我想计算最后一个的平均值,例如matlab中矩阵的每一列中的3个非零数字。这些列的末尾用零填充以创建相同长度的向量。
示例矩阵:
A = [5 6 3 5 6 8 9;
1 2 3 5 4 7 6;
0 1 2 3 4 5 6;
0 0 1 2 3 4 5;
0 0 0 1 2 3 4;
0 0 0 0 2 3 4;
0 0 0 0 2 3 4;
0 0 0 0 0 0 3]
最佳答案
可能有更有效的解决方案,但一种方法是使用 sum
来查找给定列中非零行的数量。然后通过使用 arrayfun
遍历所有列并对列中零之前的 N
行进行平均来获取 A 的平均值。
%// Number of elements to average
N = 3;
%// Last non-zero row in each column
lastrow = sum(A ~= 0, 1);
%// Ensure that we don't have any indices less than 1
startrow = max(lastrow - N + 1, 1);
%// Compute the mean for each column using the specified rows
means = arrayfun(@(k)mean(A(startrow(k):lastrow(k),k)), 1:size(A, 2));
示例
对于您的示例数据,这将产生:
3.0000 3.0000 2.0000 2.0000 2.0000 3.0000 3.6667
更新:另一种选择
另一种方法是使用卷积来实际为您解决这个问题。您可以使用卷积核计算均值。如果您想要一个矩阵的所有 3 行组合的平均值,您的内核将是:
kernel = [1; 1; 1] ./ 3;
当与感兴趣的矩阵进行卷积时,这将计算输入矩阵中所有 3 行组合的平均值。
B = [1 2 3;
4 5 6;
7 8 9];
conv2(B, kernel)
0.3333 0.6667 1.0000
1.6667 2.3333 3.0000
4.0000 5.0000 6.0000
3.6667 4.3333 5.0000
2.3333 2.6667 3.0000
在下面的示例中,我执行此操作然后仅返回我们关心的区域的值(其中平均值仅由每列中的最后一个 N
非零值组成)
%// Find the last non-zero entry in each column
lastrow = sum(A ~= 0, 1);
%// Use convolution to compute the mean for every N rows
%// This will be applied to ALL of A
convmean = conv2(A, ones(N, 1)./N);
%// Select only the means that we care about
%// Because of the padding of CONV2, these will live at the rows
%// stored in LASTROW
means = convmean(sub2ind(size(convmean), lastrow, 1:size(A, 2)));
%// Now correct for cases where fewer than N samples were averaged
means = (means * N) ./ min(lastrow, N);
同样,输出是一样的
3.0000 3.0000 2.0000 2.0000 2.0000 3.0000 3.6667
比较
我运行了一个快速测试脚本来比较这两种方法的性能。很明显,基于卷积的方法要快得多。
这是完整的测试脚本。
function benchmark()
dims = round(linspace(1, 1000, 100));
times1 = zeros(size(dims));
times2 = zeros(size(dims));
N = 3;
for k = 1:numel(dims)
A = triu(rand(dims(k)));
times1(k) = timeit(@()test_arrayfun(N, A));
A = triu(rand(dims(k)));
times2(k) = timeit(@()test_convolution(N, A));
end
figure;
plot(dims, times1);
hold on
plot(dims, times2);
legend({'arrayfun', 'convolution'})
xlabel('Dimension of A')
ylabel('Execution Time (seconds)')
end
function test_arrayfun(N, A)
%// Last non-zero row in each column
lastrow = sum(A ~= 0, 1);
%// Ensure that we don't have any indices less than 1
startrow = max(lastrow - N + 1, 1);
%// Compute the mean for each column using the specified rows
means = arrayfun(@(k)mean(A(startrow(k):lastrow(k),k)), 1:size(A, 2));
end
function test_convolution(N, A)
%// Find the last non-zero entry in each column
lastrow = sum(A ~= 0, 1);
%// Use convolution to compute the mean for every N rows
%// This will be applied to ALL of A
convmean = conv2(A, ones(N, 1)./N);
%// Select only the means that we care about
%// Because of the padding of CONV2, these will live at the rows
%// stored in LASTROW
means = convmean(sub2ind(size(convmean), lastrow, 1:size(A, 2)));
%// Now correct for cases where fewer than N samples were averaged
means = (means * N) ./ min(lastrow, N);
end
关于matlab - 如何确定最后一个的平均值matlab矩阵每列100个非零数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36288101/