Let's say I have this structure:
Results(i,j).fo
Results(i,j).co
where i=19 and j=30. How can I save in a ixj matrix all Results(i,j).fo? Or even better, How can I say to bootci to read only Results(i,j).fo
Media_tot = mean(Matrix,2)
ci = bootci(1000, #mean, Matrix');
ci = abs(ci' - repmat(Media_tot,1,2));
hE = errorbar(xdata_m, ydata_m, ci(:,1), ci(:,2));
I think this should work for your first question:
reshape([Results.fo], 19, 30)
e.g.
%// Make a 3x3 matrix of structs with 2 fields
A = [];
for k = 1:9
A(k).x = k;
A(k).y = 9-k;
end
A= reshape(A,3,3)
Now
reshape([A.x], 3,3)
ans =
1 4 7
2 5 8
3 6 9
and
reshape([A.y], 3,3)
ans =
8 5 2
7 4 1
6 3 0
Given an array of equivalent structures, e.g.
Results = [ struct('fo',1, 'co',2) struct('fo',10, 'co',20); struct('fo',100, 'co',200) struct('fo',1000, 'co',2000) ]
You can access all 'fo` using the square brackets
all_fo = [Results.fo]
% >> [1 100 10 1000]
However, they are then in a 1D-array, to get them in the original format, use
all_fo = reshape([Results.fo], size(Results))
% >> [1 10; 100 1000]
Related
I am trying to perform a row-wise "ismember" in MATLAB in order to find out where, in Set, each element of Input is.
Here is my work so far.
function lia = ismemberrow(Input, Set)
lia = false(size(Input)); % Pre-assign the output lia
for J = 1 : size(Input,1)
% The J-th row of "lia" is the list of locations of the
% elements of Input(J,:) in Set
lia(J,:) = find(ismember(Set, Input(J,:)));
end
end
For example, if the variables Input and Set are defined as follows
Input = [1 4;
4 2;
4 3;
2 4;
1 2;
3 2];
Set = [3 2 4 1];
The the output lia of lia = ismemberrow(Input,Set) will be:
lia = [4 3;
3 2;
3 1;
2 3;
4 2;
1 2];
My function works accurately so far, but this function is called many times in my project so I am thinking of that if I can reduce the for-loop so that it spends less time. May I have some opinions on it?
A single call to ismember (no loop necessary) will give you what you want in the second output argument:
>> [~, lia] = ismember(Input, Set)
lia =
4 3
3 2
3 1
2 3
4 2
1 2
I'd go with ismember as in #gnovice's answer. But here are some alternatives, just for the fun of it.
If the values in Input are guaranteed to be in Set:
[ind, ~] = find(bsxfun(#eq, Set(:), Input(:).'));
result = reshape(ind, size(Input));
If they are not guaranteed to be:
[ind, val] = max(bsxfun(#eq, Set(:), permute(Input, [3 1 2])));
result = permute(ind.*val, [2 3 1]);
If your inputs are positive integers you simply can use indexing
m(Set)=1:numel(Set);
result = m(Input)
If the range of input is large you can use sparse matrix:
s = sparse(Set,1,1:numel(Set));
result = s(Input)
Result:
4 3
3 2
3 1
2 3
4 2
1 2
I have a matrix:
X = [2,6,1; 3,8,1; 4,7,1; 6,2,1; 6,4,1; 7,3,1; 8,5,1; 7,6,1];
I want to find all row-wise combinations of X. i.e.
A(1) = [2, 6, 1; 3, 8, 1; 4, 7, 1]
A(2) = [2, 6, 1; 3, 8, 1; 6, 2, 1]
:
:
:
Here's what I've tried:
X = [2,6,1; 3,8,1; 4,7,1; 6,2,1; 6,4,1; 7,3,1; 8,5,1; 7,6,1];
p = 3
[m, n] = size(X);
comb = combnk(1:m, p);
[s, t] = size(comb);
c = [X(comb(:,1), :, :) X(comb(:,2), :, :) X(comb(:,3), :, :)];
This gives me a matrix like:
c = 2 6 1 3 8 1 4 7 1
2 6 1 3 8 1 6 2 1
2 6 1 3 8 1 6 4 1
I want to apply the concatenate matrix option to obtain c to make it dynamic depending on value of p but I'm not sure how to use it. I don't want to use For loops. Please help me out.
