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help me please. i always get a subscript indices must either be real positive integers or logical error whenever i put 0 value on my "data" how can i get rid of it, i need to have a zero on that one. whenever there is a zero Voltage(1,0) = 1. but I can't get through.
Voltage = [0 1 1 3 4 1; 1 0 5 4 5 3; 6 4 0 4 5 7; 9 3 4 0 6 4; 7 8 5 6 0 7; 4 5 6 7 3 0];
data =[0 2 3 4; 5 6 7 8; 2 3 4 5; 4 5 6 7; 3 4 5 6; 1 3 5 7; 1 2 3 4; 3 4 5 6];
Vm = data(:,1);
Vn = data(:,2);
R = data(:,3);``
X1 = data(:,4);
sz=max(Vn)
y=1:sz
for Vm=data(:,1)
if Vm==0
Voltage(y,Vm)=1
Voltage(y,Vm)=logical(Voltage(y,Vm));
Current = Voltage(y,Vm)-Voltage(y,Vn);
else Vm >= 1
Current = Voltage(y,Vm)-Voltage(y,Vn);
end
end
You are trying to reference a value in the else statement in the Voltage matrix using y but y is not an integer it is an array (or 1d matrix). If you display y you will see that it is 1 2 3 4 5 6. There are several sections of offending code, one of which is:
else Vm >= 1
disp(y) # `y` is not an integer and therefore not a valid index.
Current = Voltage(y,Vm)-Voltage(y,Vn);
To fix it, decide if y should be static or change in the loop.
Let me know if you want a further explanation.
I am trying to match 1st column of A with 1st to 3rd columns of B, and append corresponding 4th column of B to A.
For example,
A=
1 2
3 4
B=
1 2 4 5 4
1 2 3 5 3
1 1 1 1 2
3 4 5 6 5
I compare A(:,1) and B(:, 1:3)
1 and 3 are in A(:,1)
1 is in the 1st, 2nd, 3rd rows of B(:, 1:3), so append B([1 2 3], 4:end)' to A's 1st row.
3 is in the 2nd and 4th rows of B(:,1:3), so append B([2 4], 4:end)' to A's 2nd row.
So that it becomes:
1 2 5 4 5 3 1 2
3 4 5 3 6 5 0 0
I could code this using only for and if.
clearvars AA A B mem mem2 mem3
A = [1 2 ; 3 4]
B = [1 2 4 5 4; 1 2 3 5 3; 1 1 1 1 2; 3 4 5 6 5]
for n=1:1:size(A,1)
mem = ismember(B(:,[1:3]), A(n,1));
mem2 = mem(:,1) + mem(:,2) + mem(:,3);
mem3 = find(mem2>0);
AA{n,:} = horzcat( A(n,:), reshape(B(mem3,[4,5])',1,[]) ); %'
end
maxLength = max(cellfun(#(x)numel(x),AA));
out = cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),AA,'UniformOutput',false))
I am trying to make this code efficient, by not using for and if, but couldn't find an answer.
Try this
a = A(:,1);
b = B(:,1:3);
z = size(b);
b = repmat(b,[1,1,numel(a)]);
ab = repmat(permute(a,[2,3,1]),z);
row2 = mat2cell(permute(sum(ab==b,2),[3,1,2]),ones(1,numel(a)));
AA = cellfun(#(x)(reshape(B(x>0,4:end)',1,numel(B(x>0,4:end)))),row2,'UniformOutput',0);
maxLength = max(cellfun(#(x)numel(x),AA));
out = cat(2,A,cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),AA,'UniformOutput',false)))
UPDATE Below code runs in almost same time as the iterative code
a = A(:,1);
b = B(:,1:3);
z = size(b);
b = repmat(b,[1,1,numel(a)]);
ab = repmat(permute(a,[2,3,1]),z);
df = permute(sum(ab==b,2),[3,1,2])';
AA = arrayfun(#(x)(B(df(:,x)>0,4:end)),1:size(df,2),'UniformOutput',0);
AA = arrayfun(#(x)(reshape(AA{1,x}',1,numel(AA{1,x}))),1:size(AA,2),'UniformOutput',0);
maxLength = max(arrayfun(#(x)(numel(AA{1,x})),1:size(AA,2)));
out2 = cell2mat(arrayfun(#(x,i)((cat(2,A(i,:),AA{1,x},zeros(1,maxLength-length(AA{1,x}))))),1:numel(AA),1:size(A,1),'UniformOutput',0));
How about this:
%# example data
A = [1 2
3 4];
B = [1 2 4 5 4
1 2 3 5 3
1 1 1 1 2
3 4 5 6 5];
%# rename for clarity & reshape for algorithm's convenience
needle = permute(A(:,1), [2 3 1]);
haystack = B(:,1:3);
data = B(:,4:end).';
%# Get the relevant rows of 'haystack' for each entry in 'needle'
inds = any(bsxfun(#eq, haystack, needle), 2);
%# Create data that should be appended to A
%# All data and functionality in this loop is local and static, so speed
%# should be optimal.
append = zeros( size(A,1), numel(data) );
for ii = 1:size(inds,3)
newrow = data(:,inds(:,:,ii));
append(ii,1:numel(newrow)) = newrow(:);
end
%# Now append to A, stripping unneeded zeros
A = [A append(:, ~all(append==0,1))]
I have a list of numbers, [1:9], that I need to divide three groups. Each group must contain at least one number. I need to enumerate all of the combinations (i.e. order does not matter). Ideally, the output is a x by 3 array. Any ideas of how to do this in matlab?
