# Finding Middlemost row in a matrix

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mr mo on 6 Nov 2017
Commented: Jan on 9 Nov 2017
Hi. Suppose I have a (m*n) matrix A e.g.
A=[8.9505 4.8075 1.6187 0.0020
8.9755 4.7575 1.6187 0.0020
8.9755 4.7825 1.6187 0.0020
8.9755 4.8075 1.6187 0.0020
8.9755 4.8325 1.6187 0.0020
8.9755 4.8575 1.6187 0.0020
9.0005 4.7325 1.6187 0.0020
9.0005 4.7575 1.6187 0.0020
9.0005 4.7825 1.6187 0.0020
9.0005 4.8075 1.6187 0.0020
9.0005 4.8325 1.6187 0.0020
9.0005 4.8575 1.6187 0.0020
9.0255 4.7325 1.6187 0.0020
9.0255 4.7575 1.6187 0.0020
9.0255 4.7825 1.6187 0.0020
9.0255 4.8075 1.6187 0.0020
9.0255 4.8325 1.6187 0.0020
9.0255 4.8575 1.6187 0.0020
9.0505 4.7325 1.6187 0.0020
9.0505 4.7575 1.6187 0.0020
9.0505 4.7825 1.6187 0.0020
9.0505 4.8075 1.6187 0.0020
9.0505 4.8325 1.6187 0.0020
9.0755 4.7575 1.6187 0.0020
9.1255 4.6075 1.6187 0.0020
9.1505 4.5825 1.6187 0.0020
9.1505 4.6075 1.6187 0.0020
9.1755 4.5825 1.6187 0.0020
9.2005 4.5575 1.6187 0.0020];
Imagine that the first column is X coordinates and second column is Y coordinates of some points.
In the matrix A, the first column values varied form minimum value 8.9505 to maximum value 9.2005, and the second column values varied form minimum value 4.5575 to maximum value 4.8575
The middle point of column 1 is :
(8.9505 + 9.2005)/2 = 9.0755
and the middle point of column 2 is:
(4.5575 + 4.8575)/2 = 4.7075
I want to find which one of the rows of matrix A has the nearest values to these values in its first and second columns respectively. Thanks
mr mo on 6 Nov 2017
for example this row may be the answer
A(19,:) = 9.0505 4.7325 1.6187 0.0020
because the 9.0505 is near to 9.0755 where is the midpoint of first column of A, and 4.7325 is near to 4.7075 where is the midpoint of second column of A.

Jan on 6 Nov 2017
Edited: Jan on 6 Nov 2017
Perhaps - a bold guess:
meanValue = (min(A(:, 1:2)) + max(A(:, 1:2))) / 2;
dist = sum((A(:, 1:2) - meanValue).^2, 2); % >= 2016b: Auto-Expand
% With older Matlab versions:
% dist = sum(bsxfun(@minus, A(:, 1:2), meanValue).^2, 2)
[minDist, minIndex] = min(dist)
midA = A(minIndex, :)
% The center between the minimal and maximal points:
meanValue = [9.0755, 4.7075]
% (Squared) Euclidean distances of A(:, 1:2) to this point:
d = sum((A(:, 1:2) - meanValue).^2, 2)
% Minimal distance:
[minDist, minIndex] = min(d);
minIndex = 19
% Or do you want multiple outputs, if the minimal distance occurs
% multiple times?
??? minDist = min(d);
??? minIndex = find(d == minDist);
But perhaps you look for something else. Unfortunately you do not provide a mathematical definition of what you want, although you have been asked repeatedly.
Jan on 9 Nov 2017
The Euclidean distance is:
sqrt(sum(x.^2))
Then you need sqrt(minDist) to get the distance.

Image Analyst on 6 Nov 2017
Try mean() or median():
columnMedians = median(A, 1)
columnMeans = mean(A, 1)
depending on what "middlemost" means to you.
mr mo on 6 Nov 2017
Imagine that the first column is X coordinates and second column is Y coordinates of some points. The middle point of column 1 is :
(8.9505 + 9.2005)/2 = 9.0755
and the middle point of column 2 is:
(4.5575 + 4.8575)/2 = 4.7075
I want to find which one of the row or rows of matrix A has the nearest values to these values in the first and second column respectively. Thanks