In my current project, I need to find values inside of an matrix that match with individual vector values. This is an example of the process; the main program has me using lat and lon values. But I create a 20x20 matrix and then a 20x1 array of randomly placed values.
When i do the for loop, each iteration of the Leroy vector is subtracted from every value in matrix. The first min function should return the smallest value from each column and its correspoding index. The second min function should return the smallest overall value from the first min function. and which index had the smallest value.
My concern is that im not sure which integer inside the matrix returned the smallest value. Is there a way I can use the indexes or something to figure that out?
Matrix = magic(20);
Leroy = randi(20,20,1);
for i = 1:length(Leroy)
[Jenkins, J] = min(min(Leroy(i) - Matrix);
end
As Cris Luego pointed in his comment, your for loop is not required since
Leroy(i) - Matrix
translates to something like
5 - [1 2 3; 4 5 6; 7 8 9]
However, your problem to get the index of the minimum in -Matrix can be solved by using min(-Matrix(:)):
[minimum, minidx] = min(-Matrix(:));
However, you will get the linear index. If you need the index for row and column, use
[colidx,rowidx] = ind2sub(size(Matrix), minidx);
Matrix = magic(20);
Leroy = randi(20,20,1);
for i = 1:length(Leroy)
[Jenkins, J] = min((Leroy(i) - Matrix).^2);
end
Using this will help get a match between two values in two arrays or a match between array values and matrix values
Related
Say A is a 3x4x5 array. I am given a vector a, say of dimension 2 and b of dimension 2. If I do A(a,b,:) it will give 5 matrices of dimensions 2x2. I instead want the piecewise vectors (without writing a for loop).
So, I want the two vectors of A which are given by (a's first element and b's first element) and (a's second element and b's second element)
How do I do this without a for loop? If A were two dimensions I could do this using sub2ind. I don't know how to access the entire vectors.
You can use sub2ind to find the linear index to the first element of each output vector: ind = sub2ind(size(A),a,b). To get the whole vectors, you can't do A(ind,:), because the : has to be the 3rd dimension. However, what you can do is reshape A to be 2D, collapsing the first two dimensions into one. We have a linear index to the vectors we want, that will correctly index the first dimension of this reshaped A:
% input:
A = rand(3,4,5);
a = [2,3];
b = [1,2];
% expected:
B = [squeeze(A(a(1),b(1),:)).';squeeze(A(a(2),b(2),:)).']
% solution:
ind = sub2ind(size(A),a,b);
C = reshape(A,[],size(A,3));
C = C(ind,:)
assert(isequal(B,C))
You can change a and b to be 3d arrays just like A and then the sub2ind should be able to index the whole matrix. Like this:
Edit: Someone pointed out a bug. I have changed it so that a correction gets added. The problem was that ind1, which should have had the index number for each desired element of A was only indexing the first "plane" of A. The fix is that for each additional "plane" in the z direction, the total number of elements in A in the previous "planes" must be added to the index.
A=rand(3,4,5);
a=[2,3];
b=[1,2];
a=repmat(a,1,1,size(A,3));
b=repmat(b,1,1,size(A,3));
ind1=sub2ind(size(A),a,b);
correction=(size(A,1)*size(A,2))*(0:size(A,3)-1);
correction=permute(correction,[3 1 2]);
ind1=ind1+repmat(correction,1,2,1);
out=A(ind1)
I have to create a function that takes as input a vector v and three scalars a, b and c. The function replaces every element of v that is equal to a with a two element array [b,c].
For example, given v = [1,2,3,4] and a = 2, b = 5, c = 5, the output would be:
out = [1,5,5,3,4]
My first attempt was to try this:
v = [1,2,3,4];
v(2) = [5,5];
However, I get an error, so I do not understand how to put two values in the place of one in a vector, i.e. shift all the following values one position to the right so that the new two values fit in the vector and, therefore, the size of the vector will increase in one. In addition, if there are several values of a that exist in v, I'm not sure how to replace them all at once.
How can I do this in MATLAB?
Here's a solution using cell arrays:
% remember the indices where a occurs
ind = (v == a);
% split array such that each element of a cell array contains one element
v = mat2cell(v, 1, ones(1, numel(v)));
% replace appropriate cells with two-element array
v(ind) = {[b c]};
% concatenate
v = cell2mat(v);
Like rayryeng's solution, it can replace multiple occurrences of a.
