mapping of matrices with different size

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Rani V.S
Rani V.S 2015 年 1 月 21 日
編集済み: Stephen23 2015 年 1 月 23 日
i have to map the values from a matrix to another one based on the value in the first column: i used the following code
data = [0 1;0.1 2;0.2 3;0.3 4; 0.6 5 ;0.7 6;0.8 7;1 8]
[m,n]=size(data);
StopTime=1;
SampleTime=0.1;
TimeStep=(StopTime/SampleTime)+1;
initial=0
time=zeros(TimeStep,n)
for i=1:TimeStep
time(i,1)=initial;
initial=initial+SampleTime;
end
c=0;
for j=1:TimeStep
for i=1:m
if (data(i,1)==time(j,1))
time(j,2)=data(i,2);
end
end
end
the expected result is
[0 1;
0.1 2;
0.2 3;
0.3 4;
0.4 0;
0.5 0;
0.6 5;
0.7 6;
0.8 7;
0.9 0;
1 8]
but the obtained result is:
[0 1;
0.1 2;
0.2 3;
0.3 0;
0.4 0;
0.5 0;
0.6 5;
0.7 6;
0.8 0;
0.9 0;
1 0]
please hel p me to fix this issue
  1 件のコメント
Stephen23
Stephen23 2015 年 1 月 21 日
Note that you should not use i or j for the names of your loop variables, as these are the names of the inbuilt imaginary unit .

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採用された回答

Stephen23
Stephen23 2015 年 1 月 21 日
編集済み: Stephen23 2015 年 1 月 23 日
There are multiple issues with your code:
  • Using equivalence logical tests on floating point values. It is important to realize that what looks like that same number may have a slightly different binary value, and so logical tests can give erroneous results. This often happens when you calculate the two values in different ways: for example 0.1 does not equal 0.01/0.1 (try it!).
  • Using names of inbuilt functions or values as variable names: i and j.
  • Using a loop to create a vector of values, by adding one fractional value to the last at each iteration. Doing this compounds the precision errors resulting from the binary approximation of the fractional values that you believe you are adding. It is very poor coding (in any language, not just MATLAB).
The nested for loops are not a good solution to this problem: using indexing would be much neater and faster. Learn to use MATLAB indexing and inbuilt functions such as linspace for generating a vector of equally spaced fractional values. Here is one possibility that matches your desired output:
>> data = [0,1; 0.1,2; 0.2,3; 0.3,4; 0.6,5; 0.7,6; 0.8,7; 1,8];
>> time = linspace(0,1,11);
>> idx = any(1e-10 > abs(bsxfun(@minus,time,data(:,1))), 1);
>> time(2,idx) = data(:,2);
>> disp(time.')
0.0 1
0.1 2
0.2 3
0.3 4
0.4 0
0.5 0
0.6 5
0.7 6
0.8 7
0.9 0
1.0 8
Notice how I do not test for equivalency, but instead check that the difference between the values is greater than some small value (in this example, 1e-10).
  1 件のコメント
Rani V.S
Rani V.S 2015 年 1 月 22 日
Thanks for helping me. Your code failed in the case of unsorted data. So i have sorted the data according to the first column using
sortrows(data,1)
After this modification the Script handles all the data combinations..

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その他の回答 (1 件)

Thorsten
Thorsten 2015 年 1 月 21 日
Stephen made some remarks that are useful in general, but in your case you can get along with
newdata(:,1) = 0:0.1:1; % first column
newdata(10*data(:,1)+1, 2) = data(:,2); % second column
The idea is to use the first column of data as an index. Because the values run from 0 to 1 in steps of 0.1, you can convert it to a valid Matlab index by multiplication with 10 and adding 1 (Matlab's indices always start with 1). Missing values in the index (such as 4, 5 and 9) are automatically set to zero by Matlab.
  2 件のコメント
Rani V.S
Rani V.S 2015 年 1 月 22 日
this method works ony for the samle time of 0.1. if the sample time is different it will fail

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