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r - 加速匹配固定字符串 %in%/%like% 与 bool 输出

转载 作者:行者123 更新时间:2023-12-01 13:19:49 24 4
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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?

最佳答案

如果您不介意精确匹配,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

关于r - 加速匹配固定字符串 %in%/%like% 与 bool 输出,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50533906/

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