# Compare values in two matrix with different length by time in first row

조회 수: 2(최근 30일)
Nik Rocky 2021년 2월 11일
댓글: Nik Rocky 2021년 2월 17일
Hello,
I have one aim to do comparison of values in two matrices without using for loops (array are huge).
I show you example with two small arrays:
Matrix_R =
9 x 2 double
[time value]
0.020 77
0.030 98
0.040 79
0.070 95
0.080 96
0.090 99
0.100 100
0.100 200
0.120 44
Matrix_G =
13 x 2 double
[time value]
0.010 75
0.020 75
0.030 75
0.040 77
0.050 78
0.060 96
0.070 99
0.080 100
0.090 100
0.100 110
0.110 100
0.120 100
0.130 110
Matrix_R (measure) has not homogenous time values
Matrix_G (ground truth) has homogenous time values
I want compare values between two matrices (with tolerance) from second column by same time. Comparison should be work in both directions.
Example (value tolerance 10):
at time 0.010 in Matrix_G is value 75, in Matrix_R there are no value -> False negative
at time 0.020 in Matrix_G is value 75, in Matrix_R is value 77 -> True positive
at time 0.030 in Matrix_G is value 75, in Matrix_R is value 98 -> False negative and False positive
Question:
• Can I use ismembertool by matrix with different lengths?
• Maybe it makes sense to change time rows of Matrix_R to homogenous and take 0 like values?
Matrix_R_mod =
9 x 2 double
[time value]
0.010 0
0.020 77
0.030 98
0.040 79
0.050 0
0.060 0
0.070 95
0.080 96
0.090 99
0.100 100
0.100 200
0.110 0
0.120 44
what can I make with double answers of time?
I will be happy to hear some ideas from you! Thank you!

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### 채택된 답변

Abhishek Gupta 2021년 2월 15일
Hi,
You can perform the task mentioned above as follows: -
clc; clear;
%% Input variables
Matrix_R = [0.020 77; 0.030 98; 0.040 79; 0.070 95; 0.080 96; ...
0.090 99; 0.100 100; 0.100 200; 0.120 44];
Matrix_G = [0.010 75; 0.020 75; 0.030 75; 0.040 77; 0.050 78; 0.060 96; ...
0.070 99; 0.080 100; 0.090 100; 0.100 110; 0.110 100; 0.120 100; 0.130 110];
tol = 10;
%% Initialize output matrix, where
% column 1 -- all the elements in the Matrix_G with appropriate repetition
% colum 2 -- output column, where -1 -> false negative; 1 -> true positive; 0 -> False negative and False positive
freq = [Matrix_G(:,1), histc(Matrix_R(:,1),Matrix_G(:,1))]; % get frequncy of each element of Matrix_G in Matrix_R
repetitionNumber = freq(freq(:,2) > 1, 1); % get the element which repeats more than once
repetitionTimes = freq(freq(:,2) > 1, 2); % get number of time element repeats more than once
outputTime = sort([freq(:,1); repelem(repetitionNumber, repetitionTimes - 1)]); % sort w.r.t time
output = [outputTime -1*ones(size(outputTime,1), 1)]; % initialize output matrix with all time rows == -1
%%
% get common elements between Matrix_R and Matrix_G
[Lia,Locb] = ismember(Matrix_R(:,1), Matrix_G(:,1));
A = Matrix_R(Lia, 2);
B = Matrix_G(Locb, 2);
toleranceCheck = abs(A-B) <= 10; % find elements which are within the tolerance
output(ismember(output(:,1), Matrix_R(:,1)), 2) = toleranceCheck;
##### 댓글 수: 1표시숨기기 없음
Nik Rocky 2021년 2월 17일
Wow, thank you very much! Is a great solution of my case!

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