Grouping and averaging values within a matrix

Question

JIAN-HONG YE ZHU 2022-12-5

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1871582-grouping-and-averaging-values-within-a-matrix

评论： JIAN-HONG YE ZHU 2022-12-5

Hello, I have a matrix 305109x3 and I want to reduce its size by averaging similar values within the matrix.

The matrix data varies from -1.6 to 1.3 on column one and I want to for example take all the values close to -1.6 on column one and average them.

This is a section of my matrix, column 1 2 3 correspond to x y z data points:

-1.60120522975922 -0.522726416587830 1.67500007152557

-1.60120522975922 -0.518805325031281 1.67500007152557

-1.59728419780731 -0.522726416587830 1.67500007152557

-1.59650182723999 -0.551117777824402 1.66600012779236

-1.59650182723999 -0.547217786312103 1.66600012779236

-1.59650182723999 -0.543317794799805 1.66600012779236

-1.59650182723999 -0.539417743682861 1.66600012779236

-1.59650182723999 -0.535517752170563 1.66600012779236

-1.59650182723999 -0.508217751979828 1.66600012779236

-1.59650182723999 -0.504317700862885 1.66600012779236

-1.59642553329468 -0.564169049263001 1.67000007629395

-1.59642553329468 -0.560259699821472 1.67000007629395

-1.59642553329468 -0.528984725475311 1.67000007629395

-1.59642553329468 -0.525075435638428 1.67000007629395

-1.59642553329468 -0.513347327709198 1.67000007629395

-1.59336316585541 -0.522726416587830 1.67500007152557

-1.59260189533234 -0.566717803478241 1.66600012779236

-1.59260189533234 -0.555017769336700 1.66600012779236

-1.59260189533234 -0.551117777824402 1.66600012779236

-1.59260189533234 -0.547217786312103 1.66600012779236

-1.59260189533234 -0.539417743682861 1.66600012779236

-1.59260189533234 -0.531617760658264 1.66600012779236

-1.59260189533234 -0.508217751979828 1.66600012779236

-1.59260189533234 -0.504317700862885 1.66600012779236

-1.59251618385315 -0.525075435638428 1.67000007629395

-1.59251618385315 -0.517256677150726 1.67000007629395

-1.59251618385315 -0.513347327709198 1.67000007629395

So in this case I would like to average all the values close to -1.6 and all the values close to -1.59.

So instead of having the above 27x3 matrix, I would have a 2x3 matrix.

For reference the 2x3 matrix I would like after doing that would be:

-1.60121 -0.52077 1.675

-1.58829 -0.5472 1.678788

So, for the whole matrix, I would have -1.6, -1.59, -1.58 all up to 1.3 and want to be able to identify numbers close to those and average them, if that make sense?

2 个评论
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Dyuman Joshi 2022-12-5

How do you define close? What is the tolerance criteria? -1.6 +- 0.1?

"So in this case I would like to average all the values close to -1.6 and all the values close to -1.59."

Do you mean 1.3, instead of -1.59?

JIAN-HONG YE ZHU 2022-12-5

dataPoints.mat

The tolerance criteria is 0.01. My matrix contains values from -1.6 to 1.3 but the posted matrix only goes from -1.6 to -1.59 since there are too many data points.

So I would like all row values close to -1.6 to be averaged and give my one row containing that, same with -1.59, -1.58, -1.57, -1.56 and so on.

I included the variable since I might not be explaning myself right.

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Answer 1

Kevin Holly 2022-12-5

1
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https://ww2.mathworks.cn/matlabcentral/answers/1871582-grouping-and-averaging-values-within-a-matrix#answer_1120552

在 MATLAB Online 中打开

m = [-1.60120522975922 -0.522726416587830 1.67500007152557
-1.60120522975922 -0.518805325031281 1.67500007152557
-1.59728419780731 -0.522726416587830 1.67500007152557
-1.59650182723999 -0.551117777824402 1.66600012779236
-1.59650182723999 -0.547217786312103 1.66600012779236
-1.59650182723999 -0.543317794799805 1.66600012779236
-1.59650182723999 -0.539417743682861 1.66600012779236
-1.59650182723999 -0.535517752170563 1.66600012779236
-1.59650182723999 -0.508217751979828 1.66600012779236
-1.59650182723999 -0.504317700862885 1.66600012779236
-1.59642553329468 -0.564169049263001 1.67000007629395
-1.59642553329468 -0.560259699821472 1.67000007629395
-1.59642553329468 -0.528984725475311 1.67000007629395
-1.59642553329468 -0.525075435638428 1.67000007629395
-1.59642553329468 -0.513347327709198 1.67000007629395
-1.59336316585541 -0.522726416587830 1.67500007152557
-1.59260189533234 -0.566717803478241 1.66600012779236
-1.59260189533234 -0.555017769336700 1.66600012779236
-1.59260189533234 -0.551117777824402 1.66600012779236
-1.59260189533234 -0.547217786312103 1.66600012779236
-1.59260189533234 -0.539417743682861 1.66600012779236
-1.59260189533234 -0.531617760658264 1.66600012779236
-1.59260189533234 -0.508217751979828 1.66600012779236
-1.59260189533234 -0.504317700862885 1.66600012779236
-1.59251618385315 -0.525075435638428 1.67000007629395
-1.59251618385315 -0.517256677150726 1.67000007629395
-1.59251618385315 -0.513347327709198 1.67000007629395];
m(:,1) = round(m(:,1),2);
m
m = 27×3
   -1.6000   -0.5227    1.6750
   -1.6000   -0.5188    1.6750
   -1.6000   -0.5227    1.6750
   -1.6000   -0.5511    1.6660
   -1.6000   -0.5472    1.6660
   -1.6000   -0.5433    1.6660
   -1.6000   -0.5394    1.6660
   -1.6000   -0.5355    1.6660
   -1.6000   -0.5082    1.6660
   -1.6000   -0.5043    1.6660
unique_values = unique(m(:,1))
unique_values = 2×1
   -1.6000
   -1.5900
for ii = 1:length(unique_values)
    logical_array = m(:,1) == unique_values(ii);
    new(ii,:) = mean(m(logical_array,:));
end
new
new = 2×3
   -1.6000   -0.5323    1.6691
   -1.5900   -0.5318    1.6678