insert element in vector
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Hi everyone
now, how can I insert element in vector .... for example
a=[1,2,4,5]
how can I embed 3 between 2 and 4 in above vector to be
a=[1,2,3,4,5]
thank you
majid
11 件のコメント
Jan
2018 年 1 月 29 日
@Walter: Thanks for your comment concerning homework. I think Sami's point is more my "rudeness".
The admins had cleaned up this thread in Nov-2017 already. I think it is strange, that the first activity from Sami's account is flagging a 3 year old comment in a 6 year old thread as being rude.
Luck Haviland
2022 年 12 月 8 日
Not sure if I am breaking the matlab forums code of conduct by not answering OP's question but I agree with @Jonathan Campelli's way of looking at answering peoples questions. Even if the question is a homework question for one person, it can be a non homework question for another person. If the question is never answered, there will be no learning which defeats the purpose of a community forum.
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その他の回答 (6 件)
Daniel Shub
2012 年 9 月 24 日
While I think this is a homework problem ...
Function handles and cat are your friends
insert = @(a, x, n)cat(2, x(1:n), a, x(n+1:end));
insert(3, [1,2,4,5], 2)
ans = 1 2 3 4 5
0 件のコメント
Jonathan Campelli
2015 年 3 月 12 日
Here is an application specific solution:
a=[1 2 4 5] %Your predefined vector.
a=sort([a 3]) %The "[a 3]" operation adds the element 3 to the end of vector "a", creating vector [1 2 4 5 3]. "Sort()" then alligns the new vector's elements in order from least to greatest.
4 件のコメント
Andrei Bobrov
2012 年 9 月 24 日
編集済み: Andrei Bobrov
2012 年 9 月 24 日
b = 3;
i1 = 3;
a = [a(1:i1-1),b,a(i1:end)];
or
a = [1 2 4 5];
i1 = 3;
b = 3000;
n = numel(b);
out = ones(size(a) + [0 n]);
out(i1 + (0:n-1)) = 0;
out(out > 0) = a;
out(out == 0) = b;
1 件のコメント
Bernard
2019 年 7 月 29 日
a = [1 2 4 5];
b = 3;
idx = 3;
a = [a(1:length(a) < idx), b, a(1:length(a) >= idx)];
Simpler version of your second example, which I prefer because it works with idx == 1 or idx == length(a).
JMP Phillips
2021 年 4 月 20 日
編集済み: DGM
2023 年 3 月 4 日
If you are reading this in 2021 now in MATLAB you can just do something like this (no for loops, no 'cat'). It can also work with inserting more than 1 element into a vector.
y = zeros(1,length(x)+length(b)); %initialise a new vector of the appropriate size
y(a) = b; %insert the values in 'b' at the locations in 'a'
y(y==0) = x; %insert the original values in x into the new vector at their new positions.
where
- x is the existing vector to insert values into
- a is vector of indices where to put new values
- b is vector of new values
NOTE: because we use 0 we cannot insert a value of 0, this works for non-zero values only.
Example with inserting multiple values at specified locations into an array:
x = [1,2,3,4,5]
a = [3,5,1,4]
b = [5, nan, 3, 717]
Result:
y = 3 1 5 717 NaN 2 3 4 5
and the original question with inserting 1 element would be solved by
x = [1,2,4,5]
a = 3
b = 3
Result:
y = 1 2 3 4 5
1 件のコメント
Pol Cardona Rubio
2024 年 4 月 15 日
編集済み: Pol Cardona Rubio
2024 年 4 月 15 日
If instead of creating it with a zero value you create it with NaN and then index based on isnan() it can be used with any valued vector as in:
y = repmat(NaN,length(x)+length(b)); %initialise a new vector of the appropriate size
y(a) = b; %insert the values in 'b' at the locations in 'a'
y(isnan(y)) = x; %insert the original values in x into the new vector at their new positions.
Walter Roberson
2018 年 3 月 7 日
There is no MATLAB operator for inserting into a MATLAB vector. Concatenating elements as described by Wayne King, Andrei Bobrov, and Daniel Shub is the natural MATLAB solution to this task.
The only exception to this is MATLAB String objects (R2016b and later), which have insertBefore and insertAfter operations defined for them that search the input strings to match a given text and insert at that point.
I have attached code to implement insertion after a given point into generalized vectors. Generalized vectors here refers to the fact that the vectors might have 3 or more dimensions, and that the code is not restricted to numeric or char.
There are some important differences between this code and the ones posted above:
- this code handles vectors of any dimension, not just row vectors
- the output is the same class as the initial vector even if different data types are involved. For example the other implementations if asked to insert 'a' after (double) 50 would produce '2a' because [50 'a'] automatically converts the double to char because of MATLAB rules about converting to the most restrictive data type when cat() is used. The code I attached will produce [50 97] instead -- but if you ask to insert (double) 50 before 'a' then you will get '2a' because the data type of the original vector is retained
- However, an empty original array will cause the output to be in the type of the new data so that you can always use [] to indicate empty array no matter what type you are appending
There are a number of design decisions explicitly documented in the code -- this seemingly simple operation is surprisingly complex.
1 件のコメント
Manuel Infante Francés
2023 年 4 月 5 日
編集済み: Manuel Infante Francés
2023 年 4 月 5 日
Good morning, I have the following problem, in the thread of this forum:
I am trying to add an element to a column vector (B1 of m rows) that is the output of a Matlab Function block. The output vector (B) is desired to have m+1 rows. When adding the element (with value = x), the resulting output (B) is a vector of m+1 rows, with the particularity that all the rows will acquire the value (x) of the added element. Add the value you add and use the method you use (cat function, indexing etc.), the resulting vector is effectively a vector of m+1 elements, but with the same value (x) for all its elements.
Example:
function B=fcn(A,Ts)
B1 =diff(A)/Ts;
B =[1;B1];
In the example below, 1 is the value I want to index into B1 to get B. A has 501 elements and I want the output of function (B) to also have 501 elements (the value of the first row element must be 1 and from row 2 to the last one -both included- it should be the vector B1). However, all the elements of B are equal to 1. Ts comes from a constant block with scalar value (0.02).
Thank you.
Elena Fiermonte
2018 年 9 月 30 日
編集済み: DGM
2023 年 3 月 4 日
Hi everyone. I know it's a bit long, this should work with every kind of sorted vector. Feel free to refine it. Here it is:
% code
A=[1 2 4 5]; % you must predefined a sorted vector.
B=zeros(1,length(A)+1);
L=length(A);
n=input('ins num: ')
for i=1:L
for j=i:L
if n>A(i)
B(i+1)=n;
B(i)=A(i);
B(j+1)=A(j);
elseif n==A(1)
B(1:2)=A(1);
B(i+1)=A(i);
end
end
end
B
0 件のコメント
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