Is this what you want:
x = 1:9;
n = length(x);
T=3;
out = {};
%// Loop over all possible solutions
for k=1:T^n
s = dec2base(k, T, n);
out{k}{T} = [];
for p=1:n
grpIndex = str2num(s(p))+1;
out{k}{grpIndex} = [out{k}{grpIndex} x(p)];
end
end
%// Print result. size of out is the number of ways to divide the input. out{k} contains 3 arrays with the values of x
out
Maybe this is what you want. I'm assuming that the division in groups is "monotonous", that is, first come the elements of the first group, then those of the second etc.
n = 9; %// how many numbers
k = 3; %// how many groups
b = nchoosek(1:n-1,k-1).'; %'// "breaking" points
c = diff([ zeros(1,size(b,2)); b; n*ones(1,size(b,2)) ]); %// result
Each column of c gives the sizes of the k groups:
c =
Columns 1 through 23
1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 4 5
1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 1
7 6 5 4 3 2 1 6 5 4 3 2 1 5 4 3 2 1 4 3 2 1 3
Columns 24 through 28
5 5 6 6 7
2 3 1 2 1
2 1 2 1 1
This produces what I was looking for. The function nchoosekr_rec() is shown below as well.
for x=1:7
numgroups(x)=x;
end
c=nchoosekr_rec(numgroups,modules);
i=1;
d=zeros(1,modules);
for x=1:length(c(:,1))
c(x,modules+1)=sum(c(x,1:modules));
if c(x,modules+1)==length(opt_mods)
d(i,:)=c(x,1:modules);
i=i+1;
end
end
numgroups=[];
for x=1:length(opt_mods)
numgroups(x)=x;
end
count=0;
for x=1:length(d(:,1))
combos=combnk(numgroups,d(x,1));
for y=1:length(combos(:,1))
for z=1:nchoosek(9-d(x,1),d(x,2))
new_mods{count+z,1}=combos(y,:);
numgroups_temp{count+z,1}=setdiff(numgroups,new_mods{count+z,1});
end
count=count+nchoosek(9-d(x,1),d(x,2));
end
end
count=0;
for x=1:length(d(:,1))
for y=1:nchoosek(9,d(x,1))
combos=combnk(numgroups_temp{count+1},d(x,2));
for z=1:length(combos(:,1))
new_mods{count+z,2}=combos(z,:);
new_mods{count+z,3}=setdiff(numgroups_temp{count+z,1},new_mods{count+z,2});
end
count=count+length(combos(:,1));
end
end
function y = nchoosekr_rec(v, n)
if n == 1
y = v;
else
v = v(:);
y = [];
m = length(v);
if m == 1
y = zeros(1, n);
y(:) = v;
else
for i = 1 : m
y_recr = nchoosekr_rec(v(i:end), n-1);
s_repl = zeros(size(y_recr, 1), 1);
s_repl(:) = v(i);
y = [ y ; s_repl, y_recr ];
end
end
end
I tried to solve this problem, but I could not implement.
Could you help me anything for this?
Problem
Mat1 | Mat2 | Mat3
1 2 | 1 3 | 2 6
1 3 | 2 6 | 2 5
2 4 | 3 1 | 3 1
3 1 | 3 5 | 5 2
4 5 |
When there are 3 matrices(for example above), I want to get this result for the intersection rows in [column1 column2 matrixnumber] form.
The result for above example would be
1 3 1
1 3 2
2 6 2
2 6 3
3 1 1
3 1 2
3 1 3
It would be OK if the result is in the form [column1 column2 firstmatrix secondmatrix, ...]
1 3 1 2
2 6 2 3
3 1 1 2 3
For this problem, I want to use at most one for-loop.
Do you have any idea for this?
Here an alternative solution (which seems to run faster than Gunther's) using MATLAB's intersect:
Mat = {[1 2; 1 3; 2 4; 3 1; 4 5],
[1 3; 2 6; 3 1; 3 5],
[2 6; 2 5; 3 1; 5 2]};
result = zeros(sum(cellfun(#(x)size(x, 1), Mat)), 3); % # Preallocate memory
k = 1;
for cc = transpose(nchoosek(1:numel(Mat), 2))
x = intersect(Mat{cc}, 'rows'); % # Find intersection
y = ones(size(x, 1), 2) * diag(cc); % # Generate matrix indices
result(k:k + numel(y) - 1, :) = [[x; x], y(:)];
k = k + numel(y);
end
result(all(~result, 2), :) = []; % # Discard zero rows
result = unique(result, 'rows'); % # Discard repeated rows
The matrix result should now contain the unique intersection rows and their corresponding matrix indices, just like you want:
result =
1 3 1
1 3 2
2 6 2
2 6 3
3 1 1
3 1 2
3 1 3
If I understand correctly, you have a number of sets of pairs: Mat1,Mat2, Mat3, ... MatN. Now you want to find the unique pairs and then find out in which set every unique pair appears.
