Find position of the values of an array in a second array

조회 수: 3(최근 30일)
raktim banerjee
raktim banerjee 2011년 1월 25일
There is 2 strings 'dls' & 'look'. I want to find exact lower and exact higher value of every elements of 'dls' in the sorted string 'look'. I wrote this:
dls = [5 2 1 7 4];
look = [-0.6 1 2.6 4.2 5.8 7.4];
s = length(dls);
l = length(look);
dlstr = 0;
dhstr = 0;
for i = 1:s,
for j =1:l,
if dls(i)>=look(j)
continue; %j=j+1;
else
dh = look(j);
dl = look(j-1);
dlstr = [dlstr dl];
dhstr = [dhstr dh];
break;
end
end
end
dhstr(1)=[];
dlstr(1) = [];
dlstr
dhstr
The expected result is:
dlstr =
4.2000 1.0000 -0.6000 5.8000 2.6000
>> dhstr
dhstr =
5.8000 2.6000 1.0000 7.4000 4.2000
but the output is:
dlstr =
4.2000 1.0000 1.0000 5.8000 2.6000
dhstr =
5.8000 2.6000 2.6000 7.4000 4.2000
The problem occurred if any value in both strings match. '1' exists in both string. dlstr(3) should be -0.6000 & dhstr(3) should be 1.000. Please tell me how to overcome this problem?
  댓글 수: 2
raktim banerjee
raktim banerjee 2011년 1월 25일
Sorry sir! Yes, those are array.

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채택된 답변

Ashish Uthama
Ashish Uthama 2011년 1월 25일
With the above code, this correction should yield you your required output:
dls(i)>look(j)
You might be able to write this functionality more compactly by using the function find. Also: If you find the lower bound first, and the 'look' array is expected to be sorted, then wont the higher bound be the next element?

추가 답변(1개)

Todd Flanagan
Todd Flanagan 2011년 1월 25일
This:
dls(i)>=look(j)
Should be:
dls(i)>look(j)
You might be able to write this functionality more compactly by using the function find.
Note that because the array is sorted, you can find 1 bound and infer both.
  댓글 수: 3
Ashish Uthama
Ashish Uthama 2011년 1월 25일
yup, its better this way.

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