Is it possible to use Nested if any command in spss?
for example
if any(1, a to c) and if (1, s to x) xx=1.
Sure, but your syntax isn't quite right.
if (any(1, a to c) and any(1, s to x)) xx=1.
The if expression can be arbitrarily complicated, but it needs to be an expression. The if part is a statement.
HTH
Just to provide an answer that clarifies the nested part, here's an example using do if instead of if.
do if any(1, a to c).
do if any(1, s to x).
compute xx=1.
end if.
end if.
This code finds rows with at least one time a score of 1 on any of the variables a to c. Only among these rows, it finds rows with at least one time a score of 1 on any of the variables s to x. The rows that pass the first statement and subsequently pass the second statement are assigned xx=1.
It yields the same results as:
if any(1, a to c)andany(1, s to x) xx=1.
However, the following yields different results:
if any(1, a to c) xx=1.
if any(1, s to x) xx=1.
This essentially means the same as using or between 2 conditional statements
if any(1, a to x)orany(1, s to x) xx=1.
Related
Lets say we have an expression in prefix notation or(1) and A B or(2) or(3) C D E (where A, B, C, D, E are boolean values and or numbered for convenience) that we want to convert to an infix notation. In principle I have two ways to evaluate it:
(1) start at or(3) C D, then or(2), then and, then or(1)
(2) start at and A B then check or(3), or(2). Lastly check or(1)
(1) Evaluate starting from right most operator
(2) Evaluate starting from left most operator having all operands as it's direct neighbors.
Both evaluations yield (A and B) or C or D or E.
Which evaluation sequence is correct?
Will these two evaluations ever give different result for the same prefix record?
http://www.cs.man.ac.uk/~pjj/cs212/fix.html recommends the first method.
You will get the same result regarding of the order, so it is up to you.
On wikipedia it says that using call/cc you can implement the amb operator for nondeterministic choice, and my question is how would you implement the amb operator in a language in which the only support for continuations is to write in continuation passing style, like in erlang?
If you can encode the constraints for what constitutes a successful solution or choice as guards, list comprehensions can be used to generate solutions. For example, the list comprehension documentation shows an example of solving Pythagorean triples, which is a problem frequently solved using amb (see for example exercise 4.35 of SICP, 2nd edition). Here's the more efficient solution, pyth1/1, shown on the list comprehensions page:
pyth1(N) ->
[ {A,B,C} ||
A <- lists:seq(1,N-2),
B <- lists:seq(A+1,N-1),
C <- lists:seq(B+1,N),
A+B+C =< N,
A*A+B*B == C*C
].
One important aspect of amb is efficiently searching the solution space, which is done here by generating possible values for A, B, and C with lists:seq/2 and then constraining and testing those values with guards. Note that the page also shows a less efficient solution named pyth/1 where A, B, and C are all generated identically using lists:seq(1,N); that approach generates all permutations but is slower than pyth1/1 (for example, on my machine, pyth(50) is 5-6x slower than pyth1(50)).
If your constraints can't be expressed as guards, you can use pattern matching and try/catch to deal with failing solutions. For example, here's the same algorithm in pyth/1 rewritten as regular functions triples/1 and the recursive triples/5:
-module(pyth).
-export([triples/1]).
triples(N) ->
triples(1,1,1,N,[]).
triples(N,N,N,N,Acc) ->
lists:reverse(Acc);
triples(N,N,C,N,Acc) ->
triples(1,1,C+1,N,Acc);
triples(N,B,C,N,Acc) ->
triples(1,B+1,C,N,Acc);
triples(A,B,C,N,Acc) ->
NewAcc = try
true = A+B+C =< N,
true = A*A+B*B == C*C,
[{A,B,C}|Acc]
catch
error:{badmatch,false} ->
Acc
end,
triples(A+1,B,C,N,NewAcc).
We're using pattern matching for two purposes:
In the function heads, to control values of A, B and C with respect to N and to know when we're finished
In the body of the final clause of triples/5, to assert that conditions A+B+C =< N and A*A+B*B == C*C match true
If both conditions match true in the final clause of triples/5, we insert the solution into our accumulator list, but if either fails to match, we catch the badmatch error and keep the original accumulator value.
Calling triples/1 yields the same result as the list comprehension approaches used in pyth/1 and pyth1/1, but it's also half the speed of pyth/1. Even so, with this approach any constraint could be encoded as a normal function and tested for success within the try/catch expression.
Just started with Scheme. I'm having problem with printing on console.
A simple list printing example:
(define factorial
(lambda (n)
(cond
((= 0 n) 1)
(#t (* n (factorial (- n 1)))))))
I want to print n, every time the function is called. I figured that I can't do that within the same function? Do I need to call another function just so I can print?
Printing in Scheme works by calling display (and possibly, newline).
Since you want to call it sequentially before/after something else (which, in a functional (or in the case of Scheme, functional-ish) language only makes sense for the called functions side-effects), you would normally need to use begin, which evaluates its arguments in turn and then returns the value of the last subexpression. However, lambda implicitly contains such a begin-expression.
So in your case, it would go like this:
(lambda (n)
(display n) (newline)
(cond [...]))
Two remarks:
You can use (define (factorial n) [...]) as a shorthand for (define factorial (lambda (n) [...])).
The way you implement factorial forbids tail call-optimization, therefore the program will use quite a bit of stack space for larger values of n. Rewriting it into a optimizable form using an accumulator is possible, though.
