R 如何将矢量元素作为一个整体进行匹配

我想知道如何将向量作为一个整体进行匹配。我有两个向量

a，b

a <- c(5,1,2,6,3,4,8)
b <- c(1,2,3)

在

match（）

上，我获取单个向量元素的位置，对于%中的

%我获取单个向量元素的逻辑位置。但我希望将整个向量
b
立即与
a
匹配。它不应该匹配单个元素，而应该匹配整个向量，并获得匹配开始的位置

所需输出：

在上面的向量中，未找到匹配项，因为我查找的是整个向量，而不是单个向量项
 您可以强制执行此操作，只需逐个元素循环向量

a <- c(5,1,2,6,3,4,8)
b <- c(1,2,3)

matchr <- function(a,b){

    # First, loop through the a vector
    for(i in 1:(length(a)-length(b))){

        pos <- FALSE

        # Next loop through the b vector, 
        for(j in 1:length(b)){

            # as we're looping through b, check if each element matches the corresponding part of the a vector we're currently at.
            if( a[i+j-1] == b[j]){
                pos <- TRUE
            } else{
                pos <- FALSE
                break
            }
        }

        # if all the elements match, return where we are in the a vector
        if(pos == TRUE){
            return(i)
        } 
    }
    # if we finish the a vector and never got a match, return no match.
    return("No match")
}

matchr(a,b)
[1] "No match"

d <- c(7,5,4,2,1,2,3,8,5)

matchr(d,b)
[1] 5

e <- c(2,3,8)

matchr(d,e)
[1] 6

同样，通过编译函数，您将获得显著的速度优势。
这是一种方法，有几个示例：

wholematch<-function(a=c(5,1,3,2,1,2,5,6,2,6),b=c(1,2,6))
{
    for(loop.a in 1:(length(a)-length(b)))
    {
    #pmatch gives the first occurrence of each value of b in a. To be sure of finding the consecutive matches, use pmatch starting from all the possible positions of "a"
    wmatch<-(loop.a-1)+pmatch(b,a[loop.a:length(a)])
    #If at any time the number of matches is less than the length of the vector to match, we will never find a match. Return NA 
    if(length(na.omit(pmatch(b,a[loop.a:length(a)])))<length(b)) return(NA)
    #If all indices are adjacent, return the vector of indices
    if(max(diff(wmatch))==1) return(wmatch) #return(wmatch[1]) if you only want the start
    }
}

wholematch()
[1] NA

wholematch(a=c(5,1,3,2,1,2,5,6,2,6),b=c(6,2,6))
[1]  8  9 10

wholematch如果我们根据正在测试的向量检查
match（）
输出的长度（使用
na.omit
）如何

ifelse(length(na.omit(match(b, a))) == length(b), match(b, a)[1], NA)
#[1] 2
#adding a new value in b so it wont match, we get
b  <- c(1, 2, 3, 9)
ifelse(length(na.omit(match(b, a))) == length(b), match(b, a)[1], NA)
#[1] NA

ifelse（长度（na.省略（匹配（b，a）））==长度（b），匹配（b，a）[1]，na）
#[1] 2
#在b中添加一个新值使其不匹配，我们得到
b@Henrik。谢谢你找到一个可能的副本。我会看一看，可能会有帮助。@rawr感谢您引导链接。这个问题已经结束，提供的链接似乎不相关。以前，后来被删除的
Henrik提供的链接效果良好。谢谢大家@罗尔！！。我指的是顶部的链接，而不是你的链接。请不要删除链接。看起来链接有点混乱。我会重新打开，@rawr可以用他的链接复制它
wholematch<-function(a=c(5,1,3,2,1,2,5,6,2,6),b=c(1,2,6))
{
    for(loop.a in 1:(length(a)-length(b)))
    {
    #pmatch gives the first occurrence of each value of b in a. To be sure of finding the consecutive matches, use pmatch starting from all the possible positions of "a"
    wmatch<-(loop.a-1)+pmatch(b,a[loop.a:length(a)])
    #If at any time the number of matches is less than the length of the vector to match, we will never find a match. Return NA 
    if(length(na.omit(pmatch(b,a[loop.a:length(a)])))<length(b)) return(NA)
    #If all indices are adjacent, return the vector of indices
    if(max(diff(wmatch))==1) return(wmatch) #return(wmatch[1]) if you only want the start
    }
}

wholematch()
[1] NA

wholematch(a=c(5,1,3,2,1,2,5,6,2,6),b=c(6,2,6))
[1]  8  9 10

ifelse(length(na.omit(match(b, a))) == length(b), match(b, a)[1], NA)
#[1] 2
#adding a new value in b so it wont match, we get
b  <- c(1, 2, 3, 9)
ifelse(length(na.omit(match(b, a))) == length(b), match(b, a)[1], NA)
#[1] NA