# Finding three columns in one variable in another variable

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JLV on 12 Jan 2020
Commented: JLV on 30 Jan 2020 at 15:04
Dear MATLAB Community,
I was wondering if you were able to help me with the following problem. I am trying to find the following coordinates (from Columns 2-4 in CourseMeshNodeLocs) in FibreOrientationPositionElement (Columns 8-10). Columns 2-4 wont exactly match Columns 8-10, hence some of of the logic functions I have learnt over the past few weeks will not help me.
I want to find a row in FibreOrientationPositionElement (Columns 8-10) that closly represents a row in CourseMeshNodeLocs (Columns 2-4), either through averaging or finding the closest one. Once the closest co-ordinates can be found, I then want to output columns 2-7 in FibreOrientationPositionElement for each row in CourseMeshNodeLocs.
What would be the best way of doing this? I had an idea regarding if and for loops defining a range which the co-ordinates need to be in, but I believe there may be another simpler way, which I can learn for the future.
Attached are some files and a code.
CoarseMeshNodeLocs = dlmread('CoarseTjointMesh.txt','',9,0); %Fibre Orientation @ Elements. Read text file, first few lines not needed
CoarseMeshNodeLocs(:,[2 3 4]) = CoarseMeshNodeLocs(:,[2 3 4])*10^-3; %Converts mm into m
FibreOrientationPositionElement = dlmread('matrix.txt','',0,0); %Fibre Orientation and Position data at elements

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JLV on 12 Jan 2020
Yes, the smallest difference in absolute position
Mohammad Sami on 16 Jan 2020
tol = 0.001; % fine tune for your needs.
% Two values, u and v, are within tolerance if abs(u-v) <= tol*max(abs([A(:);B(:)])).
% see Matlab docs for more details
LIA = ismembertol(Col_2_4,Col_8_10,tol,'ByRows',true);
JLV on 26 Jan 2020
Interesting method, I will try this!

Stephen Cobeldick on 25 Jan 2020
Edited: Stephen Cobeldick on 25 Jan 2020
I tried a simple vectorized solution of permute and sum of squares approach, but ran into "out of memory" issues:
>> C = dlmread('CoarseTjointMesh.txt','',9,0); %Fibre Orientation @ Elements. Read text file, first few lines not needed
>> C(:,2:4) = C(:,2:4)*1e-3; % Convert mm into m
>> F = dlmread('matrix.txt','',0,0); %Fibre Orientation and Position data at elements
>> B = 3*8*size(C,1)*size(F,1)
B = 13703964384
>> num2sip(B) % Oops... I don't have 14 GBytes of memory
ans = 13.704 G
However using a cell array worked without error, and is just as simple:
% split into row vectors in cell array:
tmp = num2cell(C(:,2:4),2);
% find indices of closest set of points:
fun = @(v)min(sum((v-F(:,8:10)).^2,2)); % requires >=R2016b
[dst,idx] = cellfun(fun,tmp);
% get output rows selected by those indices:
out = F(idx,2:7);
Note that the subtraction inside the anonymous function requires scalar dimension expansion, which was introduced in R2016b. For earlier versions replace the subtraction with bsxfun.

JLV on 26 Jan 2020
I have tried going through this and it works, thank you!
A question, I understand you are trying the find the length of each co-ordinate (from the origin), and trying to find the smallest distance between them. Whichever value gives you the smallest difference is the index taken going forward. What if there are two coordinates on opposite sides of a quadrant, this means they have the same length from the origin. How does the script distinguish between this?
Stephen Cobeldick on 26 Jan 2020
@JLV: the calculation is for the (square of the) euclidean distance in 3 dimensions:
The calculation does not involve any "length" from the origin as it directly calculates the distance between any two points (where those points are relative to the origin is irrelevant).
JLV on 30 Jan 2020 at 15:04
Thank you for the clarification.

Raunak Gupta on 22 Jan 2020
Hi,
You may try finding difference between two matrices (here m x 3 matrices as m rows and 3 columns). From that you may try finding the max of the absolute of difference for each rows so it will result into a m x 1 vector. In that the minimum value will define the closest two rows in the two set of columns. You may use something like 10 times(This needs to be tuned according to what is required) of this lowest value and then find corresponding “similar” rows by using ismembertol. You may find below code useful.
% Here since the size is different for FibreOrientationPositionElement
% taking only the Number of elements that are there in CoarseMeshNodeLocs
FibreOrientationPosition_col_8_10 = FibreOrientationPositionElement(1:2393,8:10);
CoarseMeshNodeLocs_col_2_4 = CoarseMeshNodeLocs(:,2:4);
% Absolute difference between two matrices
difference = abs(FibreOrientationPosition_col_8_10 - CoarseMeshNodeLocs_col_2_4);
minimum_tol = min(max(difference,[],2));
% Here the maximum_tol can be tuned
maximum_tol = minimum_tol*10;
% Returns logical array of the similar rows
similarRowsIndex = ismembertol(FibreOrientationPositionElement(:,8:10),CoarseMeshNodeLocs(:,2:4),maximum_tol,'ByRows',true);
% Column 2-7 Values of FibreOrientationPositionElement for similar rows
similarRowsProperties = FibreOrientationPositionElement(similarRowsIndex,2:7);
Hope this helps.

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JLV on 26 Jan 2020
Thank you for the suggestion,
I have taken a look at the method and my biggest issues is
FibreOrientationPosition_col_8_10 = FibreOrientationPositionElement(1:2393,8:10);
This is because you can't simply ignore the all of the other datapoints in as they may contain relevent information required.
The two different inputs aren't necessarily in the same row to, and I believe the solution given by Stephen represents this problem accurately.
Please feel free to correct me!
Raunak Gupta on 26 Jan 2020
Hi, I also see Stephen's Method more comphrensive. I just thought in the beginning that computations will be a lot if try to rigoursly finding difference between two matrices. Anyways I guess this method will provide much exact answer.
JLV on 26 Jan 2020
Yes it is difficult to gauge the problem through text