compare every element in the first array with the second array

I have two arrays

X = [1,3,5,9]

Y = [1,2,4,9]

I want a function to compare every element in array X with the whole array Y and if this element exists in array Y then write the element and if this element doesn't exist in array Y then add 1 until it find the closest number.

If we suppose the output is Z so Z should be like that Z = [1,4,9,9]

Thanks in advance

[Jan]

Are the arrays sorted?

What is the purpose of "add 1 until it find the closest number"? The closest larger number is the closest larger one without adding 1 also.

Answer 1

[Jan]

X = [1,3,5,9];
Y = [1,2,4,9];

Z = zeros(size(X));
for k = 1:numel(X)
    Z(k) = Y(find(Y >= X(k), 1));
end
Z

Z = 1×4

1 4 9 9

This does not do, what you need and not, what I expect:

Z = discretize(X, [Y, Y(end) + 1], 'IncludedEdge', 'right')

Z = 1×4

1 2 3 3

[Ahmed Mohamed]

Thanks a million for your effort.

and for your comment

No, the array aren't sorted and the numbers are random.

"The closest larger number is the closest larger one without adding 1 also" you're right but I couldn't do it.

I have another issue in my code.

[Ahmed Mohamed]

sidb=10:1:25;
si=10.^(sidb./10);
n=3:0.1:4;
k=504;           % channel
syms i j integer
Nvalues=zeros(15);               %n values
temp= [0];
for i=0:15
    for j=0:15
        N(i+1,j+1)= i^2+i*j+j^2;
        Nvalues=N;
    end
end
PB=[0.01 .01 1];
    N_OMNI=zeros(11,16);
    N_120=zeros(11,16);
    N_60=zeros(11,16);
for i=1:16
    for x=1:11
        N_OMNI(i,x)=(1/3)*((6*si(i))^(2/n(x)));
        N_OMNI(i,x)=ceil (N_OMNI(x));
        N_120(i,x)=(1/3)*((6*si(i))^(2/n(x)));
        N_120(i,x)=ceil (N_120(x));
        N_60(i,x)=(1/3)*((6*si(i))^(2/n(x)));
        N_60(i,x)=ceil(N_60(i,x));
    end
end
Z = zeros(size(N_OMNI));
for w = 1:numel(N_OMNI)
    Z(w) = N(find(N >= N_OMNI(w), 1));
end

[Ahmed Mohamed]

I took your code and replace X,Y with N_OMNI, Nvalues

Then I run the code and Z didn't contain the whole numbers and I can't figure out what I should do to solve this issue.

Can You take my hand?

[Jan]

@Ahmed Mohamed: Why do you declare the loop counters i and j as symbolic variables?

sidb = 10:25;
si   = 10.^(sidb./10);
n    = 3:0.1:4;
k    = 504;           % channel

% No, loop counters should not be symbolic:
% syms i j integer

% Here you pre-allocate a 15x15 matrix, but Nvalues is
% overwritten repeatedly later:
Nvalues = zeros(15);  %n values
% Better pre-allocate N:
N = zeros(16, 16);

% Not used hre: temp= [0];
for i = 0:15
    for j = 0:15
        N(i+1, j+1) = i^2 + i*j + j^2;
        % Nvalues=N;  % Overwritten in each iteration
        % Simply move this line behind the loops
    end
end

% A cheaper version of the loops above:
N = (0:15) .* (0:15).';

PB     = [0.01, .01, 1];
N_OMNI = zeros(11,16);
N_120  = zeros(11,16);
N_60   = zeros(11,16);
for i=1:16
    for x=1:11
        N_OMNI(i,x) = (6*si(i))^(2/n(x)) / 3;
        N_OMNI(i,x) = ceil(N_OMNI(x));
        % Where is 2nd index      ^  do you mean N_OMNI(i,x) 
        N_120(i,x)  = (6*si(i))^(2/n(x)) / 3;
        N_120(i,x)  = ceil(N_120(x));
        % Where is 2nd index     ^  do you mean N_120(i,x) 
        N_60(i,x)   = (6*si(i))^(2/n(x)) / 3;
        N_60(i,x)   = ceil(N_60(i,x));
        
        % N_OMNI, N_120 and N_60 have the same values?!?
    end
end

Z = zeros(size(N_OMNI));
for w = 1:numel(N_OMNI)
    Z(w) = N(find(N >= N_OMNI(w), 1));
end

Answer 2

[VINAYAK LUHA]

Hi,

Your question can also be framed as -

"For every element x in X , find the least element in y in Y , such that y>=x"

This problem can also be solved using binary search in O(|X||Y|log(|Y|) time, (here |X| represents cardinality of X) ,provided we sort the Y first .

Here's the code for my approch to the problem .

ASSUMPTION

Since you did not clarify on the output when any x in X is larger than the maximum value in Y , I assume it as -1 .

    X=[5 4 -10 -2 5 8 10 88 74 5];
    Y=[2 3 4 9];
    your_ans =fun(X,Y)

your_ans = 1×10

9 4 2 2 9 9 -1 -1 -1 9

    
    function val = helper(x,Y) 
    n=length(Y);
    l=1;r=n;
    while(l<=r)
        mid =floor(l+(r-l)/2);
        if(x==Y(mid))
            val=Y(mid);
            return;
        elseif(x<Y(mid))
            r=mid-1;     
        else
            l=mid+1;
        end
    end
    if(l>length(Y))
        val=-1;
    else
        val=Y(l);
    end  
    end
    
    function ret =fun(X,Y)
    ret=zeros(1,length(X));
    sort(Y);
    for i=1:length(X)
        ret(i)=helper(X(i),Y);
    end
    return 
    end