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计算每行中一个变量与另一个变量的百分比数据表

data.table dplyr performance r

Juan Antonio Roldán Díaz · 技术社区 · 6 年前

我正在寻找一种方法来优化聚合数据表,我有几百万个数据,而我当前的实现速度很慢。

library(data.table)
df <- data.table(Factor = as.factor(rep(LETTERS[1:3], 3)),
                 Variable = 1:9)

当前实施:

aux <- df[, .(sumVar = sum(Variable)/sum(df$Variable)), by = .(Factor)]
df[aux, sumVar := sumVar, on = .(Factor = Factor)]

期望输出:

> df
   Factor Variable    sumVar
1:      A        1 0.2666667
2:      B        2 0.3333333
3:      C        3 0.4000000
4:      A        4 0.2666667
5:      B        5 0.3333333
6:      C        6 0.4000000
7:      A        7 0.2666667
8:      B        8 0.3333333
9:      C        9 0.4000000

我想我的问题在于 merge ,但我不知道如何提高它,我不熟悉 dplyr 我还没有找到任何方法可以一步一步地完成手术 data.table .

感谢您的帮助!

3 回复 | 直到 6 年前

fidelin 6 年前

像这样的

df[ , ':='(sumVar = sum(Variable)/sum(df$Variable)), by = .(Factor)]

Hugh 6 年前

你的例子中有很多重复的地方,所以我不确定我解释的是否正确。尽管如此,最好只计算一次分母并使用 gsum :

BigTotal <- df[, sum(Variable)]
df[, sumVar1 := sum(Variable), by = .(Factor)][, propVar := sumVar1 / BigTotal]

df <- data.table(
  Factor = as.factor(sample(LETTERS, size = 10^8, replace = T)),
  Variable = sample(10^3, size = 10^8, replace = T)
)

microbenchmark::microbenchmark(dt1 = {
  aux <- df[, .(sumVar = sum(Variable)/sum(df$Variable)), keyby = .(Factor)]
  df[aux, sumVar := sumVar, on = .(Factor = Factor)]
},
dt2 = {
BigTotal <- df[, sum(Variable)]
df[, sumVar1 := sum(Variable), by = .(Factor)][, propVar := sumVar1 / BigTotal]
}, 
times = 2)


Unit: seconds
 expr      min       lq     mean   median       uq      max neval cld
  dt1 9.523696 9.523696 9.567555 9.567555 9.611414 9.611414     2   b
  dt2 3.996581 3.996581 4.521274 4.521274 5.045967 5.045967     2  a

Ben 6 年前

你有什么样的数据和时间安排?在下面100米行的例子中,我得到了以下计时

library(data.table)

df <- data.table(
  Factor = as.factor(sample(LETTERS, size = 10^8, replace = T)),
  Variable = sample(10^3, size = 10^8, replace = T)
)

# data.table solution 1
system.time({
  aux <- df[, .(sumVar = sum(Variable)/sum(df$Variable)), by = .(Factor)]
  df[aux, sumVar := sumVar, on = .(Factor = Factor)]
})  # ~10.5 seconds

# data.table solution 2
system.time({
  df[, sumVar := sum(Variable)/sum(df$Variable), by = Factor]
})  # ~8.3 seconds

# dplyr solution 1
system.time({
  df %>% dplyr::group_by(Factor) %>% dplyr::mutate(A=sum(Variable)/sum(df$Variable))
})  # ~10.0 seconds

df <- data.table(
  Factor = as.factor(sample(as.character(10^6), size = 10^8, replace = T)),
  Variable = as.numeric(sample(10^3, size = 10^8, replace = T))
)

# data.table solution 1
system.time({
  aux <- df[, .(sumVar = sum(Variable)/sum(df$Variable)), by = .(Factor)]
  df[aux, sumVar := sumVar, on = .(Factor = Factor)]
})  # ~5.0 seconds

# data.table solution 2
system.time({
  df[, sumVar := sum(Variable)/sum(df$Variable), by = Factor]
})  # ~3.1 seconds

# dplyr solution 1
system.time({
  df %>% dplyr::group_by(Factor) %>% dplyr::mutate(A=sum(Variable)/sum(df$Variable))
})  # ~6.9 seconds