映射2个向量 - 帮助向量化 [英] Mapping 2 vectors - help to vectorize

问题描述

在Matlab中工作我有两个不同长度的x坐标向量。例如：

  xm = [15 20 24 25 26 35 81 84 93]; 
 xn = [14 22 26 51 55 59 70 75 89 96]; 
  

我需要将xm映射到xn，换句话说，找出xn中的哪个坐标最接近XM。所以如果我有与这些坐标相关联的值，我可以使用这个地图作为索引，并将这些值相关联。

两个向量都被排序，每个向量中都没有重复。

我用for-loop写了一个简单的函数：

  function xmap = vectors_map（xm，xn）
 xmap = zeros（size（xm））; 
 for k = 1：numel（xm）
 [〜，ind] = min（abs（xm（k）-xn））; 
 xmap（k）= ind（1）; 
 end 
  

以上示例返回

  xmap = 
 1 2 2 3 3 3 8 9 10 
  

它可以正常工作，但需要一段时间用长向量（超过100,000点）。

任何想法如何向量化此代码？ p>

解决方案

哦！另一个选择：由于您正在寻找两个排序列表之间的紧密对应关系，您可以同时使用它们，使用类似合并的算法。这应该是O（max（length（xm），length（xn））） - ish。

  
 match_for_xn =零（长（xn），1）; 
 last_M = 1; 
 N = 1：length（xn）
％搜索M直到找到匹配项。 
 for M = last_M：length（xm）
 dist_to_curr = abs（xm（M） -  xn（N））; 
 dist_to_next = abs（xm（M + 1） -  xn（N））; 
 
如果dist_to_next> dist_to_curr 
 match_for_xn（N）= M; 
 last_M = M; 
 break 
 else 
 continue 
 end 
 
 end％M 
 end％N 
  / pre> 
 
 
编辑：
看到@ yuk的评论，上面的代码不完全正确！
 
Working in Matlab I have 2 vectors of x coordinate with different length. For example:
xm = [15 20 24 25 26 35 81 84 93];
xn = [14 22 26 51 55 59 70 75 89 96];
I need to map xm to xn, or in other words to find which coordinates in xn are closest to xm. So if I have values associated with those coordinates, I can use this map as index and correlate those values.

Both vectors are sorted and there are no duplicates in each vector.

I wrote a simple function with for-loop:
function xmap = vectors_map(xm,xn)
xmap = zeros(size(xm));
for k=1:numel(xm)
    [~, ind] = min(abs(xm(k)-xn));
    xmap(k) = ind(1);
end
For the above example is returns
xmap =
    1     2     2     3     3     3     8     9    10
It works ok, but takes a while with long vectors (over 100,000 points).

Any ideas how to vectorize this code?
解决方案
Oh!  One other option: since you're looking for close correspondences between two sorted lists, you could go through them both simultaneously, using a merge-like algorithm.  This should be O(max(length(xm), length(xn)))-ish.

match_for_xn = zeros(length(xn), 1);
last_M = 1;
for N = 1:length(xn)
  % search through M until we find a match.
  for M = last_M:length(xm)
    dist_to_curr = abs(xm(M) - xn(N));
    dist_to_next = abs(xm(M+1) - xn(N));

    if dist_to_next > dist_to_curr
      match_for_xn(N) = M;
      last_M = M;
      break
    else
      continue
    end

  end % M
end % N
EDIT:
See @yuk's comment, the above code is not totally correct!

                        
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