The problem mentioned by siliconwafer, that the array changes size, is here solved by intermediately keeping the partial arrays in cells of a cell array. Converting back to an array concenates these parts.
Something I would do is to first find the values of v that are equal to a which we will call ind. Then, create a new output vector that has the output size equal to numel(v) + numel(ind), as we are replacing each value of a that is in v with an additional value, then use indexing to place our new values in.
Assuming that you have created a row vector v, do the following:
%// Find all locations that are equal to a
ind = find(v == a);
%// Allocate output vector
out = zeros(1, numel(v) + numel(ind));
%// Determine locations in output vector that we need to
%// modify to place the value b in
indx = ind + (0:numel(ind)-1);
%// Determine locations in output vector that we need to
%// modify to place the value c in
indy = indx + 1;
%// Place values of b and c into the output
out(indx) = b;
out(indy) = c;
%// Get the rest of the values in v that are not equal to a
%// and place them in their corresponding spots.
rest = true(1,numel(out));
rest([indx,indy]) = false;
out(rest) = v(v ~= a);
The indx and indy statements are rather tricky, but certainly not hard to understand. For each index in v that is equal to a, what happens is that we need to shift the vector over by 1 for each index / location of v that is equal to a. The first value requires that we shift the vector over to the right by 1, then the next value requires that we shift to the right by 1 with respect to the previous shift, which means that we actually need to take the second index and shift by the right by 2 as this is with respect to the original index.
The next value requires that we shift to the right by 1 with respect to the second shift, or shifting to the right by 3 with respect to the original index and so on. These shifts define where we're going to place b. To place c, we simply take the indices generated for placing b and move them over to the right by 1.
What's left is to populate the output vector with those values that are not equal to a. We simply define a logical mask where the indices used to populate the output array have their locations set to false while the rest are set to true. We use this to index into the output and find those locations that are not equal to a to complete the assignment.
Example:
v = [1,2,3,4,5,4,4,5];
a = 4;
b = 10;
c = 11;
Using the above code, we get:
out =
1 2 3 10 11 5 10 11 10 11 5
This successfully replaces every value that is 4 in v with the tuple of [10,11].
I think that strrep deserves a mention here.
Although it's called string replacement and warns for non-char input, it still works perfectly fine for other numbers as well (including integers, doubles and even complex numbers).
v = [1,2,3,4]
a = 2, b = 5, c = 5
out = strrep(v, a, [b c])
Warning: Inputs must be character arrays or cell arrays of strings.
out =
1 5 5 3 4
You are not attempting to overwrite an existing value in the vector. You're attempting to change the size of the vector (meaning the number of rows or columns in the vector) because you're adding an element. This will always result in the vector being reallocated in memory.
Create a new vector, using the first and last half of v.
Let's say your index is stored in the variable index.
index = 2;
newValues = [5, 5];
x = [ v(1:index), newValues, v(index+1:end) ]
x =
1 2 5 5 3 4
I'm writing a function that requires some values in a matrix of arbitrary dimansions to be dropped in a specified dimension.
For example, say I have a 3x3 matrix:
a=[1,2,3;4,5,6;7,8,9];
I might want to drop the third element in each row, in which case I could do
a = a(:,1:2)
But what if the dimensions of a are arbitrary, and the dimension to trim is defined as an argument in the function?
Using linear indexing, and some carefully considered maths is an option but I was wondering if there is a neater soltion?
For those interested, this is my current code:
...
% Find length in each dimension
sz = size(dat);
% Get the proportion to trim in each dimension
k = sz(d)*abs(p);
% Get the decimal part and integer parts of k
int_part = fix(k);
dec_part = abs(k - int_part);
% Sort the array
dat = sort(dat,d);
% Trim the array in dimension d
if (int_part ~=0)
switch d
case 1
dat = dat(int_part + 1 : sz(1) - int_part,:);
case 2
dat = dat(:,int_part + 1 : sz(2) - int_part);
end
end
...
