元素智能阵列复制根据计数 [英] Element-wise array replication according to a count
问题描述
我的问题是类似于此 之一,但我想根据在相同尺寸的第二阵列指定的计数复制的每个元素
My question is similar to this one, but I would like to replicate each element according to a count specified in a second array of the same size.
这样的一个例子,说我有一个数组 V = [3 1 9 4]
,我想用代表= [2 3 1 5]
来复制所述第一元件的2倍,二三次,依此类推,以获得 [3 3 1 1 1 9 4 4 4 4 4]
。
An example of this, say I had an array v = [3 1 9 4]
, I want to use rep = [2 3 1 5]
to replicate the first element 2 times, the second three times, and so on to get [3 3 1 1 1 9 4 4 4 4 4]
.
到目前为止,我用一个简单的循环来完成这项工作。这就是我开始:
So far I'm using a simple loop to get the job done. This is what I started with:
vv = [];
for i=1:numel(v)
vv = [vv repmat(v(i),1,rep(i))];
end
我设法preallocating空间来改善:
I managed to improve by preallocating space:
vv = zeros(1,sum(rep));
c = cumsum([1 rep]);
for i=1:numel(v)
vv(c(i):c(i)+rep(i)-1) = repmat(v(i),1,rep(i));
end
不过,我还是觉得必须有一个更聪明的方式来做到这一点...谢谢
However I still feel there has to be a more clever way to do this... Thanks
推荐答案
下面是我想做到这一点的一种方法:
Here's one way I like to accomplish this:
>> index = zeros(1,sum(rep));
>> index(cumsum([1 rep(1:end-1)])) = 1;
index =
1 0 1 0 0 1 1 0 0 0 0
>> index = cumsum(index)
index =
1 1 2 2 2 3 4 4 4 4 4
>> vv = v(index)
vv =
3 3 1 1 1 9 4 4 4 4 4
这通过首先创建零的索引向量相同的长度的所有值的最后计数。通过执行代表的累加值
与放置在开始的1去掉最后一个元素和载体,我得到指数的载体导入指数
表示将复制值的组将开始。这些点都标有的。当首页
执行的累加值,我得到一个最终的指数矢量,我可以使用索引到 v
来创建多相复制的价值观的载体。
This works by first creating an index vector of zeroes the same length as the final count of all the values. By performing a cumulative sum of the rep
vector with the last element removed and a 1 placed at the start, I get a vector of indices into index
showing where the groups of replicated values will begin. These points are marked with ones. When a cumulative sum is performed on index
, I get a final index vector that I can use to index into v
to create the vector of heterogeneously-replicated values.
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