SPSS Statistics

Why doesn't this throw an error:

IF ANY (q0050_0001,q0051_0001,q0052_0001,q0053_0001,1) q5053_0001=1.

vs.

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

?

Cheers,

Bob

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

Why doesn't this throw an error:

IF ANY (q0050_0001,q0051_0001,q0052_0001,q0053_0001,1) q5053_0001=1.

vs.

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

?

Cheers,

Bob

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

Thank you Jon, as always. I didn't understand what was happening to my syntax since no error was triggered. I put both instructions into ChatGPT and, at first, it told me that this instruction:

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

would indeed throw an error – but it did not in my command syntax. When I challenged it on this fact, it then told me the following:

Ah, thank you for clarifying - that's a great observation, and it leads to a really subtle but important point about how SPSS interprets ANY(). SPSS is interpreting the first argument, q0050_0001, as the value to test, and the rest (q0051_0001, q0052_0001, q0053_0001, and 1) as the values to test it against. So this code says:

"If the value of q0050_0001 is equal to any of the following: q0051_0001, q0052_0001, q0053_0001, or 1, then set q5053_0001 = 1."

SPSS is syntactically permissive with ANY(). It allows both variable names and constants in the argument list, and it simply evaluates the function as ANY(value, list_of_comparisons), as in:

"Does the value of q0050_0001 match any of [q0051_0001, q0052_0001, q0053_0001, 1]?"

It's not an error, just a different logic - and that's why you didn't get the results you were hoping for.

40 years later and I'm still learning things about basic functions!

Cheers,

 

Bob

 

 

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

 

 

Why doesn't this throw an error:

IF ANY (q0050_0001,q0051_0001,q0052_0001,q0053_0001,1) q5053_0001=1.

vs.

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

?

Cheers,

Bob

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

Thank you Jon, as always. I didn't understand what was happening to my syntax since no error was triggered. I put both instructions into ChatGPT and, at first, it told me that this instruction:

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

would indeed throw an error – but it did not in my command syntax. When I challenged it on this fact, it then told me the following:

Ah, thank you for clarifying - that's a great observation, and it leads to a really subtle but important point about how SPSS interprets ANY(). SPSS is interpreting the first argument, q0050_0001, as the value to test, and the rest (q0051_0001, q0052_0001, q0053_0001, and 1) as the values to test it against. So this code says:

"If the value of q0050_0001 is equal to any of the following: q0051_0001, q0052_0001, q0053_0001, or 1, then set q5053_0001 = 1."

SPSS is syntactically permissive with ANY(). It allows both variable names and constants in the argument list, and it simply evaluates the function as ANY(value, list_of_comparisons), as in:

"Does the value of q0050_0001 match any of [q0051_0001, q0052_0001, q0053_0001, 1]?"

It's not an error, just a different logic - and that's why you didn't get the results you were hoping for.

40 years later and I'm still learning things about basic functions!

Cheers,

 

Bob

 

 

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

 

 

Why doesn't this throw an error:

IF ANY (q0050_0001,q0051_0001,q0052_0001,q0053_0001,1) q5053_0001=1.

vs.

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

?

Cheers,

Bob

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

Thank you Jon, as always. I didn't understand what was happening to my syntax since no error was triggered. I put both instructions into ChatGPT and, at first, it told me that this instruction:

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

would indeed throw an error – but it did not in my command syntax. When I challenged it on this fact, it then told me the following:

Ah, thank you for clarifying - that's a great observation, and it leads to a really subtle but important point about how SPSS interprets ANY(). SPSS is interpreting the first argument, q0050_0001, as the value to test, and the rest (q0051_0001, q0052_0001, q0053_0001, and 1) as the values to test it against. So this code says:

"If the value of q0050_0001 is equal to any of the following: q0051_0001, q0052_0001, q0053_0001, or 1, then set q5053_0001 = 1."

SPSS is syntactically permissive with ANY(). It allows both variable names and constants in the argument list, and it simply evaluates the function as ANY(value, list_of_comparisons), as in:

"Does the value of q0050_0001 match any of [q0051_0001, q0052_0001, q0053_0001, 1]?"

It's not an error, just a different logic - and that's why you didn't get the results you were hoping for.

40 years later and I'm still learning things about basic functions!

Cheers,

 

Bob

 

 

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting

 

 

Why doesn't this throw an error:

IF ANY (q0050_0001,q0051_0001,q0052_0001,q0053_0001,1) q5053_0001=1.

vs.

IF ANY (1,q0050_0001,q0051_0001,q0052_0001,q0053_0001) q5053_0001=1.

?

Cheers,

Bob

Robert W. Walker

CEO & Founder

23 S. Main St.·  Norwalk, CT 06854

Tel. +1.203.255.0505

https://usemotion.com/meet/rwwalker/meeting