If you have a large number of sets, I suggest you start using a cell array to hold them all, makes things a lot easier:
N = 3; % total number of data sets
Mat = cell(N,1);
Mat{1} = [1 2;
1 3;
2 4;
3 1;
4 5];
Mat{2} = [1 3;
2 6;
3 1;
3 5];
Mat{3} = [2 6;
2 5;
3 1;
5 2];
% etc.
First let's find the unique pairs:
uniq_pairs = unique(cat(1,Mat{:}),'rows');
M = size(uniq_pairs ,1);
Then use ismember to check which sets contain which pairs:
matcontpair = false(M,N); %preallocate
for ii=1:N % unavoidable loop
matcontpair(:,ii) = ismember(uniq_pairs,Mat{ii},'rows');
end
To translate this intersection matrix to a set of matrix numbers for each pair, loop through it again and store the final result in a cell array (you can't use an array, because they might not be of same size (some pairs only found once, other twice, other three times ...)
pair_occurence= cell(M,1);
d=1:N;
for jj=1:M
pair_occurence{jj} = d(matcontpair(jj,:));
end
Now you have a matrix uniq_pairs of size Mx2 containing the unique pairs, and a occurence cell array pair_occurence of size Mx1: each cell corresponds to a pair and contains a list of matrices where the pair is present.
If you want to remove pairs from the list which are only present in one matrix, use the following:
% find them
lonely_pairs = cellfun(#numel,pair_occurence)<2;
% and destroy them
uniq_pairs(lonely_pairs,:) = [];
pair_occurence(lonely_pairs) = [];
I have a matrix which has the following form:
M =
[1 4 56 1;
1 3 5 1;
1 3 6 4;
2 3 5 0;
2 0 0 0;
3 1 2 3;
3 3 3 3]
I want to split this matrix based on the number given in the first column. So I want to split the matrix into this:
A =
[1 4 56 1;
1 3 5 1;
1 3 6 4]
B =
[2 3 5 0;
2 0 0 0]
C =
[3 1 2 3;
3 3 3 3]
I tried this by making the following loop, but this gave me the desired matrices with rows of zeros:
for i = 1:length(M)
if (M(i,1) == 1)
A(i,:) = M(i,:);
elseif (M(i,1) == 2)
B(i,:) = M(i,:);
elseif (M(i,1) == 3)
C(i,:) = M(i,:);
end
end
The result for matrix C is then for example:
C =
[0 0 0 0;
0 0 0 0;
0 0 0 0;
2 3 5 0;
2 0 0 0]
How should I solve this issue?
Additional information:
The actual data has a date in the first column in the form yyyymmdd. The data set spans several years and I want to split this dataset in matrices for each year and after that for each month.
You can use arrayfun to solve this task:
M = [
1 4 56 1;
1 3 5 1;
1 3 6 4;
2 3 5 0;
2 0 0 0;
3 1 2 3;
3 3 3 3]
A = arrayfun(#(x) M(M(:,1) == x, :), unique(M(:,1)), 'uniformoutput', false)
The result A is a cell array and its contents can be accessed as follows:
>> a{1}
ans =
1 4 56 1
1 3 5 1
1 3 6 4
>> a{2}
ans =
2 3 5 0
2 0 0 0
>> a{3}
ans =
3 1 2 3
3 3 3 3
To split the data based on an yyyymmdd format in the first column, you can use the following:
yearly = arrayfun(#(x) M(floor(M(:,1)/10000) == x, :), unique(floor(M(:,1)/10000)), 'uniformoutput', false)
monthly = arrayfun(#(x) M(floor(M(:,1)/100) == x, :), unique(floor(M(:,1)/100)), 'uniformoutput', false)
If you don't know how many outputs you'll have, it is most convenient to put the data into a cell array rather than into separate arrays. The command to do this is MAT2CELL. Note that this assumes your data is sorted. If it isn't use sortrows before running the code.
%# count the repetitions
counts = hist(M(:,1),unique(M(:,1));
%# split the array
yearly = mat2cell(M,counts,size(M,2))
%# if you'd like to split each cell further, but still keep
%# the data also grouped by year, you can do the following
%# assuming the month information is in column 2
yearByMonth = cellfun(#(x)...
mat2cell(x,hist(x(:,2),unique(x(:,2)),size(x,2)),...
yearly,'uniformOutput',false);
You'd then access the data for year 3, month 4 as yearByMonth{3}{4}
EDIT
If the first column of your data is yyyymmdd, I suggest splitting it into three columns yyyy,mm,dd, like below, to facilitate grouping afterward:
ymd = 20120918;
yymmdd = floor(ymd./[10000 100 1])
yymmdd(2:3) = yymmdd(2:3)-100*yymmdd(1:2)