### 计算机代写|并行计算作业代写Parallel Computing代考|Use parfor-Loops for Reduction Assignments

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 计算机代写|并行计算作业代写Parallel Computing代考|Use parfor-Loops for Reduction Assignments

These two examples show parfor-loops using reduction assignments. A reduction is an accumulation across iterations of a loop. The example on the left uses $x$ to accumulate a sum across 10 iterations of the loop. The example on the right generates a concatenated array, $1: 10$. In both of these examples, the execution order of the iterations on the workers does not matter: while the workers calculate individual results for each iteration, the client properly accumulates and assembles the final loop result.

If the loop iterations operate in a nondeterministic sequence, you might expect the concatenation sequence in the example on the right to be nonconsecutive. However, MATLAB recognizes the concatenation operation and yields deterministic results.
The next example, which attempts to compute Fibonacci numbers, is not a valid parfor-loop because the value of an element of $f$ in one iteration depends on the values of other elements of $f$ calculated in other iterations.
$f=$ zeros $(1,50)$;
$f(1)=1$;
$f(2)=2$;
$f=$ zeros $(1,50)$
$f(1)=1$
$f(2)=2$
parfor $n=3: 50$
$\quad f(n)=f(n-1)+f(n-2) ;$
end
parfor $n=3: 50$
$f(n)=f(n-1)+f(n-2)$;
end
When you are finished with your loop examples, clear your workspace and delete your parallel pool of workers:
clear
delete (gcp)

## 计算机代写|并行计算作业代写Parallel Computing代考|Handle Classes

You can send handle objects as inputs to the body of a parfor-loop. However, any changes that you make to handle objects on the workers during loop iterations are not automatically propagated back to the client. That is, changes made inside the loop are not automatically reflected after the loop. output variables of the parfor-loop. In the following example, maps is a sliced input/output variable.
maps $={$ containers $\cdot$ Map (), containers $\cdot \operatorname{Map}()$, containers $\cdot$ Map ()$}$;
parfor ii $=1:$ numel (maps)
mymap $=\operatorname{maps}{i i} ;$ \&o input slice assigned to local copy
for $j j=1: 1000$
maps = {containers. Map(), containers. Map( ), containers. Map()};
parfor ii $=1:$ numel $($ maps)
mymap $=$ maps ${i i} ; ~ \%$ input slice assigned to local copy
for jj $=1: 1000$
mymap(num2str $(j j))=$ rand;
end
maps ${i i}=$ mymap; $\quad \%$ modified local copy assigned to output slice
end
mymap $($ num2str $(j j))=$ rand;
end
maps ${$ ii $}=$ mymap; \% modified local copy assigned to output slice
end

## 计算机代写|并行计算作业代写Parallel Computing代考|Loop Variables

The loop variable defines the loop index value for each iteration. You set it in the first line of a parfor statement.
parfor $p=1: 12$
For values across all iterations, the loop variable must evaluate to ascending consecutive integers. Each iteration is independent of all others, and each has its own loop index value.
Required (static): Assignments to the loop variable are not allowed.
This restriction is required, because changing $p$ in the parfor body cannot guarantee the independence of iterations.
This example attempts to modify the value of the loop variable $p$ in the body of the loop, and thus is invalid.
parfor $p=1: n$
$p=p+1$
$a(p)=i$;
end
Required (static): You cannot index or subscript the loop variable in any way.
This restriction is required, because referencing a field of a loop variable cannot guarantee the independence of iterations.
The following code attempts to reference a field (b) of the loop variable ( $p$ ) as if it were a structure. Both lines within the loop are invalid.
parfor $p=1: n$
$p \cdot b=3$
$x(p)=\operatorname{fun}(p \cdot b)$
$p \cdot b=3$
end $x(p)=\operatorname{fun}(p \cdot b)$
Similarly, the following code is invalid because it attempts to index the loop variable as a 1-by-1 matrix:
parfor $p=1: n$
$x=p(1)$
end
Required (static): You cannot use a range increment in for-loops nested inside a par for-loop.
Consider the following example:
\begin{aligned} N &=10 ; \ T &=3 ; \ A &=\operatorname{zeros}(N, T) ; \ B &=\operatorname{zeros}(N, T) ; \end{aligned}
The following code is invalid.
parfor $i=1: 1: \mathrm{N}$
for $\mathrm{t}=1: 1: \mathrm{T}$

## 计算机代写|并行计算作业代写Parallel Computing代考|Use parfor-Loops for Reduction Assignments

F=零(1,50);
F(1)=1;
F(2)=2;
F=零(1,50)
F(1)=1
F(2)=2

F(n)=F(n−1)+F(n−2);

_n=3:50
F(n)=F(n−1)+F(n−2);

clear
delete (gcp)

## 计算机代写|并行计算作业代写Parallel Computing代考|Handle Classes

parfor ii=1:数字（地图）
mymap=地图⁡一世一世;\&o 分配给本地副本

parfor ii=1:取名字(地图）

end
mymap的已修改本地副本(num2str(jj))=兰特；

## 计算机代写|并行计算作业代写Parallel Computing代考|Loop Variables

p=p+1

end

p⋅b=3
X(p)=乐趣⁡(p⋅b)
p⋅b=3

parforp=1:n
X=p(1)
end

ñ=10; 吨=3; 一种=零⁡(ñ,吨); 乙=零⁡(ñ,吨);

## 有限元方法代写

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## MATLAB代写

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