How to implement this logic

[tintin_99]

Hi,

Could you please let me know how can I implemet this logic

if  2.39<x<2.4  then y=4.8

if  2.42<x<2.45  then y=5.0

if  2.47<x<2.50  then y=5.2

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.

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All the values are hard coded

Thanks

[billko]

As usual, show us what you got.  🙂

[RavensFan]

And what if x <= 2.39?

2.4 <= x <=2 .42?

2.45 <= x <=2 .47?

Or x >= 2.50?

There are holes in your ranges.

[tintin_99]

Then the value is invalid,

it should be actually like this

2.39<=x<=2.4 then y=4.8

2.42<=x<=2.45 then y=5

.

.

[RavensFan]

I would multiply the value by 100 then wire that into a case structure selector terminal.

The case structure would have 4 cases defined.

239 .. 240    it outputs a 4.8 to the tunnel

242 .. 245    it outputs a 5.0 to the tunnel

247 .. 250    it outputs a 5.2 to the tunnel

Default        it outputs a 0 or some other value that you define that you want to represent as an invalid value.

[aputman]

You could use a formula node from the Structures menu.  It uses expressions similar to C.

[Ranjeet_Singh]

Multiplying by 100 and then comparing integer is always better then comparing floating value. Opt for 'X100'

[billko]

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@Ranjeet_Singh wrote:

Multiplying by 100 and then comparing integer is always better then comparing floating value. Opt for 'X100'

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So... what happens if the number has three decimal places?  😉

More correctly stated, I guess you could say "multiply the numbers to be compared by 10^x, where x equals the precision of your most precise number.

[crossrulz]

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@billko wrote:

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@Ranjeet_Singh wrote:

Multiplying by 100 and then comparing integer is always better then comparing floating value. Opt for 'X100'

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So... what happens if the number has three decimal places?  😉

More correctly stated, I guess you could say "multiply the numbers to be compared by 10^x, where x equals the precision of your most precise number.

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This is an excellent point.  Given the example, multiply by 100 would work fine.  But you also need to be aware of rounding, which is a whole other issue you probably don't want to think about.  When converting to an integer, the number will round toward 0.

[billko]

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@crossrulz wrote:

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@billko wrote:

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@Ranjeet_Singh wrote:

Multiplying by 100 and then comparing integer is always better then comparing floating value. Opt for 'X100'

---

So... what happens if the number has three decimal places?  😉

More correctly stated, I guess you could say "multiply the numbers to be compared by 10^x, where x equals the precision of your most precise number.

---

This is an excellent point.  Given the example, multiply by 100 would work fine.  But you also need to be aware of rounding, which is a whole other issue you probably don't want to think about.  When converting to an integer, the number will round toward 0.

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But by multiplying as I have shown, that should result in all numbers being integers... or are you saying that if you specified 0.5 was internally represented as 0.499 you are in danger of coming up with 4 as your integer, not 5?