12

我想要一个函数来返回向量匹配子序列的初始索引。例如：

y <- c("a","a","a","b","c")

multi_match(c("a","a"), y)
# [1] 1 2

multi_match(c("a","b"), y)
# [1] 3

我有一个粗略的实现，但我觉得我必须重新发明轮子，而且有点笨拙。有没有更好的方法来实现这一点，或者在某个地方是否存在具有类似功能的预先存在的功能？

multi_match <- function(x, table){
    # returns initial indicies of all substrings in table which match x
    if(length(table) < length(x)){
        return(NA)
    }else{
        check_mat <- matrix(nrow = length(x), ncol = length(table))
        for(i in 1:length(x)){
            check_mat[i,] <- table %in% x[i]
        }
        out <- vector(length = length(table))
        for(i in 1:(length(table)-(length(x)-1))){
            check <- vector(length=length(x))
            for(j in 1:length(x)){
                check[j] <- check_mat[j,(i+(j-1))]
            }
            out[i] <- all(check)
        }
        if(length(which(out))==0){
            return(NA)
        }else{
            return(which(out))
        }
    }
}
4

2 回答 2

19

rollapply在动物园尝试：

library(zoo)
which(rollapply(y, 2, identical, c("a", "a")))
## [1] 1 2
which(rollapply(y, 2, identical, c("a", "b")))
## [1] 3
于 2013-04-26T19:45:18.040 回答
6 
set.seed(0)
a <- sample(1:6,12000, TRUE)
b <- 2:4

vecIn <- function(a,b){
which(
Reduce('+', lapply(seq_along(y <- lapply(b, '==', a)), function(x){
                                            y[[x]][x:(length(a) - length(b) +x)]
                                           }
                  )
      ) == length(b)
     )
} 

> vecIn(a,b)
 [1]     2   154   986  1037  1046  1257  1266  1750  2375  2677  3184  3206
[13]  3499  3526  3882  4238  4311  4388  4437  4580  4714  4766  4827  5046
[25]  5279  5629  6153  6842  6856  6919  7200  7516  7520  7707  7824  7859
[37]  8140  8191  8687  9208  9281  9313 10022 10320 10617 10720 10958 11179
[49] 11567 11591 11698 11811

library(zoo)
library(rbenchmark)

func1 <- function(a,b){
 gregexpr(paste0(b,collapse=""),paste0(a,collapse=""))
}

func2 <- function(a,b){
 which(rollapply(a, length(b), identical, b))
}

func3 <- vecIn

一些基准 

benchmark(func1(a,b), func2(a,b), func3(a,b))
         test replications elapsed relative user.self sys.self user.child
1 func1(a, b)          100   0.673    5.904     0.680    0.000          0
2 func2(a, b)          100  28.808  252.702    28.198    0.672          0
3 func3(a, b)          100   0.114    1.000     0.116    0.000          0
  sys.child
1         0
2         0
3         0
于 2013-05-14T07:03:23.020 回答