MatLab function for each two values in struct inside loop
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Hello,
I have a struct with some fileds, (attached)
something like exp = {'Names', 'Date', 'X' , 'Y'}
Names = ['John', 'Adam', 'Markus',.....]
X = [ X1, X2, X3, X4,.....]
Y = [ Y1, Y2, Y3, Y4,.....]
I want to run an equation to calculate the sqrt for each two values depending on the same field (Date)
For Date (Date 1)
s12 = sqrt((X2-X1)^2+(Y2-Y1)^2);
then
s13= sqrt((X3-X1)^2+(Y3-Y1)^2);
and
s14= sqrt((X4-X1)^2+(Y4-Y1)^2);
and
s23 = sqrt((X3-X2)^2+(Y3-Y2)^2);
and
s24 = sqrt((X4-X2)^2+(Y4-Y2)^2);
and
s34 = sqrt((X4-X3)^2+(Y4-Y3)^2);
And the same for (Date2)
So I wrote a loop as follows:
for i= 1:length(example)
for j = i+1;
for u = 1:length(unique(example.dates,'rows'));
s(i) = sqrt((example(j).x-example(i).x)^2+(example(j).y-example(i).y)^2);
abr{i} = sprintf('%s-%s',strtrim(example(j).name),strtrim(example(i).name));
end
end
end
%%%%% Export results to an array (Date, Name i_Name j, S)
result = [];
for n = 1:length(unique(example.dates,'rows'));
result = [result; dates(n) abr(n) d(n)];
end
But I am sure there is something wrong because I should get s for all the pairs of the values and just for the same date each time.
The results should be something like that:
Please help me to figure out how to do that.
4 个评论
Stephen23
2019-3-29
"I have a struct with many fileds,"
"something like struct = {'Names', 'A' , 'B', 'C', ....}"
What you have shown is a cell array, not a structure. You need to learn the difference:
"But I am sure there is something wrong..."
Possibly, but because you have not provided a complete working example, nor any data that matches your description, it is hard for us to know exactly what you are doing or what you want.
It would help if you simply uploaded sample data in one .mat file by clicking the paperclip button, and giving a clear example of the expected output.
Mohammed Hammad
2019-3-29
Luna
2019-3-29
You have the same names and also same dates, are you sure that your name list will be unique for each date? Otherwise you will get a row like Ronny vs Ronny for example.
Mohammed Hammad
2019-3-29
编辑：Mohammed Hammad
2019-3-29
采纳的回答
load example.mat
C = {example.Name};
D = {example.dates};
M = [example.x;example.y];
[U,~,idu] = unique(D);
N = max(idu);
fprintf('%-12s %-10s %-10s %s\n','Date','Name1','Name2','Result')
for k = 1:N
idf = find(idu==k);
P = nchoosek(1:numel(idf),2);
V = sqrt(sum((M(:,idf(P(:,1)))-M(:,idf(P(:,2)))).^2,1));
T = [repmat(U(k),size(V));C(idf(P(:,1)));C(idf(P(:,2)));num2cell(V)];
fprintf('%-12s %-10s %-10s %#.5g\n',T{:});
end
Which prints this in the command window:
Date Name1 Name2 Result
02-Apr-2002 John Adam 4.8237e+06
02-Apr-2002 John Markus 6.7899e+06
02-Apr-2002 John Ronny 7.4859e+06
02-Apr-2002 John Michael 3.8794e+06
02-Apr-2002 Adam Markus 5.9770e+06
02-Apr-2002 Adam Ronny 6.6450e+06
02-Apr-2002 Adam Michael 4.7198e+06
02-Apr-2002 Markus Ronny 7.3046e+05
02-Apr-2002 Markus Michael 2.9522e+06
02-Apr-2002 Ronny Michael 3.6236e+06
03-Apr-2002 Ronny Markus 5.9770e+06
03-Apr-2002 Ronny Michael 6.6450e+06
03-Apr-2002 Ronny John 4.8237e+06
03-Apr-2002 Ronny Adam 4.7198e+06
03-Apr-2002 Ronny Romio 5.8499e+05
03-Apr-2002 Ronny Sebastian 7.4353e+06
03-Apr-2002 Markus Michael 7.3046e+05
03-Apr-2002 Markus John 6.7899e+06
03-Apr-2002 Markus Adam 2.9522e+06
03-Apr-2002 Markus Romio 6.1503e+06
03-Apr-2002 Markus Sebastian 1.0072e+07
03-Apr-2002 Michael John 7.4859e+06
03-Apr-2002 Michael Adam 3.6236e+06
03-Apr-2002 Michael Romio 6.8414e+06
03-Apr-2002 Michael Sebastian 1.0753e+07
03-Apr-2002 John Adam 3.8794e+06
03-Apr-2002 John Romio 4.3132e+06
03-Apr-2002 John Sebastian 3.3172e+06
03-Apr-2002 Adam Romio 4.6077e+06
03-Apr-2002 Adam Sebastian 7.1300e+06
03-Apr-2002 Romio Sebastian 6.8562e+06
>>
And checking the last one:
>> sqrt((example(11).x-example(12).x).^2 + (example(11).y-example(12).y).^2)
ans =
6.8562e+06
>> example(11:12).Name
ans =
Romio
ans =
Sebastian
>> example(11:12).dates
ans =
03-Apr-2002
ans =
03-Apr-2002
7 个评论
Luna
2019-3-29
+1 :)
Mohammed Hammad
2019-3-29
编辑：Mohammed Hammad
2019-3-29
"I changed"
idf = find(idu==N); %It seems to me correct now.
