036.561    0.03962035

以下是我尝试使用data.table(未成功)

的内容
  IData[, as.list(unlist(lapply(.SD, function(x) list(mean=mean(x),median=median(x),sum=sum(x))))), by=let, .SDcols=c("numbers1","numbers2")] [,.(Perc = numbers1.sum/sum(numbers1.sum)),by=let]

我有两个问题:

a)如何使用data.table解决此问题?

b)我已经看到上面的线程使用了prop.table。有人可以指导我如何使用这个功能吗?

我真诚地感谢任何指导。

1 个答案:

答案 0 :(得分:1)

我们可以使用与data.table

类似的方法
res <- IData[, .(numbers1.mean = mean(numbers1),
          numbers1.median = median(numbers1),
          numbers2.mean=mean(numbers2),
          sum.numbers1.n = sum(numbers1)), let
          ][, perc.numbers1 := sum.numbers1.n/sum(sum.numbers1.n)
           ][, c("let", "numbers1.mean",  "numbers1.median", 
                        "numbers2.mean", "perc.numbers1"), with = FALSE]

head(res)
#    let numbers1.mean numbers1.median numbers2.mean perc.numbers1
#1:   N     10320.951         10473.0      9374.435    0.03567927
#2:   H      9683.590          9256.5      9328.035    0.03648391
#3:   L     10223.322         10226.0      9806.210    0.04005400
#4:   S      9922.486          9618.0     10233.849    0.03678742
#5:   C      9592.620          9226.0      9791.221    0.03517997
#6:   F     10323.867         10382.0     10036.561    0.03962035
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