delete element from vector

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Majid Al-Sirafi
Majid Al-Sirafi 2012 年 9 月 24 日
コメント済み: Sibghat 2024 年 3 月 2 日
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 件のコメント
Rosie
Rosie 2017 年 7 月 5 日
編集済み: Walter Roberson 2017 年 7 月 5 日
Hi majed
You can use the follwoing
a(index)=[]
a(3)=[]
the number will delete
Good luck
Hamna Ameer
Hamna Ameer 2017 年 9 月 29 日
編集済み: Hamna Ameer 2017 年 9 月 29 日
a(3)=[] how can i directly store this in a new vector say b?

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採用された回答

Daniel Shub
Daniel Shub 2012 年 9 月 24 日
編集済み: MathWorks Support Team 2018 年 11 月 9 日
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 件のコメント
Walter Roberson
Walter Roberson 2017 年 7 月 5 日
b = a(a >= 2 & a <= 4); %keep 2 to 4
Rik
Rik 2021 年 3 月 31 日
@Anthony Dave Flags are not for personal bookmarks. Please remove your flag.

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その他の回答 (7 件)

Jan
Jan 2012 年 9 月 24 日
編集済み: Jan 2012 年 9 月 24 日
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.
  3 件のコメント
Joel Bay
Joel Bay 2019 年 6 月 28 日
"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
irvin rynning 2021 年 12 月 6 日
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

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masoud sistaninejad
masoud sistaninejad 2021 年 8 月 23 日
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 件のコメント
Andy Rojas
Andy Rojas 2021 年 11 月 24 日
Thank you!
Emma Fickett
Emma Fickett 2022 年 10 月 29 日
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
Andrei Bobrov 2012 年 9 月 24 日
a = a(abs(a - 3) > eps(100))
  1 件のコメント
Majid Al-Sirafi
Majid Al-Sirafi 2012 年 9 月 24 日
than you very much

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Will Reeves
Will Reeves 2022 年 2 月 15 日
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

Elias Gule
Elias Gule 2015 年 12 月 1 日
% Use logical indexing
a = a(a~=3)
  2 件のコメント
denny
denny 2017 年 8 月 31 日
I like this answer.
Ntsakisi Kanyana
Ntsakisi Kanyana 2020 年 3 月 31 日
Does it work on strings?

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Abdul samad
Abdul samad 2023 年 8 月 4 日
編集済み: Abdul samad 2023 年 8 月 4 日
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

Sibghat
Sibghat 2024 年 3 月 2 日
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
  1 件のコメント
Sibghat
Sibghat 2024 年 3 月 2 日
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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