How to take a value between two values

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?

 Accepted Answer

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 Comments

this is what i got.
Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values.
Use the ANY or ALL functions to reduce operands to logical scalar values.
d = 21;
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
r = 2
This runs fine. I tried d = 11, 15, and 21. No error in any case. Is your d not a scalar, but a vector or matrix or some other kind of variable?
i input the 'd' value from editfield.
btw, i don't know whether it is a scalar or a vector. how can i get some informations of it? i mean, how is the way to identify the property of a variable to know it is a scalar or not?
thanks
"how is the way to identify the property of a variable to know it is a scalar or not?"
isscalar(d)
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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More Answers (2)

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 Comments

Stephen23
Stephen23 on 15 Sep 2023
Edited: Stephen23 on 15 Sep 2023
"Alternately, you could use the discretize function to discretize your data"
Considering the r output values, how would that work?
@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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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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