If you only want to print n once, when the user calls the function, you will indeed need to write a wrapper, like this:
(define (factorial n)
(display n) (newline)
(inner-factorial n))
And then rename your function to inner-factorial.
Let's say you have a Boolean rule/expression like so
(A OR B) AND (D OR E) AND F
You want to convert it into as many AND only expressions as possible, like so
A AND D AND F
A AND E AND F
B AND D AND F
B AND E AND F
You are just reducing the OR's so it becomes
(A AND D AND F) OR (A AND E AND F) OR (...)
Is there a property in Boolean algebra that would do this?
Take a look at DeMorgan's theorem. The link points to a document relating to electronic gates, but the theory remains the same.
It says that any logical binary expression remains unchanged if we
Change all variables to their complements.
Change all AND operations to ORs.
Change all OR operations to ANDs.
Take the complement of the entire expression.
(quoting from the above linked document)
Your example is exploiting the the distributivity of AND over OR, as shown here.
All you need to do is apply that successively. For example, using x*(y+z) = (x*y)+(x*z) (where * denotes AND and + denotes OR):
0. (A + B) * (D + E) * F
1. Apply to the first 2 brackets results in ((A+B)*D)+((A+B)*E)
2. Apply to content of each bracket results in (A*D+B*D)+(A*E+B*E)
3. So now you have ((A*D+B*D)+(A*E+B*E))*F
4. Applying the law again results in (A*D+B*D)*F+(A*E+B*E)*F
5. Apply one more time results in A*D*F+B*D*F+A*E*F+B*E*F, QED
You may be interested in reading about Karnaugh maps. They are a tool for simplifying boolean expressions, but you could use them to determine all of the individual expressions as well. I'm not sure how you might generalize this into an algorithm you could write a program for though.
You might be interested in Conjunctive Normal form or its brother, Disjunctive normal form.
As far as I know boolean algebra can not be build only with AND and OR operations.
If you have only this two operation you are not able to receive NOT operation.
You can convert any expression to the full set of boolean operations.
Here is some full sets:
AND and NOT
OR and NOT
Assuming you can use the NOT operation, you can rewrite any Boolean expression with only ANDs or only ORs. In your case:
(A OR B) AND (D OR E) AND F
I tend to use engineering shorthand for the above and write:
AND as a product (. or nothing);
OR as a sum (+); and
NOT as a single quote (').
So:
(A+B)(D+E)F
The corollary to arithmetic is actually quite useful for factoring terms.
By De Morgan's Law:
(A+B) => (A'B')'
So you can rewrite your expression as:
(A+B)(D+E)F
(A'B')'(D'E')'F
I've just found something I'd call a quirk in F# and would like to know whether it's by design or by mistake and if it's by design, why is it so...
If you write any range expression where the first term is greater than the second term the returned sequence is empty. A look at reflector suggests this is by design, but I can't really find a reason why it would have to be so.
An example to reproduce it is:
[1..10] |> List.length
[10..1] |> List.length
The first will print out 10 while the second will print out 0.
Tests were made in F# CTP 1.9.6.2.
EDIT: thanks for suggesting expliciting the range, but there's still one case (which is what inspired me to ask this question) that won't be covered. What if A and B are variables and none is constantly greater than the other although they're always different?
Considering that the range expression does not seem to get optimized at compiled time anyway, is there any good reason for the code which determines the step (not explicitly specified) in case A and B are ints not to allow negative steps?
As suggested by other answers, you can do
[10 .. -1 .. 1] |> List.iter (printfn "%A")
e.g.
[start .. step .. stop]
Adam Wright - But you should be able
to change the binding for types you're
interested in to behave in any way you
like (including counting down if x >
y).
Taking Adam's suggestion into code:
let (..) a b =
if a < b then seq { a .. b }
else seq { a .. -1 .. b }
printfn "%A" (seq { 1 .. 10 })
printfn "%A" (seq { 10 .. 1 })
This works for int ranges. Have a look at the source code for (..): you may be able to use that to work over other types of ranges, but not sure how you would get the right value of -1 for your specific type.
What "should" happen is, of course, subjective. Normal range notation in my mind defines [x..y] as the set of all elements greater than or equal to x AND less than or equal to y; an empty set if y < x. In this case, we need to appeal to the F# spec.
Range expressions expr1 .. expr2 are evaluated as a call to the overloaded operator (..), whose default binding is defined in Microsoft.FSharp.Core.Operators. This generates an IEnumerable<_> for the range of values between the given start (expr1) and finish (expr2) values, using an increment of 1. The operator requires the existence of a static member (..) (long name GetRange) on the static type of expr1 with an appropriate signature.
Range expressions expr1 .. expr2 .. expr3 are evaluated as a call to the overloaded operator (.. ..), whose default binding is defined in Microsoft.FSharp.Core.Operators. This generates an IEnumerable<_> for the range of values between the given start (expr1) and finish (expr3) values, using an increment of expr2. The operator requires the existence of a static member (..) (long name GetRange) on the static type of expr1 with an appropriate signature.
The standard doesn't seem to define the .. operator (a least, that I can find). But you should be able to change the binding for types you're interested in to behave in any way you like (including counting down if x > y).
In haskell, you can write [10, 9 .. 1]. Perhaps it works the same in F# (I haven't tried it)?
edit:
It seems that the F# syntax is different, maybe something like [10..-1..1]
Ranges are generally expressed (in the languages and frameworks that support them) like this:
low_value <to> high_value
Can you give a good argument why a range ought to be able to be expressed differently? Since you were requesting a range from a higher number to a lower number does it not stand to reason that the resulting range would have no members?