This is fully vectorized, so it should be fast:
n = 3; %// number of rows to pick each time
ind = reshape(nchoosek(1:size(X,1), n).', [], 1); %'// indices of combinations
A = permute(reshape(X(ind,:).', size(X,2), n, []), [2 1 3]);
The result is
A(:,:,1)
ans =
2 6 1
3 8 1
4 7 1
A(:,:,2)
ans =
2 6 1
3 8 1
6 2 1
etc.
Should you need the result in the form of a cell array, you can convert A from 3D-array to cell array this way:
A = mat2cell(A, size(A,1), size(A,2), ones(1,size(A,3)));
Your thinking is pretty close. This code does the job. I put comments in code, which should be easy to read.
X = [2,6,1; 3,8,1; 4,7,1; 6,2,1; 6,4,1; 7,3,1; 8,5,1; 7,6,1];
p = 3;
%// List all combinations choosing 3 out of 1:8.
v = nchoosek(1:size(X,1), p);
%// Use each row of v to create the matrices, and put the results in an cell array.
%// This is the A matrix in your question.
A = arrayfun(#(k)X(v(k,:), :), 1:size(v,1), 'UniformOutput', false);
%// And you can concatenate A vertically to get c.
flatA = cellfun(#(x)reshape(x, 1, []), A, 'UniformOutput', false);
c = vertcat(flatA{:});
PS: From my understanding I thought the result you wanted was A, which is an easy to use cell array. But I added an extra step to get c exactly as in your question just in case.
Disclaimer: arrayfun and cellfun are pretty much equivalent to for loop in terms of performance.
You can do it using reshape and a bunch of transposes since Matlab is column-major ordered:
c = reshape(X(comb',:)',9,[])'
or if you want a 3D matrix:
A = permute(reshape(X(comb',:)',3,3,[])', [2,1,3])
The code below produces a cell array containing n (=210) 2x3x4-shaped n-d arrays.
n = prod(5:7);
makendarray = #(i) reshape(1:24, 2:4) + (i * 1000);
ndarrays = cellfun(makendarray, num2cell(1:n), 'un', 0);
For each i ∈ {1, … ,n}, the contents of the i-th n-d array are all integers whose initial digits match those of i. For example, for i = 123:
>> ndarrays{123}
ans(:,:,1) =
123001 123003 123005
123002 123004 123006
ans(:,:,2) =
123007 123009 123011
123008 123010 123012
ans(:,:,3) =
123013 123015 123017
123014 123016 123018
ans(:,:,4) =
123019 123021 123023
123020 123022 123024
I want to assemble all these n-d arrays into a larger [5x6x7x2x3x4 double] n-d array X in such a way that expressions of the form X(i,j,k,:,:,:) will correspond to one of the original smaller n-d arrays. This last requirement is the one that's giving me difficulty.
I have no problem producing the larger [5x6x7x2x3x4 double] n-d array from the smaller ones, but expressions of the form X(i,j,k,:,:,:) do not produce one of the original smaller n-d arrays.
Below is an example of one of the things I've tried; the last output should match below should match the output shown for ndarrays{123} above, but doesn't:
>> X = reshape(cell2mat(ndarrays), [5:7 2:4]);
>> [idx{1:3}] = ind2sub(5:7, 123);
>> squeeze(X(idx{:}, :, :, :))
ans(:,:,1) =
62001 62003 62005
167001 167003 167005
ans(:,:,2) =
62007 62009 62011
167007 167009 167011
ans(:,:,3) =
62013 62015 62017
167013 167015 167017
ans(:,:,4) =
62019 62021 62023
167019 167021 167023
EDIT: OK, by (pretty much blind) trial-and-error I found that the following does the trick:
tmp = cellfun(#(c) reshape(c, [1 24]), ndarrays, 'un', 0);
X = reshape(cat(1, tmp{:}), [5:7 2:4]);
IOW: linearize the subarrays before passing them to cat(1, ...). I'm surprised that it's necessary to explicitly perform this linearization step: it's what I expect MATLAB to do by default (in cell2mat, for example).
Be that as it may, is there a better (or at least clearer/easier to understand) way to achieve the same effect?
(BTW, for this problem, the initial shape of ndarray is given and not subject to change. IOW, solutions that require modifying the makesubarray function in the example do not fit the situation I'm dealing with.)
