While loop for the elements of an array

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I have an array:
a=[1 1 1 1 1 1 1 10 1 1 1 1 1 1 12 1 1 1 1 3];
I want to make a while loop that does the following
enas=0;
while a(i)==1 %
enas=enas+1;
end
But I don't know how to express it in matlab. Can you help me please?

采纳的回答

Image Analyst
Image Analyst 2013-5-27
Here's how you'd do it:
a=[1 1 1 1 1 1 1 10 1 1 1 1 1 1 12 1 1 1 1 3];
enas=0;
k = 1;
while a(k)==1 %
enas=enas+1
k = k + 1
end
But here's how a real MATLAB programmer would do it:
enas = find(a~=1, 1, 'first')-1
  3 个评论
Image Analyst
Image Analyst 2013-5-27
If you need to count the length of each stretch of 1's in your array, and if you have the Image Processing Toolbox, you'd do this:
measurements = regionprops(a==1, 'Area');
allLengths = [measurements.Area]; % Get lengths of all stretches of 1s.
If you don't have the Image Processing Toolbox, it's more difficult - let me know if you have that unfortunate case.
Giorgos Papakonstantinou
Unfortunately I don't. I just do:
a=[1 1 1 1 1 1 1 10 1 1 1 1 1 1 12 1 1 1 1 3];
enas = find(a~=1)-1
segments=zeros(length(enas),1);
segments(1,1)=enas(1);
segments(2:end,1)=diff(enas)-1;
segments
which a bit complicated but it does the job... If you have better ideas tell me. thank you!

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更多回答(1 个)

Jason Nicholson
Jason Nicholson 2013-5-27
See the lines below. This will work.
a=[1 1 1 1 1 1 1 10 1 1 1 1 1 1 12 1 1 1 1 3];
i = 1;
enas=0;
while a(i)==1 %
enas=enas+1;
i = i +1;
end
  4 个评论
Matt Kindig
Matt Kindig 2013-5-27
编辑:Matt Kindig 2013-5-27
This should do it:
b = [0, a, 0]; %ensure that ends are not 1
edges = find(b~=1); %location elements that are not 1
spans = diff(edges)-1; %distance between edges is span of 1's
enas = spans(spans~=0) %should output 7 6 4
Giorgos Papakonstantinou
Suppose that I want find the edges of ones and of not ones. Suppose the array is:
a=[1 1 1 1 1 1 12 10 1 1 1 1 1 11 12 1 1 1 2 3]
a(1:6) -->area1 of ones
a(7:8) -->area1 of not ones
a(9:a13) -->area2 of ones
a(14:15)-->area2 of not ones
a(16:19)-->area3 of ones
a(20) -->area3 of not ones
and so on..

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