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短语快速区分值法

distinct-values distinct r

Przemyslaw Remin · 技术社区 · 5 年前

What is the fastest way to get a vector of sorted unique values from a data.table?

test_df <-
  data.frame(
    company = c(1, 1,  2, 2, 3)
  )

unique_values = df[,logical(1), keyby = company]$company

但我一直在犯错误:

错误 [.data.frame (df,逻辑(1),keyby=company):未使用参数(keyby=company)

编辑请注意,我的问题的重点是让这个特定的方法发挥作用。对于其他实现目标的方法的建议,请遵循我提到的帖子。

0 回复 | 直到 5 年前

GKi 4 年前

以防你正在寻找一个快速的 unique 看看 kit::funique collapse::funique :

setDTthreads(1)
microbenchmark::microbenchmark(
dt = y[,logical(1), keyby = company]$company,
base = unique(x$company),
collapse = collapse::funique(x$company),
kit = kit::funique(x$company))
#Unit: milliseconds
#     expr       min        lq      mean    median        uq       max neval
#       dt 12.862388 13.575131 14.759180 14.248541 14.945780 49.930937   100
#     base 12.939646 13.505176 14.734066 14.773846 15.415468 18.256204   100
# collapse  3.302862  3.589133  3.685685  3.692886  3.773045  4.063564   100
#      kit  1.903043  2.433478  2.963308  2.882986  3.076537  6.183840   100

setDTthreads(4)
microbenchmark::microbenchmark(
dt = y[,logical(1), keyby = company]$company,
base = unique(x$company),
collapse = collapse::funique(x$company),
kit = kit::funique(x$company))
#Unit: milliseconds
#     expr       min        lq      mean    median        uq       max neval
#       dt  5.480513  7.384032  7.873730  7.569420  8.346282 11.193741   100
#     base 12.998406 13.295775 14.464446 13.736353 14.856721 47.320488   100
# collapse  3.333292  3.549712  3.655851  3.645528  3.737236  4.325676   100
#      kit  1.881232  2.825040  2.959422  2.917149  3.004288  5.281440   100

数据和图书馆:

set.seed(42)
n <- 1e6
company <- c("A", "S", "W", "L", "T", "T", "W", "A", "T", "W")
item <- c("Thingy", "Thingy", "Widget", "Thingy", "Grommit", 
          "Thingy", "Grommit", "Thingy", "Widget", "Thingy")
sales <- c(120, 140, 160, 180, 200, 120, 140, 160, 180, 200)

x <- data.frame(company = sample(company, n, TRUE), 
                      item = sample(item, n, TRUE), 
                sales = sample(sales, n, TRUE))

library(data.table)
y <- as.data.table(x)

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