EDIT2: In reference to Luis Mendo's answer, here's the output (copy-pasted verbatim from my MATLAB workspace) of a small script (with echo on), showing the sizes of various items:
n = prod(5:7);
makendarray = #(i) reshape(1:24, 2:4) + (i * 1000);
ndarrays = cellfun(makendarray, num2cell(1:n), 'un', 0);
size(ndarrays)
ans =
1 210
size(permute(ndarrays, [2 3 4 1]))
ans =
210 1
size(cell2mat(permute(ndarrays, [2 3 4 1])))
ans =
420 3 4
echo off;
You can achieve it with a little bit of permute and reshape. The logic of this approach can be followed by observing size(X) at the end of each step (indicated in the comments):
X = cell2mat(permute(ndarrays(:), [2 3 4 1])); %// size [2 3 4 210]
X = permute(X, [4 5 6 1 2 3]); %// size [210 1 1 2 3 4]
X = reshape(X, [7 6 5 2 3 4]); %// size [7 6 5 2 3 4]
X = permute(X, [3 2 1 4 5 6]); %// size [5 6 7 2 3 4]
I have a cell array of size 1*n (X=cell(1, n)). every cell in this cell array(X), has a different size. So, to be clear, here is an example:
X = {X1, X2, X3} (n=3)
X1 = [1 3 5 9]
X2 = [1 2 3 6 7]
X3 = [1 7 8 9]
I have a vector with length of m. for example:
Y = [0 3 6 9] (m=4)
I want to divide cells in my cell array according to this vector in this way:
X1-1=[1] (because 1>=Y(1) & 1<Y(2)), X1-2=[3 5] (because 3 and 5>=Y(2) & 1<Y(3)), X1-3=[9] (because 9>=Y(3) & 9<Y(4))
and so on for X2 and X3...
and finally I have:
Z = {X1-1 X2-1 X3-1
X1-2 X2-2 X3-2
X1-3 X2-3 X3-3}
Here is my code with for-loops:
Z = cell(size(X,2), size(Y,2));
for i = 1:size(X,2)
T = cell2mat(X(i));
for j = 1:size(Y,2)-1
idx = (T > Y(j) & T < Y(j+1));
Z{i,j} = {T(idx)'};
end
end
Z = cellfun(#(z) cell2mat(z), Z, 'UniformOutput', false);
I want to get the same results without using for-loops. Any vectorization ideas?
You can replace the two for loops with one cellfun and one arrayfun. Not necessarily faster, though:
Z = cellfun(#(Xi) arrayfun(#(k) Xi(Y(k)<=Xi&Xi<Y(k+1)), 1:numel(Y)-1, 'uni', 0).', X, 'uni', 0);
The desired results are stored in Z{1}{1} (corresponding to your X1-1), Z{1}{2} (corresponding to X1-2), etc.
Example: I'm using your data but changing Y to [0 3 6 9.1] (note that with Y = [0 3 6 9] the X1-3 in your example is not correct):
>> Z{1}{1}
ans =
1
>> Z{1}{2}
ans =
3 5
>> Z{1}{3}
ans =
9
>> Z{2}{1}
ans =
1 2
How to repeat
A = [ 1 2 ;
3 4 ]
repeated by
B = [ 1 2 ;
2 1 ]
So I want my answer like matrix C:
C = [ 1 2 2;
3 3 4 ]
Thanks for your help.
Just for the fun of it, another solution making use of arrayfun:
res = cell2mat(arrayfun(#(a,b) ones(b,1).*a, A', B', 'uniformoutput', false))'
This results in:
res =
1 2 2
3 3 4
To make this simple, I assume that you're only going to add more columns, and that you've checked that you have the same number of columns for each row.
Then it becomes a simple combination of repeating elements and reshaping.
EDIT I've modified the code so that it also works if A and B are 3D arrays.
%# get the number of rows from A, transpose both
%# A and B so that linear indexing works
[nRowsA,~,nValsA] = size(A);
A = permute(A,[2 1 3]);
B = permute(B,[2 1 3]);
%# create an index vector from B
%# so that we know what to repeat
nRep = sum(B(:));
repIdx = zeros(1,nRep);
repIdxIdx = cumsum([1 B(1:end-1)]);
repIdx(repIdxIdx) = 1;
repIdx = cumsum(repIdx);
%# assemble the array C
C = A(repIdx);
C = permute(reshape(C,[],nRowsA,nValsA),[2 1 3]);
C =
1 2 2
3 3 4