Aquí es un punto de referencia de la solución de @ David Arenburg there, así como un resumen de algunas de las soluciones incluidas en este documento (@mnel, @Sven Hohenstein, @Henrik):
library(dplyr)
library(data.table)
library(microbenchmark)
library(tidyr)
library(ggplot2)
df <- mtcars
DT <- as.data.table(df)
DT_32k <- rbindlist(replicate(1e3, mtcars, simplify = FALSE))
df_32k <- as.data.frame(DT_32k)
DT_32M <- rbindlist(replicate(1e6, mtcars, simplify = FALSE))
df_32M <- as.data.frame(DT_32M)
bench <- microbenchmark(
base_32 = aggregate(hp ~ cyl, df, function(x) length(unique(x))),
base_32k = aggregate(hp ~ cyl, df_32k, function(x) length(unique(x))),
base_32M = aggregate(hp ~ cyl, df_32M, function(x) length(unique(x))),
dplyr_32 = summarise(group_by(df, cyl), count = n_distinct(hp)),
dplyr_32k = summarise(group_by(df_32k, cyl), count = n_distinct(hp)),
dplyr_32M = summarise(group_by(df_32M, cyl), count = n_distinct(hp)),
data.table_32 = DT[, .(count = uniqueN(hp)), by = cyl],
data.table_32k = DT_32k[, .(count = uniqueN(hp)), by = cyl],
data.table_32M = DT_32M[, .(count = uniqueN(hp)), by = cyl],
times = 10
)
Resultados:
print(bench)
# Unit: microseconds
# expr min lq mean median uq max neval cld
# base_32 816.153 1064.817 1.231248e+03 1.134542e+03 1263.152 2430.191 10 a
# base_32k 38045.080 38618.383 3.976884e+04 3.962228e+04 40399.740 42825.633 10 a
# base_32M 35065417.492 35143502.958 3.565601e+07 3.534793e+07 35802258.435 37015121.086 10 d
# dplyr_32 2211.131 2292.499 1.211404e+04 2.370046e+03 2656.419 99510.280 10 a
# dplyr_32k 3796.442 4033.207 4.434725e+03 4.159054e+03 4857.402 5514.646 10 a
# dplyr_32M 1536183.034 1541187.073 1.580769e+06 1.565711e+06 1600732.034 1733709.195 10 b
# data.table_32 403.163 413.253 5.156662e+02 5.197515e+02 619.093 628.430 10 a
# data.table_32k 2208.477 2374.454 2.494886e+03 2.448170e+03 2557.604 3085.508 10 a
# data.table_32M 2011155.330 2033037.689 2.074020e+06 2.052079e+06 2078231.776 2189809.835 10 c
Parcela:
as_tibble(bench) %>%
group_by(expr) %>%
summarise(time = median(time)) %>%
separate(expr, c("framework", "nrow"), "_", remove = FALSE) %>%
mutate(nrow = recode(nrow, "32" = 32, "32k" = 32e3, "32M" = 32e6),
time = time/1e3) %>%
ggplot(aes(nrow, time, col = framework)) +
geom_line() +
scale_x_log10() +
scale_y_log10() + ylab("microseconds")
información de la sesión:
sessionInfo()
# R version 3.4.1 (2017-06-30)
# Platform: x86_64-pc-linux-gnu (64-bit)
# Running under: Linux Mint 18
#
# Matrix products: default
# BLAS: /usr/lib/atlas-base/atlas/libblas.so.3.0
# LAPACK: /usr/lib/atlas-base/atlas/liblapack.so.3.0
#
# locale:
# [1] LC_CTYPE=fr_FR.UTF-8 LC_NUMERIC=C LC_TIME=fr_FR.UTF-8
# [4] LC_COLLATE=fr_FR.UTF-8 LC_MONETARY=fr_FR.UTF-8 LC_MESSAGES=fr_FR.UTF-8
# [7] LC_PAPER=fr_FR.UTF-8 LC_NAME=C LC_ADDRESS=C
# [10] LC_TELEPHONE=C LC_MEASUREMENT=fr_FR.UTF-8 LC_IDENTIFICATION=C
#
# attached base packages:
# [1] stats graphics grDevices utils datasets methods base
#
# other attached packages:
# [1] ggplot2_2.2.1 tidyr_0.6.3 bindrcpp_0.2 stringr_1.2.0
# [5] microbenchmark_1.4-2.1 data.table_1.10.4 dplyr_0.7.1
#
# loaded via a namespace (and not attached):
# [1] Rcpp_0.12.11 compiler_3.4.1 plyr_1.8.4 bindr_0.1 tools_3.4.1 digest_0.6.12
# [7] tibble_1.3.3 gtable_0.2.0 lattice_0.20-35 pkgconfig_2.0.1 rlang_0.1.1 Matrix_1.2-10
# [13] mvtnorm_1.0-6 grid_3.4.1 glue_1.1.1 R6_2.2.2 survival_2.41-3 multcomp_1.4-6
# [19] TH.data_1.0-8 magrittr_1.5 scales_0.4.1 codetools_0.2-15 splines_3.4.1 MASS_7.3-47
# [25] assertthat_0.2.0 colorspace_1.3-2 labeling_0.3 sandwich_2.3-4 stringi_1.1.5 lazyeval_0.2.0
# [31] munsell_0.4.3 zoo_1.8-0
es que hay una manera de hacer lo mismo que el uso de SQL en R? – user3581800
@ user3581800 Con el paquete 'sqldf' puede hacer' sqldf ("SELECT nombre, COUNT (distinct (order_no)) FROM myvec GROUP BY nombre") ' – jogo