"parfor中的变量无法分类." Matlab中的错误 [英] "the variable in a parfor cannot be classified." error in Matlab

问题描述

我正在尝试使用parfor实现一个非常简单的程序，但是出现一些错误.我看到几乎所有的SO问题都有可能重复，但是没有一个与我的问题情况相似.我得到的错误是:

I am trying to implement a very simple program with parfor but I get some errors. I saw nearly all of the SO questions for a possible duplication but non of them was similar to my question situation. The error I get is :

错误:变量 log_likelihood_II_with_entropy 在parfor中不能是 分类.

Error: The variable log_likelihood_II_with_entropy in a parfor cannot be classified.

我的代码写在下面:

em_iterations=10;

users=5;
log_likelihood_II_with_entropy=zeros(users,em_iterations);

parfor u = 1:1:users
      for current_iter=1:1:em_iterations
           log_likelihood_II_with_entropy(u,current_iter)=rand();               
      end
end

推荐答案

由于log_likelihood_II_with_entropy依赖于parfor索引(u)和内部索引"(current_iter)，因此无法对其进行分类.每个parfor迭代均独立于其他迭代，并且不会按顺序执行(即u不一定会按1,2,3,4，...，users >).

Since log_likelihood_II_with_entropy relies on both the parfor index (u) and an "inside index" (current_iter) it cannot be classified. Every parfor iteration is independent from the others and they are not executed in order (that is, u will not necessarily go from 1 to users in order 1,2,3,4,...,users).

我的建议是让单个parfor迭代(工作人员)构建整个log_likelihood_II_with_entropy行.

My suggestion is to let the single parfor iteration (worker) build an entire row of log_likelihood_II_with_entropy.

parfor u=1:users
    single_row=zeros(1,em_iterations);
    for current_iter=1:1:em_iterations
        single_row(current_iter)=rand();
    end
    log_likelihood_II_with_entropy(u,:)=single_row;
end

以这种方式，每个parfor任务(parfor主体本身)将预分配并评估单个行，而不管u的值是多少.然后它将替换/连接log_likelihood_II_with_entropy矩阵中的该值.

In this manner every parfor task (the parfor body itself) will preallocate and evaluate a single row, no matter what the u value is. And then it will replace/concatenate such value in the log_likelihood_II_with_entropy matrix.

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