You just made that loop totally pointless. Lets have a look at why...
Each k value matches groups of the same value in idu, where each group in idu corresponds to one date U{k}. Thus in my code each loop iteration gets the group of values for one date.
In contrast your code change will repeatedly get the values for the last date U{N}.
"Can be the code like following (is it right ?):"
This line is correct:
M = [example.x;example.y;example.z]; % correct
It defines one matrix with three rows which correspond to x, y, and z, while its columns correspond to the elements of example. So far so good!
But the rest of your changes are not correct, and seem to be based on an incorrect understanding of the matrix P: its purpose is to select the pairs of names (it does not select between x, y, and z as you seem to think), so it should actually be defined just as I wrote it in my answer:
P = nchoosek(1:numel(idf),3); % incorrect
P = nchoosek(1:numel(idf),2); % correct: combinations of TWO names.
The two columns of P provide all combinations of two names, without repetition. Its columns are used as indices to select the required names. The next line uses the columns of P to select the combinations of one date group (given as indices in idf), and uses those as date group indices to select the columns of M:
V = sqrt(sum((M(:,idf(P(:,1)))-M(:,idf(P(:,2)))-M(:,idf(P(:,3)))).^2,1)); % incorrect
V = sqrt(sum((M(:,idf(P(:,1)))-M(:,idf(P(:,2)))).^2,1)); % correct
% ^ colon! ^ colon! ^ sum down columns!
Remember that the rows of M correspond to x, y, and z: we do not need to select them individually, all of the rows are automatically selected using the colon indexing (which also means that my code will work for any number of dimensions, not just two or three). I specified the sum with the second "dimension" argument so that it sums along the first dimension, which automatically takes care of the x, y, z dimensions (or however many there may be).
So actually the only change you need to make in order to add a third dimension, is this line:
M = [example.x;example.y;example.z];
Mohammed Hammad
2019-3-30
"what is the best method (high performance) to save the results to one array (Date, Name1, Name2, V)?"
Preallocate a cell array before the loop, add the required data, then concatenate the contents after the loop. Something like this (untested, but should get you started):
Z = cell(1,N); % preallocate
for k = 1:N
...
Z{k} = T; % add data
end
Z = [Z{:}] % concatenate
"and if the equation changed to something else (any mathematical equation for the same x,y,z), does the structure of the code will be changed or just V!"
You would need to change the line that defines V:
V = ... new calculation here!
Just keep in mind that the calculation needs to accept a dims*columns matrix, and it needs to return a 1*columns row vector (where dims represent x, y, z, etc, and columns represents the number of instances of that date). Most likely you would write vectorized code:
Mohammed Hammad
2019-3-30
编辑：Mohammed Hammad
2019-3-30
1. This loop:
for i=1:length(theFiles)
eval(['load ' theFiles(i).name]);
array = cell2struct(data, Headings, 2);
end
serves absolutely no purpose whatsoever, because you do not use any indexing (or concatenation) to store the output array, and so you simply overwrite array except for the one defined on the last iteration. So every previous iteration is just a waste of time. Also that eval usage should be avoided, which you can do very simply:
load(theFiles(i).name)
or even better:
S = load(theFiles(i).name);
array = cell2struct(S.data, Headings, 2);
2. You never use the variable array for anything anyway: once you have imported that data and created array you just ignore it. Seems like a waste of time to me!
3. For some reason you are still accessing a variable named example, although this is not defined anywhere in your code.
In any case that error message is very easy to understand: you will get that error whenever numel(idf)<2, i.e. there is only one name in the entire date group. If there is only one name (for a particular date) then clearly it is not possible to generate all combinations of two names from it! You will need to decide what to do in those cases (I cannot decide how your algorithm should work), but most likely you will need to add an IF condition, something like this:
Z = cell(1,N); % preallocate. Why did you remove this??? Very bad idea!
for k = 1:N
idf = find(idu==N);
if numel(idf)>1
... your code
Z{k} = T; % add data
else
... decide what you want to do with zero/one names.
end
end
Z = [Z{:}] % concatenate
Depending on how you decide to handle the one-name groups, you might need to use indexing for the concatenation as well.
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