# delete element from vector

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Majid Al-Sirafi on 24 Sep 2012
Edited: Rik on 12 Jun 2024
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
##### 7 CommentsShow 5 older commentsHide 5 older comments
Rosie on 5 Jul 2017
Edited: Walter Roberson on 5 Jul 2017
Hi majed
You can use the follwoing
a(index)=[]
a(3)=[]
the number will delete
Good luck
Hamna Ameer on 29 Sep 2017
Edited: Hamna Ameer on 29 Sep 2017
a(3)=[] how can i directly store this in a new vector say b?

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### Accepted Answer

Daniel Shub on 24 Sep 2012
Edited: MathWorks Support Team on 9 Nov 2018
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) = []
##### 6 CommentsShow 4 older commentsHide 4 older comments
Walter Roberson on 5 Jul 2017
b = a(a >= 2 & a <= 4); %keep 2 to 4
Rik on 31 Mar 2021
@Anthony Dave Flags are not for personal bookmarks. Please remove your flag.

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### More Answers (7)

Jan on 24 Sep 2012
Edited: Jan on 24 Sep 2012
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.
##### 5 CommentsShow 3 older commentsHide 3 older comments
Keanu on 12 Jun 2024
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 on 12 Jun 2024
Edited: Rik on 12 Jun 2024
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.

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masoud sistaninejad on 23 Aug 2021
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
##### 2 CommentsShow NoneHide None
Andy Rojas on 24 Nov 2021
Thank you!
Emma Fickett on 29 Oct 2022
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!!

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Andrei Bobrov on 24 Sep 2012
a = a(abs(a - 3) > eps(100))
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Majid Al-Sirafi on 24 Sep 2012
than you very much

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Will Reeves on 15 Feb 2022
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
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Elias Gule on 1 Dec 2015
% Use logical indexing
a = a(a~=3)
##### 2 CommentsShow NoneHide None
denny on 31 Aug 2017
I like this answer.
Ntsakisi Kanyana on 31 Mar 2020
Does it work on strings?

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Abdul samad on 4 Aug 2023
Edited: Abdul samad on 4 Aug 2023
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
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Sibghat on 2 Mar 2024
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
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Sibghat on 2 Mar 2024
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

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