3

我有一个模式向量,需要agrep在它们上使用。问题是agrep似乎一次只采用一种模式。

patt <- c("test","10 Barrel")
lut  <- c("1 Barrel","10 Barrel Brewing","Harpoon 100 Barrel Series","resr","rest","tesr")

for (i in 1:length(patt)) {
  print(agrep(patt[i],lut,max=1,v=T))
}

结果:

[1] "rest" "tesr"
[1] "1 Barrel"                  "10 Barrel Brewing"         "Harpoon 100 Barrel Series"

for在长模式上很慢,因此尝试以矢量化形式进行:

VecMatch1 = function(string, stringVector){
  stringVector[agrep(string, stringVector, max = 1)]
}
a = VecMatch1(patt,lut)

Warning message:
In agrep(string, stringVector, max = 1) :
  argument 'pattern' has length > 1 and only the first element will be used

可能是lapply等功能可以提供帮助吗?谢谢!!

4

1 回答 1

6

使用lapply

lapply(patt, agrep, x=lut, max.distance=c(cost=1, all=1), value=TRUE)

[[1]]
[1] "rest" "tesr"

[[2]]
[1] "1 Barrel"                  "10 Barrel Brewing"         "Harpoon 100 Barrel Series"

您可能可以使用 dplyr 或 data.table 获得更快的性能。

于 2015-07-15T16:12:27.363 回答