It doesn't get any neater than this:
function A = trim(A, n, d)
%// Remove n-th slice of A in dimension d
%// n can be vector of indices. d needs to be scalar
sub = repmat({':'}, 1, ndims(A));
sub{d} = n;
A(sub{:}) = [];
This makes use of the not very well known fact that the string ':' can be used as an index. With due credit to this answer by #AndrewJanke, and to #chappjc for bringing it to my attention.
a = a(:, 1:end-1)
end, used as a matrix index, always refers to the index of the last element of that matrix
if you want to trim different dimensions, the simplest way is using and if/else block - as MatLab only supports 7 dimensions at most, you wont need an infinite number of these to cover all bases
The permute function allows to permute the dimension of an array of any dimension.
You can place the dimension you want to trim in a prescribed position (the first, I guess), trim, and finally restore the original ordering. In this way you can avoid running loops and do what you want compactly.
Let's say, I have an M x N matrix. Now, I want to insert a constant M x 1 column vector (say all 1's) in between each of the N columns. Therefore, my resulting matrix would be of dimension (M x (2*N-1)), with every other column being 1's.
Is there an easy way to do that?
Vertically concatenate a matrix of ones, reshape, and cut off the last column of ones. For a matrix A:
B = reshape([A; ones(size(A))],size(A,1),[]);
B(:,end)=[]
Here is another way to do it, using the possibility of out of bounds indexing in assignments:
M(:,1:2:end*2)=M;
M(:,2:2:end)=1
If you don't mind creating a temporary matrix, one way to do it would be to do the following:
old_matrix = rand(M,N); % Just for example
new_matrix = ones(M,2*N-1);
new_matrix(:,1:2:end) = old_matrix;
Note that for an arbitrary constant matrix, you could replace the second line with the following:
new_matrix = repmat(const_array,1,2*N-1);
I have this loop which generates a vector "Diff". How do I place the values of Diff in an array that records all the Diff's generated? The problem is that the length of Diff should be a fixed length (36) which is the width of the table "CleanPrice". But because col_set varies in length (according to the number of NaNs in the data it is reading), then Diff also varies in length. What I need it to do is assign the answers generated according to their appropriate column number. i.e. row(i) of diff should contain col(i) where all other rows in Diff should be assigned a "0" or "NaN". Basically I need DiffArray to be a (nTrials x 36) array where each row is the (36 x 1) DiffArray generated. At the moment though, each time the length of col changes, I get the following error:
??? Subscripted assignment dimension mismatch.
Error in ==> NSSmodel
at 41 DiffMatrix(end+1,:)=Diff
This is my code:
DiffArray=[];
StartRow=2935;
EndRow=2940;
nTrials=EndRow-StartRow;
for row=StartRow:EndRow;
col_set=find(~isnan(gcm3.data.CleanPrice(row,1:end)));
col=col_set(:,2:end);
CleanPrices=transpose(gcm3.data.CleanPrice(row,col));
Maturity=gcm3.data.CouponandMaturity(col-1,2);
SettleDate=gcm3.data.CouponandMaturity(row,3);
Settle = repmat(SettleDate,[length(Maturity) 1]);
CleanPrices =transpose(gcm3.data.CleanPrice(row,col));
CouponRate = gcm3.data.CouponandMaturity(col-1,1);
Instruments = [Settle Maturity CleanPrices CouponRate];
PlottingPoints = gcm3.data.CouponandMaturity(1,2):gcm3.data.CouponandMaturity(36,2);
Yield = bndyield(CleanPrices,CouponRate,Settle,Maturity);
SvenssonModel = IRFunctionCurve.fitSvensson('Zero',SettleDate,Instruments)
ParYield=SvenssonModel.getParYields(Maturity);
[PriceActual, AccruedIntActual] = bndprice(Yield, CouponRate, Settle, Maturity);
[PriceNSS, AccruedIntNSS] = bndprice(ParYield, CouponRate, Settle, Maturity);
Diff=PriceActual-PriceNSS
DiffArray(end+1,:)=Diff
end
I looked at num2cell in this post but wasn't sure how to apply it correctly and started getting errors relating to that instead.
Is it correct to say you want to add an 'incomplete' row to DiffArray? If you know exactly where each element should go you could maybe do something like this:
indices = [1:7; 2:8; 3:9; [1 2 3 6 7 8 10]];
Diff = rand(4, 7);
DiffArray = zeros(4, 10) * NaN;
for row = 1:4
DiffArray(row, indices(row, :)) = Diff(row,:);
end
of course in your case you would be calculating Diff and Index (a row vector) inside the loop and not using preassigned arrays. The above is just to illustrate how to use an indexing vector to position a short row in a matrix.