MATLAB: How do i change the length of a vector in a for loop without it causing an index out of bounds error

for loopMATLABvector

for k = length(otherHand)
        for p = 1:length(possible)
            if otherHand(k) == possible(p)
                currentHand(end+1) = otherHand(k);
                otherHand(k) = [];
            end
        end
end

The length of otherHand starts at 5, but is shortened within the if statement.

If this is not possible, is there a way i can transfer one element of one vector to the end of another?

Thanks in advance.

Best Answer

You can't delete elements of an array while you're iterating over it unless you work in reverse and are extremely careful.

However, you need to learn to use matlab as it's meant to:

ismatch = ismember(otherhand, possible);
currenthand = otherhand(ismatch);
otherhand = otherhand(~ismatch);

No loop needed and much simpler and understandable code.

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MATLAB: How to initialize a variable to zero in userdefined matlab function, which should be executed only once then the values should change dynamically

Such cases are handled using persistent variables.

function [ia_avr] = fcn(ia)
    persistent iasum ia_num ia_avr_old
    
    if isempty(iasum) % only runs once
        iasum = 0;
        ia_num = 0;
        ia_avr_old = 0;
    end
    
    iasum = iasum+Ia; 
    if xxxx
        ia_num=ia_num+1;
        ia_avr=ia_avr_old;
    elseif xxxx
        ia_num=ia_num+1;
        ia_avr=iasum/ia_num;
        ia_avr_old=ia_avr;
    else            
        ia_avr=ia_avr_old;
        iasum=0;
        ia_num=0;
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MATLAB: Compare two numeric cell arrays

Once I got rid of that unnecessary conversion to numeric value then this task becomes a lot easier. I also preallocated the cell array hand to significantly speed the code up.

myAletters = 'DIRNEEP';
hand = cell(1,floor(exp(1)*factorial(numel(myAletters)))-1);
hand_count=0;
for i=1:length(myAletters)
      hand_interim_combination=nchoosek(myAletters,i);
      for j=1:length(hand_interim_combination(:,1))
          hand_interim_permutation=perms(hand_interim_combination(j,:));
           for k=1:length(hand_interim_permutation(:,1))
               hand_count=hand_count+1;
               hand{hand_count}=hand_interim_permutation(k,:);
           end
      end
end
%load('myDict.mat');
myDict = {'IRE','PEER','XYZ'};
word_plays = ismember(hand,myDict);

and checking the output:

>> hand(word_plays) 
ans = 
  'IRE'    'IRE'    'PEER'    'PEER'

I also tidied up the formatting (crtl+I), replaced unpredictable length with reliable size, and renamed variables to make it more readable (for me):

vec = 'DIRNEEP';
hand = cell(1,floor(exp(1)*factorial(numel(vec)))-1);
itr = 0;
for m = 1:numel(vec)
    cmb = nchoosek(vec,m);
    for n = 1:size(cmb,1)
        prm = perms(cmb(n,:));
        for k = 1:size(prm,1)
            itr = itr+1;
            hand{itr} = prm(k,:);
        end
    end
end

Perhaps the simplest solution would be to use cellstr instead of the innermost loop:

vec = 'DIRNEEP';
hand = cell(1,2^numel(vec)-1);
k = 0;
for m = 1:numel(vec)
    cmb = nchoosek(vec,m);
    for n = 1:size(cmb,1)
        k = k+1;
        hand{k} = cellstr(perms(cmb(n,:)));
    end
end
hand = vertcat(hand{:});

Note that these are all just improvements on your algorithm. With a large dictionary a more efficient algorithm might use the small number of available letters to iteratively select smaller sets of the dictionary, until matches are achieved.

Best Answer

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