A Paradoxical problem with the SIMPLEST conditional statement !!

Replies to This Discussion

Permalink Reply by Emilio Morello on June 27, 2010 at 9:30am

Hi Mehrad,

might the numbers be approximated somehow ?
I mean for example 4.80000000001 instead of 4.8 ?

Cheers
Emilio

Permalink Reply by Mehrad Mahnia on June 28, 2010 at 1:31am

Thanx Emilio

most of all i was confused by the "why" of this occurrence, but i haven't noticed this subtle point . . .

Permalink Reply by Andrew Heumann on June 27, 2010 at 10:34am

I too have noticed a lot of inconsistencies with the equality operator. I often round my inputs to a fixed number of decimal places or use the "similarity" operator component to work around it.

Permalink Reply by Mehrad Mahnia on June 28, 2010 at 2:02am

thanx Andrew

I've replaced the "=" with "≈" in the function's Expression section and it worked . . .do you know how much the "≈" tolerates around the specified number? for example if we describe " x ≈ 4.8 " as " x= 4.8 ± a ", do you know what is the min and max for "a"?

and second, i had tried the "round" function, but i faced problem with it too! for example:

if the input is a series as {0.0, 0.5, 1, 1.5, 2, 2.5, ...}
the output for Round(x, 0.5) is : {0, 0, 1, 2, 2, 2, 3, 4, 4, 4, 5, 6, 6, 6, ... }

and for Round(x, 2) the output is : {0.0, 0.5, 1, 1.5, 2, 2.5, ... }

i can't understand the logic that lies behind this function, i think
for Round(x, 0.5) the output must be {0.0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, ... }
and for Round(x, 2) it must be {0.0, 0.0, 2, 2, 2, 2, 4, 4, 4, 4, ... }

so, is there any problem with it, or I misendestood the logic ?

Permalink Reply by taz on June 28, 2010 at 10:47am

In the expression Round(x, [d]) the optional input d is the number of decimal places for the rounding operation so your results are as expected...

d can only be an integer so your expression value of 0.5 is being interpreted as 0. For your expression value of 2 the results are also correct since your input values aren't more than two decimal places long.

Permalink Reply by David Rutten on June 28, 2010 at 5:57am

Equality shouldn't be used on floating point numbers, unless you can be absolutely certain that they can indeed be identical. Whenever the floating point number is the result of a mathematical operation (addition, multiplication, division and -especially- subtraction) then you should expect there to be binary noise in the least significant digits.

Equality can be used on Integers without problems, but for floating points you should compare the difference to a threshold value. For example:

If (Math.Abs(A - B) < 1e-16) Then

Where 1e-16 is a value you need to pick. If you are dealing with very big numbers, you'll need a much larger equality threshold.

--
David Rutten
david@mcneel.com
Turku, Finland