如何通过MATLAB中的特定值增加数组中的某些元素

Suppose we have an array

假设我们有一个数组

A = zeros([1,10]);

We have several indexes with possible duplicate say:

我们有几个可能重复的索引:

indSeq = [1,1,2,3,4,4,4];

How can we increase A(i) by the number of i in the index sequence i.e. A(1) = 2, A(2) = 1, A(3) = 1, A(4) = 3?

如何通过索引序列中的i增加A(i)，即A(1) = 2, A(2) = 1, A(3) = 1, A(4) = 3?

The code A(indSeq) = A(indSeq)+1 does not work.

代码A(indSeq) = A(indSeq)+1不起作用。

I know that I can use the following for loop to achieve the goal, but I wonder if there is anyway that we can avoid for-loop? We can assume that the indSeq is sorted.

我知道我可以使用下面的for循环来实现目标，但是我想知道我们是否可以避免for循环?我们可以假设indSeq是有序的。

A for-loop solution:

一个for循环的解决方案:

for i=1:length(indSeq)
  A(indSeq(i)) = A(indSeq(i))+1;
end;

3 个解决方案

#1

You can use accumarray for such a label based counting job, like so -

您可以使用accumarray为这样一个基于标签的计数工作，如so -

accumarray(indSeq(:),1)

Benchmarking

基准测试

As suggested in the other answer, you can also use hist/histc. Let's benchmark these two for a large datasize. The benchmarking code I used had -

如另一个答案所示，您还可以使用hist/histc。让我们为这两个数据进行基准测试。我使用的基准代码是

%// Create huge random array filled with ints that are duplicated & sorted
maxn = 100000;
N = 10000000;
indSeq = sort(randi(maxn,1,N));

disp('--------------------- With HISTC')
tic,histc(indSeq,unique(indSeq));toc

disp('--------------------- With ACCUMARRAY')
tic,accumarray(indSeq(:),1);toc

Runtime output -

运行时输出-

--------------------- With HISTC
Elapsed time is 1.028165 seconds.
--------------------- With ACCUMARRAY
Elapsed time is 0.220202 seconds.

#2

This is run-length encoding, and the following code should do the trick for you.

这是一个运行长度的编码，下面的代码应该为您完成这个技巧。

A=zeros(1,10);
indSeq = [1,1,2,3,4,4,4,7,1];
indSeq=sort(indSeq); %// if your input is always sorted, you don't need to do this
pos = [1; find(diff(indSeq(:)))+1; numel(indSeq)+1];
A(indSeq(pos(1:end-1)))=diff(pos)

which returns

它返回

A =
     3     1     1     3     0     0     1     0     0     0

This algorithm was written by Luis Mendo for MATL.

这个算法是由Luis Mendo为MATL编写的。

#3

I think what you are looking for is the number of occurences of unique values of the array. This can be accomplished with:

我认为你要找的是数组中唯一值出现的次数。这可以通过以下方式完成:

[num, val] = hist(indSeq,unique(indSeq));

the output of your example is:

您的示例的输出是:

num = 2 1 1 3
val = 1 2 3 4

so num is the number of times val occurs. i.e. the number 1 occurs 2 times in your example

所以num是val发生的次数。例如，数字1在你的例子中出现了2次

#1