Question

假设我有一个维度为P的位置矩阵10x2，其中第一列包含x值，第二列包含相应的y值。我想要平均长度的平均值。到目前为止，我的方法是使用以下代码：

avg = sum( sqrt( P(:,1).^2 + P(:,2).^2))/10);

我被告知，积分函数integral2对于此任务来说更快更精确。如何使用integral2计算平均值？

Answer 1

就是这样，这个问题仍然没有答案：

function q42372466(DO_SUM)
if ~nargin % nargin == 0
  DO_SUM = true;
end

% Generate some data:
P = rand(2E7,2);

% Correctness:
R{1} = m1(P);
R{2} = m2(P);
R{3} = m3(P);
R{4} = m4(P);
R{5} = m5(P);
R{6} = m6(P);

for ind1 = 2:numel(R)
  assert(abs(R{1}-R{ind1}) < 1E-10);
end

% Benchmark:
t(1) = timeit(@()m1(P));
t(2) = timeit(@()m2(P));
t(3) = timeit(@()m3(P));
t(4) = timeit(@()m4(P));
t(5) = timeit(@()m5(P));
t(6) = timeit(@()m6(P));

% Print timings:
disp(t);


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Original method:
function out = m1(P)
  if DO_SUM
    out = sum( sqrt( P(:,1).^2 + P(:,2).^2))/max(size(P));
  else
    out = mean( sqrt( P(:,1).^2 + P(:,2).^2));
  end
end

% pdist2 method:
function out = m2(P)
  if DO_SUM
    out = sum(pdist2([0,0],P))/max(size(P));
  else
    out = mean(pdist2([0,0],P));
  end
end

% Shortened method #1:
function out = m3(P)
  if DO_SUM  
    out = sum(sqrt(sum(P.*P,2)))/max(size(P));
  else
    out = mean(sqrt(sum(P.*P,2)));
  end    
end

% Shortened method #2:
function out = m4(P)
  if DO_SUM  
    out = sum(sqrt(sum(P.^2,2)))/max(size(P));
  else
    out = mean(sqrt(sum(P.^2,2)));
  end    
end

% hypot
function out = m5(P)
  if DO_SUM
    out = sum(hypot(P(:,1),P(:,2)))/max(size(P));
  else
    out = mean(hypot(P(:,1),P(:,2)));
  end
end

% (a+b)^2 formula , Divakar's idea
function out = m6(P)
  % Since a^2 + b^2 = (a+b)^2 - 2ab, 
  if DO_SUM
    out = sum(sqrt(sum(P,2).^2 - 2*prod(P,2)))/max(size(P));
  else
    out = mean(sqrt(sum(P,2).^2 - 2*prod(P,2)));
  end
end

end

我的R2016b + Win10 x64的典型结果：

>> q42372466(0) % with mean()
    0.1165    0.1971    0.2167    0.2161    0.1719    0.2375

>> q42372466(1) % with sum()
    0.1156    0.1979    0.2233    0.2181    0.1610    0.2357

这意味着您的方法实际上是上述最好的方法，相当可观！
（老实说 - 没想到！）

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1 个答案: