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How to take a value between two values

17 vues (au cours des 30 derniers jours)
Amron
Amron le 15 Sep 2023
Commenté : Dyuman Joshi le 27 Sep 2023
hello, please help me.
i couldn't use "if" statement to take 'r' value. The 'r' value depends on 'd' value. but i don't know how to make the code for this.
lets take an example. when i input 'd' value as 15, the 'r' value that i got is 1.6 instead of 1.8. what should i do?

Réponse acceptée

kintali narendra
kintali narendra le 15 Sep 2023
make small changes in the if statement. The code given below is the right way to use if.
if (10.2 <= d) && (d < 13)
r = 1.6
elseif (13.5 <= d) && (d < 20.0)
r = 1.8
elseif (20.0 <= d) && (d < 22.4)
r = 2.0
end
  5 commentaires
Stephen23
Stephen23 le 15 Sep 2023
"how is the way to identify the property of a variable to know it is a scalar or not?"
isscalar(d)
Amron
Amron le 27 Sep 2023
thankyou kintali narendra, your code works. futhermore, in order to solve the issue that i got when i use that code, i use the "spinner" value rather than "edit field". As the result, the value of 'd' is detected as scalar.

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Plus de réponses (2)

Steven Lord
Steven Lord le 15 Sep 2023
If you hover over the underlined <= operators in the code in the MATLAB Editor, you will see a Code Analyzer message explaining why that code doesn't do what you think it does and suggest how you can modify the code to do what you likely want to do.
Alternately, you could use the discretize function to discretize your data without a (potentially lengthy) if / elseif / elseif / elseif ... / end statement.
  2 commentaires
Stephen23
Stephen23 le 15 Sep 2023
Modifié(e) : Stephen23 le 15 Sep 2023
"Alternately, you could use the discretize function to discretize your data"
Considering the r output values, how would that work?
Dyuman Joshi
Dyuman Joshi le 27 Sep 2023
@Stephen23 like this -
d = randi([5 25],1,8)
d = 1×8
12 18 21 5 11 9 19 17
X = [10.2,13,13.5,20,22.4];
Y = [1.6,1.8,2,Inf];
out = discretize(d,X,Y)
out = 1×8
1.6000 2.0000 Inf NaN 1.6000 NaN 2.0000 2.0000

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Stephen23
Stephen23 le 15 Sep 2023
d = [11,15,21];
X = [10.2,13.5,20,22.4];
Y = [1.6,1.8,2,Inf];
Z = interp1(X,Y,d, 'previous')
Z = 1×3
1.6000 1.8000 2.0000

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