Bug in input program?

[SD]

Hello again,

if I run the following code in v20:

input program.
loop a=1.01 to 2.01 by 0.01.
end case.
end loop.
end file.
end input program.
execute.

compute test1_12 = (a=1.12).
compute test1_13 = (a=1.13).
execute.

test1_12 is exactly 1 where it should be, but test1_13 is 0 where a=1.13.
I already got a solution (compute test1_13 = (rnd(a,0.01).) to work around it, but I'm curious why it occurs or if I'm doing something wrong what I simply don't see.

Thanks in advance.

[Jon K Peck]

Run this program.

input program.
loop a=1.01 to 2.01 by 0.01.
end case.
end loop.
end file.
end input program.
compute  z = a- 1.13.
format z(E22.15).
execute.

You will see that the values are very slightly different from what you expect.  This is the nature of floating point arithmetic.  You should never assume that fractional parts of floating point numbers are exactly equal to what you usually see.  You can read an explanation here: http://en.wikipedia.org/wiki/Floating_point.

Statistical procedures do their calculations in carefully chosen round about ways in order to address this problem, but in simple calculations there will generally be small differences between, say, 1.01 + 0.01 repeated times and the value 1.13.  In this case, the difference is 2.220446049250313 x 10**-16 where your code expects a zero.


Jon Peck (no "h") aka Kim
Senior Software Engineer, IBM
[hidden email]
phone: [phone]




From:        SD <[hidden email]>
To:        [hidden email],
Date:        06/14/2013 05:19 AM
Subject:        [SPSSX-L] Bug in input program?
Sent by:        "SPSSX(r) Discussion" <[hidden email]>

---

Hello again,

if I run the following code in v20:

input program.
loop a=1.01 to 2.01 by 0.01.
end case.
end loop.
end file.
end input program.
execute.

compute test1_12 = (a=1.12).
compute test1_13 = (a=1.13).
execute.

test1_12 is exactly 1 where it should be, but test1_13 is 0 where a=1.13.
I already got a solution (compute test1_13 = (rnd(a,0.01).) to work around
it, but I'm curious why it occurs or if I'm doing something wrong what I
simply don't see.

Thanks in advance.






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[David Marso]

You should NEVER assume exact equality when comparing floating point numbers.
The following approach will keep your code from exploding!
You may need to adjust the value of #eps, but this should suffice.

COMPUTE #eps=.00000000000001.
COMPUTE test1_12 = RANGE(a,1.12-#eps,1.12+#eps).
COMPUTE test1_13 = RANGE(a,1.13-#eps,1.13+#eps).

SD wrote

Hello again,

if I run the following code in v20:

input program.
loop a=1.01 to 2.01 by 0.01.
end case.
end loop.
end file.
end input program.
execute.

compute test1_12 = (a=1.12).
compute test1_13 = (a=1.13).
execute.

test1_12 is exactly 1 where it should be, but test1_13 is 0 where a=1.13.
I already got a solution (compute test1_13 = (rnd(a,0.01).) to work around it, but I'm curious why it occurs or if I'm doing something wrong what I simply don't see.

Thanks in advance.

[Morris, Ramona (MCSCS)]

I'm sorry I can't reply to you.  I will be out of the office until Monday June 24, 2013

For assistance please contact my colleague  Brian McNair: [hidden email] or by phone [phone].

Thank you
Ramona Morris
R&E Ontario Police College

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[SD]

Ok, thank you both. Good to know.