[英]How to determine the mean of the last e.g. 100 non-zero numbers in each column of a matrix in matlab
I would like to calculate the mean of the last eg 3 non-zero numbers in each column of a matrix in matlab. 我想在matlab中计算矩阵每列中最后三个非零数的平均值。 The columns were filled with zeros at the end to create vectors of the same length. 在末尾用列填充列以创建相同长度的向量。
example matrix: 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]
There may be a more efficient solution but one way is to use sum
to find the number of non-zero rows in a given column. 可能有一个更有效的解决方案,但一种方法是使用sum
来查找给定列中的非零行数。 Then grab average the values of A by looping through all columns with arrayfun
and averaging the N
rows before the zero in the column. 然后通过使用arrayfun
循环遍历所有列并在列中的零之前平均N
行来获取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));
Example 例
For your example data this would yield: 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
An alternate approach would be to use convolution to actually solve this for you. 另一种方法是使用卷积来实际为您解决此问题。 You can compute a mean using a convolution kernel. 您可以使用卷积内核计算均值。 If you want the mean of all 3-row combinations of an matrix, your kernel would be: 如果你想要矩阵的所有3行组合的平均值,你的内核将是:
kernel = [1; 1; 1] ./ 3;
When convolved with the matrix of interest, this will compute the average of all 3-row combinations within the input matrix. 当与感兴趣的矩阵卷积时,这将计算输入矩阵内所有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
In the example below, I do this and then only return the values at the regions we care about (where the average is only composed of the last N
non-zeros in each column) 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N
非零组成)
%// 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);
And the output, again, is the same 而输出又是一样的
3.0000 3.0000 2.0000 2.0000 2.0000 3.0000 3.6667
I ran a quick test script to compare the performance between these two methods. 我运行了一个快速测试脚本来比较这两种方法之间的性能。 It is clear that the convolution-based approach is much faster. 很明显,基于卷积的方法要快得多。
Here is the full test script. 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
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