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How to interpolate based on given column

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Mekala balaji
Mekala balaji 2018-11-7
编辑： madhan ravi 2018-11-7
Hi,
I have below array(double), and want to assign 50 for maximum value and 0 for minimum value, and for the rest of the values between maxima and minima interpolation.
0.2
0.8
1.0
1.3
0.8
2.3
4.9
10.0
0.2
5
2.3
8.7
10
In this case,
0.2 -->50
10.0 -->0
for rest of the values, to interpolate between 50& 0.

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采纳的回答

Stephen23
Stephen23 2018-11-7
编辑：Stephen23 2018-11-7
No need to use interpolation, just some simple, efficient arithmetic:
>> V = [0.2;0.8;1.0;1.3;0.8;2.3;4.9;10.0;0.2;5;2.3;8.7;10]
V =
0.20000
0.80000
1.00000
1.30000
0.80000
2.30000
4.90000
10.00000
0.20000
5.00000
2.30000
8.70000
10.00000
>> Vmin = min(V);
>> Vmax = max(V);
>> 50*(Vmax-V)/(Vmax-Vmin)
ans =
50.00000
46.93878
45.91837
44.38776
46.93878
39.28571
26.02041
0.00000
50.00000
25.51020
39.28571
6.63265
0.00000

更多回答（1 个）

madhan ravi
madhan ravi 2018-11-7
编辑：madhan ravi 2018-11-7
see interp1()
a=[0.2
0.8
1.0
1.3
0.8
2.3
4.9
10.0
0.2
5
2.3
8.7
10]
[~,x1]= max(a)
[~,x2]=min(a)
a(x1)=50
a(x2)=0
Interpolated_values=interp1(a,linspace(a(1),a(end))); %interpolated between 0 and 50

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