delete element from vector

Hi everyone

how can I delete element from vector .... for example

a=[1,2,3,4,5]

how can I delete 3 from above vector to be

a=[1,2,4,5]

thank you

majid

[Hassan AL Dawood]

What I know is that you can replace the place with a 0 and then run an If statement inside a for loop to create a new Array without that 0

[Hassan AL Dawood]

Or you can set it equal to a(1,3)=[]

[SUBROTA HALDER]

a=setdiff(a, a(1,3))

[Walter Roberson]

yugandar sooraz comments to Hassan AL Dawood:

??? Subscripted assignment dimension mismatch.

[Walter Roberson]

a(3) = [];

[Rosie]

Hi majed

You can use the follwoing

a(index)=[]
a(3)=[]

the number will delete

Good luck

[Hamna Ameer]

a(3)=[] how can i directly store this in a new vector say b?

Answer 1

[Daniel Shub]

I can think of three ways that are all slightly different

a=[1,2,3,4,5];

If you want to get rid of all cases where a is exactly equal to 3

b = a(a~=3);

If you want to delete the third element

b = a;
b(3) = [];

or on a single line

b = a([1:2, 4:end]);

Or, as Jan suggests:

a = [2,3,1,5,4]
a(a == 3) = []

[Majid Al-Sirafi]

thank you very much

[Daniel Shub]

@C Zeng I have removed your "Good answer" flag. Flags are to call the attention of the moderation. Votes are to say good answer.

[Walter Roberson]

Mustafa Uslu comments,

"Practical, fast and accurate!"

[kwabena boafo-mensah]

how does this work when i need to delete a range of row elements from a vector

[Walter Roberson]

b = a(a >= 2 & a <= 4);   %keep 2 to 4

[Rik]

@Anthony Dave Flags are not for personal bookmarks. Please remove your flag.

Answer 2

[Jan]

a = [1,2,3,4,5]
a(3) = []

Or:

a = [2,3,1,5,4]
a(a == 3) = []

These methods are explained exhaustively in the "Getting Started" chapters of the documentation. It is strongly recommended to read them completely. The forum is not though to explain the fundamental basics. Thanks.

[Majid Al-Sirafi]

thank you very much

[Joel Bay]

"These methods are explained exhaustively in the "Getting Started" chapters of the documentation."

Wrong, definetely not exhaustively after comparing Daniel's answer and the documentation. Logical indexing is not even mentioned. The answers to this question is still useful in 2019.

[irvin rynning]

unfortunately some of us prefer to use Matlab to solve problems in a timely manner, and cannot always engage in stackover-flow style plaudits on criticizing one's peers

[Keanu]

A point of clarification for anyone who may be confused:

Consider the two arrays p = [10;20;30;40] and b = [10,20,30,40] (note the semicolon vs. comma) as an example. In this case, p(3) = [] and b(3) = [] will remove the third element from the array entirely, leaving p = [10;20;40] and b = [10,20,40].

If we were to mistakenly say p(3,1) = [] or b(1,3) = [], MATLAB will throw an error: "A null assignment can have only one non-colon index." Of course, this minor distinction will not be immediately clear to a beginner. Moreover, I do not expect anyone to understand this distinction from reading the "exhaustive" documentation.

The help forums are a guide to anyone with a legitimate question. To this day, I am puzzled by responses that jab at the author for merely asking.

[Rik]

I'm surprised that is the error message you get, since it doesn't (at first glance at least) match the cause of the error, and yet:

p = [10;20;30;40];p(3,1) = []

A null assignment can have only one non-colon index.

But your comparison is strained, since your code has in indexing error, which is only superficially related to the deletion of array elements.

The only problem with this question is that it should be covered by any half-decent tutorial, perhaps in the first 15 minutes even. In addition to this, you can find extra information in the documentation. My personal bar is that you shouldn't be able to enter the question in Google and get the solution in the first result.

Answer 3

[masoud sistaninejad]

A = [ 1 2 3 4 5 6 7]

A = 1×7

1 2 3 4 5 6 7

B = [1 3 6]

B = 1×3

1 3 6

C = setdiff(A,B)

C = 1×4

2 4 5 7

[Andy Rojas]

Thank you!

[Emma Fickett]

I've scoured through so many forums trying to remove a vector of values from another vector and setdiff does exactly what I needed, thank you so much!!

Answer 4

[Andrei Bobrov]

a = a(abs(a - 3) > eps(100))

[Majid Al-Sirafi]

than you very much

Answer 5

[Will Reeves]

really crude, but if you wanted to remove a row defined by and index, rather than a value, you could do something like this:

function out=removeRow(in,index)
% removes a row from an matrix

[~,n]=size(in);

if index>n || index<0
    error('index needs to be within the range of the data')
else

    if n==1
        out=[]; % you've removed the last entry
    else
        % strip out the required entry
        if index==1
            out=in(2:end);
        elseif index==n
            out=in(1:end-1);
        else
            out=in([1:index-1 index+1:n]);
        end
    end

end

Answer 6

[Elias Gule]

    % Use logical indexing
    a = a(a~=3)

[denny]

I like this answer.

[Ntsakisi Kanyana]

Does it work on strings?

Answer 7

[Abdul samad]

Yes , you can delete 3 from the given array by assigning the null matrix, like this .

In the command window do like this.

>> a=[1,2,3,4,5];

>> a(3) = [ ];

>>a

This will delete the 3 from the array a = [1,2,3,4,5];

Thank You

Answer 8

[Sibghat]

The removal of the element at the 3rd index has already been addressed. However, if you want to remove all occurences of the number '3' from the array 'a', you can use the following code (with and without using the find method).

% For instance, let's modify the array 'a'
a = [1, 3, 2, 3, 4, 3, 5, 3];
b = find(a == 3);  % Find the index of the element to delete
% The above line-of-code will also work without using the find keyword...
a(b) = [];   % Delete the element(s)
a

a = 1×4

1 2 4 5

[Sibghat]

And if you want to store the removed values in another variable and display the the exact position of the value. You can do it by either replacing the other values with zeroes or by replacing the desired value with zeroes. Hopefully, the following code will help.

a = [1, 3, 2, 3, 4, 3, 5, 3];

indices_of_3 = find(a == 3);    % Find indices of elements equal to 3

removed_values = a(a == 3);     % Store the removed values in another variable named 'removed_values'


% Create a vector with zeroes where the number is 3
b = zeros(size(a));
b(a ~= 3) = a(a ~= 3);

% Create a vector with zeroes where the number is not 3
c = zeros(size(a));
c(indices_of_3) = a(indices_of_3);


% Remove all occurrences of 3 from 'original_vector'
a(a == 3) = [];

% Display the results
% Modified vector after removal of all occurrences of 3
a

a = 1×4

1 2 4 5

% Removed values
removed_values

removed_values = 1×4

3 3 3 3

% Displaying zero where values is 3
b

b = 1×8

1 0 2 0 4 0 5 0

% Displaying zero where value is not 3
c

c = 1×8

0 3 0 3 0 3 0 3