# Fill a vector based on another vector

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Pilar Jiménez on 21 Oct 2019
Edited: Pilar Jiménez on 21 Oct 2019
I know the title is somewhat confusing, but I hope you can help me. I need to make a code for a specific application, I am confused on how to do a part. I have the following:
Nmax=25;
A=ones(1:Nmax);
B=[1;1;6;12,1,2,5,2];
What I need to do is fill vector A with the number of zeros indicated by vector B, that is, vector A changes from ones to zeros based on vector B in the order in which it is found. For example:
The first and second values of B indicate that it is 1 zero, therefore, vector A = [00111 ...]
The third value of B indicates that there are 6 zeros, therefore, vector A = [0000000011111 ...]
And so on.
If the number of zeros exceeds the value of Nmax, save it in another vector and cut the count to the integers of B.
The logic of the code should be like this, I was already trying index by index with a for cycle, but it makes errors.

Yuan Li on 21 Oct 2019
A(1:sum(B(:))) = 0;

#### 1 Comment

Pilar Jiménez on 21 Oct 2019
Sorry, I edited the question. It happens that the sum of B exceeds the size of A, so I must work in this case index by index.
When exceeding I must keep the last 5 and 2 zeros in another vector, so I will also add a conditional.