How can I vectorize this find function?
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for i=1:length(x)
ind_1 = find(b <x(i), 1, 'last');
ind_2 = ind_1+1;
x1 = b(ind_1);
x2 = b(ind_2);
y1 = a(ind_1);
y2 = a(ind_2);
y(i) = (y2-y1)/(x2-x1)*(x(i)-x1) + y1;
end
where b =
0
2.49999993684469e-06
4.99999987368938e-06
7.49999981053406e-06
9.99999974737875e-06
1.24999996842234e-05
1.49999996210681e-05
1.74999986484181e-05
1.99999994947575e-05
2.25000003410969e-05
a = -3.90885305404663
-3.90826201438904
-3.90855741500855
-3.90796637535095
-3.90885305404663
-3.90855741500855
-3.90826201438904
-3.90826201438904
-3.90767097473145
-3.90826201438904
x = 0.500000000000000
1
1.50000000000000
2
2.50000000000000
3
3.50000000000000
4
4.50000000000000
5
3 Comments
madhan ravi
on 20 Nov 2018
Edited: madhan ravi
on 20 Nov 2018
provide the values , did you see the answer below?
Original question retrieved from Google cache, as posted by doyi joo:
for i=1:length(x)
ind_1 = find(b <x(i), 1, 'last');
ind_2 = ind_1+1;
x1 = b(ind_1);
x2 = b(ind_2);
y1 = a(ind_1);
y2 = a(ind_2);
y(i) = (y2-y1)/(x2-x1)*(x(i)-x1) + y1;
end
where
b =
0
2.49999993684469e-06
4.99999987368938e-06
7.49999981053406e-06
9.99999974737875e-06
1.24999996842234e-05
1.49999996210681e-05
1.74999986484181e-05
1.99999994947575e-05
2.25000003410969e-05
a = -3.90885305404663
-3.90826201438904
-3.90855741500855
-3.90796637535095
-3.90885305404663
-3.90855741500855
-3.90826201438904
-3.90826201438904
-3.90767097473145
-3.90826201438904
x = 0.500000000000000
1
1.50000000000000
2
2.50000000000000
3
3.50000000000000
4
4.50000000000000
5
Rena Berman
on 22 Jul 2020
(Answers Dev) Restored edit
Accepted Answer
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