从具有起始和结束位置的向量创建序列 [英] Create a sequence from vectors with start and end positions

问题描述

给出两个相等长度的独立向量:f.start和f.end，我想构造一个序列(乘以1)，从f.start[1]:f.end[1]到f.start[2]:f.end[2]，...，到f.start[n]:f.end[n].

Given two separate vectors of equal length: f.start and f.end, I would like to construct a sequence (by 1), going from f.start[1]:f.end[1] to f.start[2]:f.end[2], ..., to f.start[n]:f.end[n].

这里是一个只有6行的示例.

Here is an example with just 6 rows.

   f.start  f.end
[1,]   45739 122538
[2,]  125469 202268
[3,]  203563 280362
[4,]  281657 358456
[5,]  359751 436550
[6,]  437845 514644

粗略地讲，循环可以做到这一点，但是对于较大的数据集(行> 2000)而言，循环非常慢.

Crudely, a loop can do it, but is extremely slow for larger datasets (rows>2000).

f.start<-c(45739,125469,203563,281657,359751,437845)
f.end<-c(122538,202268,280362,358456,436550,514644)
f.ind<-f.start[1]:f.end[1]
for (i in 2:length(f.start))
{
 f.ind.temp<-f.start[i]:f.end[i]
 f.ind<-c(f.ind,f.ind.temp)
}

我怀疑这可以通过apply()来完成，但是我还没有弄清楚如何在apply中包含两个单独的参数，因此希望能提供一些指导.

I suspect this can be done with apply(), but I have not worked out how to include two separate arguments in apply, and would appreciate some guidance.

推荐答案

您可以尝试使用mapply或Map，它们同时对两个向量进行迭代.您需要提供该函数作为第一个参数:

You can try using mapply or Map, which iterates simultaneously on your two vectors. You need to provide the function as first argument:

vec1 = c(1,33,50)
vec2 = c(10,34,56)

unlist(Map(':',vec1, vec2))
# [1]  1  2  3  4  5  6  7  8  9 10 33 34 50 51 52 53 54 55 56

只需将vec1和vec2替换为f.start和f.end(只要提供all(f.start<=f.end)

Just replace vec1 and vec2 by f.start and f.end provided all(f.start<=f.end)

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