compare every element in the first array with the second array

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Ahmed Mohamed
Ahmed Mohamed 2022-6-2
评论： Jan 2022-6-3
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
  1 个评论
Jan
Jan 2022-6-2
编辑：Jan 2022-6-2
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.

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Jan
Jan 2022-6-2
编辑：Jan 2022-6-2
Ran in:
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
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Ahmed Mohamed
Ahmed Mohamed 2022-6-2
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
Jan 2022-6-3
@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

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更多回答（1 个）

VINAYAK LUHA
VINAYAK LUHA 2022-6-2
Ran in:
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

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