Input value and compare to values in an array

2018 11 3
3 个回答
回答已采纳
更新时间：2018 11 3
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0 个投票

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.

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Star Strider
Star Strider 2018-11-3

1 个投票

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.

更多回答（2 个）

madhan ravi
madhan ravi 2018-11-3
编辑：madhan ravi 2018-11-3

0 个投票

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 个评论
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Stephan
Stephan 2018-11-3
I think your code gives wrong result for input = 3
Stephan
Stephan 2018-11-3
编辑：Stephan 2018-11-3
Wrong result for 1.15 i think - OP want to rounding down

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Stephan
Stephan 2018-11-3

0 个投票

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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