Input value and compare to values in an array

clc, clear, clf
M1=input('What is the Mach number? ');
Mach=[1.050,1.1,1.2,1.3,1.4,1.6,1.8,2,2.5,3,4,5,6,7,8,10];
Maxd=[0.558,1.515,3.944,6.662,9.427,14.652,19.183,22.974,29.797,34.073,38.774,41.118,42,440,43.791,44.429];
So I have this as my code. What I want to do is take an input mach number, find the closest value in the Mach array (rounding down) and then use the corresponding index of the Maxd array for calculations. I'm sure this is simpler than I think it is but I would very much appreciate the help.

 Accepted Answer

Using the interp1 (link) function:
M1=input('What is the Mach number? ');
Mach=[1.050,1.1,1.2,1.3,1.4,1.6,1.8,2,2.5,3,4,5,6,7,8,10];
Maxd=[0.558,1.515,3.944,6.662,9.427,14.652,19.183,22.974,29.797,34.073,38.774,41.118,42,440,43.791,44.429];
Result = interp1(Mach, Maxd, M1, 'previous')
M1 = 1.5
Result =
9.427
You can easily change the interpolation method if you want a different result.

3 Comments

Thank you so much! this works like a charm!
+1 cool method @star strider
As always, my pleasure!

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More Answers (2)

M1=input('What is the Mach number? ');
Mach=[1.050,1.1,1.2,1.3,1.4,1.6,1.8,2,2.5,3,4,5,6,7,8,10];
Maxd=[0.558,1.515,3.944,6.662,9.427,14.652,19.183,22.974,29.797,34.073,38.774,41.118,42,440,43.791,44.429];
[~,index]=min(abs(floor((Mach-(M1))))) %floor rounds down towards negative infinity
Maxd(index) %the value to be used in calculations
command window:
>> COMMUNITY
What is the Mach number? 3
index =
10
ans =
34.0730

4 Comments

I think your code gives wrong result for input = 3
Stephan
Stephan on 3 Nov 2018
Edited: Stephan on 3 Nov 2018
Wrong result for 1.15 i think - OP want to rounding down
Thank you for the help!

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M1=input('What is the Mach number? ');
Mach=[1.050,1.1,1.2,1.3,1.4,1.6,1.8,2,2.5,3,4,5,6,7,8,10];
Maxd=[0.558,1.515,3.944,6.662,9.427,14.652,19.183,22.974,29.797,34.073,38.774,41.118,42,440,43.791,44.429];
Value = Maxd(find((sort([Mach, M1]))==M1,1,'last')-1)

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