How to check if a generated number is between 2 values

I am trying to program multiple steps for a percentage value.

I am using rand() to generate a random number and I want to use the output to define a range.

Example:

Var = rand();

if Var >= 0.9

xxx

elseif Var < 0.9 and >= 0.8

yyy

elseif Var < 0.8 and >= 0.7

zzz

end

I am tring to get 90th percentile, 80th percentile, 70th percentile.

Answer 1

[Benjamin Thompson]

You can use the logical operator & to combine multiple tests. You can either get an index vector with the results of all tests as I am showing here, or you can check single values one by one and use this test in an if statement.

» rand(2,1)

ans =

0.814723686393179

0.905791937075619

» var = rand(2,1)

var =

0.126986816293506

0.913375856139019

» I = (var > 0.1) & (var < 0.5)

I =

2x1 logical array

1

0

Answer 2

[Image Analyst]

Not sure what you're asking. Of course for a known distribution, like rand() which does the uniform distribution, you can do as you have done (except use "&&" instead of "and" and you'd need to use Var twice) to determine if a single variable is in a certain percentile range, because the percentile ranges are known in advance for that distribution.

Actually if you know the ranges, you can just drop down from the max to 0 and not bother checking the lower limit, though it's perhaps more readable and understandable if you do

Var = rand();
if Var >= 0.9
    xxx
elseif Var >= 0.8
    yyy
elseif Var >= 0.7
    zzz
%etc. for other numbers
end

If you don't know the distribution, you can use histogram or histcounts to find the actual percentiles of your data.

Answer 3

[Steven Lord]

Other options that may be able to do what you want include discretize and interp1.

x = rand(5, 1)

x = 5×1

0.6115 0.7907 0.6304 0.2460 0.1239

p = 0:10:100;
y1 = discretize(x, p/100, p(2:end)) % or

y1 = 5×1

70 80 70 30 20

y2 = interp1(p/100, p, x, 'next')

y2 = 5×1

70 80 70 30 20