加速匹配固定字符串 %in% / %like% 与布尔输出

Question

我想要一个逻辑向量来指示第二个列表中是否存在匹配项。如果您需要完全匹配，可以使用 %in% 运算符，但我对任何匹配都感兴趣，所以我创建了 %like% 运算符：

table <- rownames(mtcars) 
table
#>  [1] "Mazda RX4"           "Mazda RX4 Wag"       "Datsun 710"         
#>  [4] "Hornet 4 Drive"      "Hornet Sportabout"   "Valiant"            
#>  [7] "Duster 360"          "Merc 240D"           "Merc 230"           
#> [10] "Merc 280"            "Merc 280C"           "Merc 450SE"         
#> [13] "Merc 450SL"          "Merc 450SLC"         "Cadillac Fleetwood" 
#> [16] "Lincoln Continental" "Chrysler Imperial"   "Fiat 128"           
#> [19] "Honda Civic"         "Toyota Corolla"      "Toyota Corona"      
#> [22] "Dodge Challenger"    "AMC Javelin"         "Camaro Z28"         
#> [25] "Pontiac Firebird"    "Fiat X1-9"           "Porsche 914-2"      
#> [28] "Lotus Europa"        "Ford Pantera L"      "Ferrari Dino"       
#> [31] "Maserati Bora"       "Volvo 142E"

x <- c('Porsche', 'Porsche 914-2', 'Porsche 911', 'Volvo')

x %in% table
#> [1] FALSE  TRUE FALSE FALSE

"%like%" <- function(x, table) sapply(x, function(x) 
  sum(grepl(pattern = x, x = table))>0, USE.NAMES = FALSE)

x %like% table
#> [1]  TRUE  TRUE FALSE  TRUE

不幸的是，%like% 运算符非常慢：

library(microbenchmark)

x1 <- c('Porsche', 'Porsche 914-2', 'Porsche 911', 'Volvo')
x2 <- rep(x1, 10)
x3 <- rep(x1, 100)
table <- rownames(mtcars) 

"%like%" <- function(x, table) sapply(x, function(x) 
  sum(grepl(pattern = x, x = table))>0, USE.NAMES = FALSE)

microbenchmark(x1 %in% table, x1 %like% table, times = 1000)
#> Unit: microseconds
#>             expr    min      lq      mean  median      uq     max neval
#>    x1 %in% table  1.549  1.8635  2.248905  2.2545  2.5000   7.331  1000
#>  x1 %like% table 69.697 71.2110 73.235948 72.6555 74.0835 149.087  1000
microbenchmark(x2 %in% table, x2 %like% table, times = 1000)
#> Unit: microseconds
#>             expr     min       lq       mean   median       uq      max
#>    x2 %in% table   2.327   2.8795   3.330329   3.3055   3.6515    7.539
#>  x2 %like% table 573.005 581.0885 590.760082 584.2270 588.2580 1624.687
#>  neval
#>   1000
#>   1000
microbenchmark(x3 %in% table, x3 %like% table, times = 1000)
#> Unit: microseconds
#>             expr      min       lq       mean   median        uq       max
#>    x3 %in% table    9.195    9.950   11.79078   10.923   12.5675    36.341
#>  x3 %like% table 5612.931 5707.168 5973.83801 5737.892 5823.7875 11868.495
#>  neval
#>   1000
#>   1000

如何加速 %like% 运算符？

Answer 1

如果您不介意完全匹配，可以在 grepl 中使用 fixed = T 来加快速度

"%birger%" <- function(x, table) sapply(x, function(x) 
  sum(grepl(pattern = x, x = table))>0, USE.NAMES = FALSE)

'%birger.fixed%' <- function(x, table) sapply(x, function(x) 
  any(grepl(pattern = x, x = table, fixed = T)), USE.NAMES = FALSE)

all.equal(x %birger.fixed% table, x %birger% table)
# [1] TRUE

microbenchmark(x %birger.fixed% table, x %birger% table, times = 1000, unit = 'relative')

# Unit: relative
#                    expr      min       lq     mean   median       uq      max neval
#  x %birger.fixed% table 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000  1000
#        x %birger% table 2.059546 2.011009 1.903589 1.913446 1.857798 1.336424  1000

加速匹配固定字符串 %in% / %like% 与布尔输出

Speedup matching fixed strings %in% / %like% with boolean output

r

match

microbenchmark