Dynamically check the condition for iteration in for loop - MATLAB

Question

Example of my code: (Actual code is very long. Could be found in edit history)

X = {'a','b','c','d'}

for i = 1:length(X)
     if X(i) == 'a'    %// for example
          X(i)=[];
     end
end

Why didn't the counter stop at 3rd iteration? It tried to continue till 4th iteration and generated the following error :

Index exceeds matrix dimensions.

But as the first element 'a' was deleted, The actual size of array has become 3 (instead of 4). Shouldn't the 'looping' be stopped after the 3rd iteration?

I don't have matlab now, but i will check that tonight , interesting ... ;) — user813853
– user813853, Commented Jun 1, 2015 at 14:58

Luis Mendo · Accepted Answer · 2015-06-02 09:16:28Z

3

You probably mean

X = ['a','b','c','d']

(with square brackets), so X is a char array (string), not a cell array

The problem is that within the loop you remove one entry of X, so X is left with 3 entries. Thus, when you try to access its 4th entry (at iteration i=4) you get an error.

This happens because the for exit condition is not re-evaluated at each iteration. At the for statement you say that i has to run from 1 to 4 (4 is the value of length(X) at that time), and that's what happens.

To achieve what you want you probably need a while loop. The while-loop condition is evaluated after each iteration, using the current values for the variables, to determine whether a new iteration should take place or not. So in the following code only 3 iterations occur, and you get no errors:

i = 1;
while i<=length(X)
    if X(i) == 'a'
        X(i) = [];
    else
        i = i + 1;
    end
end

The counter i should be incremented only if no element of X has been removed. Thanks to @matlabgui for the catch.

edited Jun 2, 2015 at 9:16

answered Jun 1, 2015 at 15:10

Luis Mendo

113k13 gold badges80 silver badges154 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

matlabgui Over a year ago

@LuisMendo you need to add a continue after the X(i) = [] - consider the case where X = ['a' 'a' 'c' 'd']

Luis Mendo Over a year ago

@matlabgui Thanks! You are right, I didn't think about that. I've edited the answer. Now I increment the counter i only if no element has been removed. I think that's the same as continue but I find it clearer.

matlabgui Over a year ago

@LuisMendo - no problem! :) Yes counting on the else has the same desire - I have used both methods - depends on the bigger picture of your code!

matlabgui · Accepted Answer · 2015-06-01 15:26:10Z

1

I suspect your code is:

X={'a','b','c','d'}

for i = 1:length(X)
     if X{i} == 'a'    %for example
          X(i)=[]
     end
end

When X{i} == 'a' you are removing X(i) -> i.e. X becomes 3 long instead of 4. But you loop is programmed to go to the length of X before you started (i.e. 4).

In this type of situation you can do the loop in reverse:

X={'a','b','c','d'}

for i = length(X):-1:1
     if X{i} == 'a'    %for example
          X(i)=[]
     end
end

Another method is to store an inedx in the loop and then remove at the end:

X={'a','b','c','d'}

index = false(length(X),1);
for i = 1:length(X)
     if X{i} == 'a'    %for example
          index(i)= true;
     end
end
X(index) = [];

edited Jun 1, 2015 at 15:26

answered Jun 1, 2015 at 15:11

matlabgui

5,6821 gold badge15 silver badges16 bronze badges

1 Comment

Devel Over a year ago

yes that's exactly the problem but I have to start my loop at the right direction : for i = 1 : length(X)

scmg · Accepted Answer · 2015-06-01 15:03:25Z

0

You will get error since Matlab's index of any array starts at 1, and you did the for-loop starting from 0.

answered Jun 1, 2015 at 15:03

scmg

1,9041 gold badge16 silver badges24 bronze badges

Comments

Kami Kaze · Accepted Answer · 2015-06-01 15:16:16Z

0

MATLAB Arrays start with 1 and then you have to decrease the end of the loop because your vektor got smaller or get a second variable.

And I wasn't able to compare chars.. but with numbers it worked

so it would be

j=0
for i = 1:length(X)
     j=j+1;
     if X(j) == 'a'    %for example
          X(j)=[];
          j=j-1;
     end
end

edited Jun 1, 2015 at 15:16

answered Jun 1, 2015 at 14:59

Kami Kaze

2,08017 silver badges28 bronze badges

Comments

GameOfThrows · Accepted Answer · 2015-06-01 15:17:46Z

0

I cannot recreate the problem, the code uses cells, and you cannot index the 0th cell. So I edited to code to look like this:

for i = 1:length(X)
     if X{i} == 'a'    %for example
      X{i}=[];
     end
end

>>X = 

[]    'b'    'c'    'd'

No problems detected

EDIT: based on your code, you are deleting the cells while running this loop; try deleting the value in your cell instead of deleting the cell,you want to use curly brackets {} instead of [], and if you want to delete the empty cells, simply do it at the end using:

X(~cellfun('isempty',X))

edited Jun 1, 2015 at 15:17

answered Jun 1, 2015 at 15:04

GameOfThrows

4,5202 gold badges30 silver badges44 bronze badges

2 Comments

Devel Over a year ago

You're right, but i have more complex code I have tryed to explain with simple code ( I will edit and put my code maybe you will see the problem ) thank you

GameOfThrows Over a year ago

@Devel I have edited my answer, please check if it makes any difference

Collectives™ on Stack Overflow

Dynamically check the condition for iteration in for loop - MATLAB

5 Answers 5

3 Comments

1 Comment

Comments

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

3 Comments

1 Comment

Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related