485814_en-US

liliconcom

Difference between FOR and FOR EACH loops

Can somebody explain why this:

dt = Open( "$SAMPLE_DATA/Fitness.jmp" );
col = { :Runtime, :RunPulse, :RstPulse };

y expr  = Expr( Y() );
For( c = 1, c <= N Items( col ), c++,
 Insert Into( y expr, col[c] );
);

Name Expr(y expr);

produces expected output

Y( :Runtime, :RunPulse, :RstPulse )

but this:

dt = Open( "$SAMPLE_DATA/Fitness.jmp" );
col = { :Runtime, :RunPulse, :RstPulse };

y expr  = Expr( Y() );
For each({coll, index},col,
 Insert Into( y expr, coll );
);

Name Expr(y expr);

produces this:

Y( ., ., . )

I would expect it to give the same result? How should I change FOR EACH loop to give the same result?

 

 

Consumer and Market ResearchSharing and Communicating ResultsRe: Difference between FOR and FOR EACH loops

It is not a bug -- in this instance it is actually JMP being consistent.  When indexing a list the result is returned unevaluated.  When storing the result of an indexed item into a variable it then evaluates the result.  Remember that For Each( {value, index}, list, expression... ) is equivalent to For( index = 1, index <= N Items( list ), index++, value = list[index]; expression... )

 

A = {1 + 2, 2 + 3, 3 + 4};
b = {};
res = A[1];
Insert Into( b, A[1] );
Insert Into( b, res );
b
/*:

{1 + 2, 3}

 

Re: Difference between FOR and FOR EACH loops

Yes, this is it.

It is a strange behavior, since 1. we have For Each Row statement for by row evaluation, and 2. For Each should treat a list (and I'm trying to work with a list of references to columns) same as For loop.

Is this a bug or is it intended behavior?

Re: Difference between FOR and FOR EACH loops

My guess is that the columns get evaluated into current rows values in For Each (you can test this by adding Row() = 1; for example). You might be able to avoid this by using Name Expr()

Names Default To Here(1);

dt = Open("$SAMPLE_DATA/Fitness.jmp");
col = {:Runtime, :RunPulse, :RstPulse};

y expr = Expr(Y());
For(c = 1, c <= N Items(col), c++,
 Insert Into(y expr, col[c]);
);
Show(Name Expr(y expr));

Row() = 1;
y expr = Expr(Y());
For Each({coll, index}, col, 
 Insert Into(y expr, coll);
);
Show(Name Expr(y expr));

y expr = Expr(Y());
For Each({coll, index}, col, 
 Insert Into(y expr, Name Expr(coll));
);
Show(Name Expr(y expr));