在列表中获取匹配索引的快捷方法

Given a list a containing vectors of unequal length and a vector b containing some elements from the vectors in a, I want to get a vector of equal length to b containing the index in a where the element in b matches (this is a bad explanation I know)...

给定一个列表包含不等长度的向量和包含a中向量的一些元素的向量b，我想得到一个长度相等的向量，包含b中元素匹配的索引（这是一个不好的解释）我知道）...

The following code does the job:

以下代码完成了这项工作：

a <- list(1:3, 4:5, 6:9)
b <- c(2, 3, 5, 8)

sapply(b, function(x, list) which(unlist(lapply(list, function(y, z) z %in% y, z=x))), list=a)
[1] 1 1 2 3

Replacing the sapply with a for loop achieves the same of course

用for循环替换sapply当然也可以实现相同的目的

The problem is that this code will be used with list and vectors with a length above 1000. On a real life set the function takes around 15 seconds (both the for loop and the sapply).

问题是这个代码将用于长度大于1000的列表和向量。在现实生活中，该函数大约需要15秒（for循环和sapply）。

Does anyone have an idea how to speed this up, safe for a parallel approach? I have failed to see a vectorized approach (and I cannot program in C, though that would probably be the fastest).

有没有人知道如何加快速度，对并行方法安全？我没有看到矢量化方法（我不能用C编程，尽管这可能是最快的）。

Edit:

编辑：

Will just emphasize Aaron's elegant solution using match() which gave a speed increase in the order of 1667 times (from 15 to 0.009)

将使用match（）强调Aaron的优雅解决方案，其速度提升1667次（从15到0.009）

I expanded a bit on it to allow multiple matches (the return is then a list)

我在它上面扩展了一下以允许多个匹配（返回是一个列表）

a <- list(1:3, 3:5, 3:7)
b <- c(3, 5)
g <- rep(seq_along(a), sapply(a, length))
sapply(b, function(x) g[which(unlist(a) %in% x)])
[[1]]
[1] 1 2 3

[[2]]
[1] 2 3

The runtime for this was 0.169 which is arguably quite slower, but on the other hand more flexible

这个的运行时间是0.169，这可能相当慢，但另一方面更灵活

2 个解决方案

#1

Here's one possibility using match:

这是使用匹配的一种可能性：

> a <- list(1:3, 4:5, 6:9)
> b <- c(2, 3, 5, 8)
> g <- rep(seq_along(a), sapply(a, length))
> g[match(b, unlist(a))]
[1] 1 1 2 3

findInterval is another option:

findInterval是另一种选择：

> findInterval(match(b, unlist(a)), cumsum(c(0,sapply(a, length)))+1)
[1] 1 1 2 3

For returning a list, try this:

要返回列表，请尝试以下操作：

a <- list(1:3, 4:5, 5:9)
b <- c(2,3,5,8,5)
g <- rep(seq_along(a), sapply(a, length))
aa <- unlist(a)
au <- unique(aa)
af <- factor(aa, levels=au)
gg <- split(g, af)
gg[match(b, au)]

#2

As a comment to your post suggests, it depends on what you want to do if/when the same element appears in multiple vectors in a. Assuming that you want the lowest index you could do:

正如您对帖子的评论所暗示的那样，如果/当相同元素出现在a中的多个向量中时，它取决于您想要做什么。假设您想要最低的索引，您可以这样做：

apply(sapply(a, function(vec) {b %in% vec}), 1, which.max)

#1