how to use for loop?
显示 更早的评论
if we have an array a= [1,2,3,1,2,3,4,1,2] and second array b=[1,2]
we want to know what is the number of occurrences that b are in a
which means the answer should be 3.
then how to write that using for loop?
采纳的回答
Unless you absolutely must use a loop, use the strfind (link) function:
a = [1,2,3,1,2,3,4,1,2];
b = [1,2];
c = strfind(a,b);
Result = numel(c)
producing:
Result =
3
12 个评论
if I want to use the "for" is it possible?
because when I tried to do so but matlab warned that the matrix deminisions are not true
I did not understand why
It is. You must compare 2 consecutive elements of ‘a’ to ‘b’ in each iteration of the loop.
Try this:
a = [1,2,3,1,2,3,4,1,2];
b = [1,2];
for k = 1:numel(a)-1
z(k,:) = a(k:k+1) == b;
end
Result = nnz(z(:,1)) % Using A ‘for’ Loop
producing:
Result =
3
Note that you can replace the nnz call with a sum call.
thank you fis of all;
for i=1:numel(x)-1;
z(i:1)=x(i:i+1)==y;
end
r=sum(z(:,1));
I entered x=[1,2,1,2,1,2,1,2,1,2,1] and y=[1,2,1,2,1]
the answer should be 4
but matlab said:
Matrix dimensions must agree
I don't understand
You must change my code to accommodate the length of the vector you want to compare, here ‘y’.
This works:
x=[1,2,1,2,1,2,1,2,1,2,1];
y=[1,2,1,2,1];
for i=1:numel(x)-numel(y)+1 % Limits Must Accommodate The Length Of ‘y’
z(i,:)=x(i:i+numel(y)-1)==y; % The Comparison Must Consider Segments Of ‘x’ Equal To The Length Of ‘y’
end
r=sum(z(:,1))
producing:
r =
4
why did you put (i+4) what the 4 mean I don't understand
I edited my previous Comment to make it more general. The current code should address your concerns.
thank you, but the code did not work it gives me an error answer.
As always, my pleasure.
What error did it throw?
It worked when I ran it. Please post the code you used. Also, what MATLAB version (release) are you using?
aya qassim
2019-1-2
编辑:aya qassim
2019-1-2
I have the version of matlab R2017b
for i=1:numel (x)-numel(y)+1
z(i:1)=(x(i:i+numel(y)-1)==y);
end
a=sum(z(:,1));
x=[1,2,1,2,1,2,1,2,1,2,1]
y=[1,2,1,2,1]
matlab answer:
error, subscripted assignment dimensions mismatch.
thank you,
Your ‘z’ assignment subscrpting is incorrect.
Your code:
z(i:1)=(x(i:i+numel(y)-1)==y);
Your code will either assign the right-hand-side of the equation (a logical vector) to a single scalar (if i=1) or to an empty array (if i>1), neither of which will work.
My code:
z(i,:)=x(i:i+numel(y)-1)==y;
My code assigns the right-hand-side to a matrix row. It works. Use my code.
Please see the documentation section on Matrix Indexing (link) to understand the difference between my code (that works) and your code (that errors), and how to index a matrix correctly.
Thank you, I understood my mistake.
As always, my pleasure.
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