the new Where?

hogi — Thu, 27 Jun 2024 18:00:03 GMT

Hi,

how can I use the new Where function with lists of lists or matrices?

xs = {10, 20, 30, 50};
Where( Print(xs); xs ==10 );

xs = {[10 20], [30 50]};
Where( Print(xs); xs ==[10 20] );

Re: the new Where?

txnelson — Thu, 27 Jun 2024 18:29:36 GMT

on the first one

xs = {10, 20, 30, 50};
v=Where( xs ==10 );
print(v);

I am not sure of the second one

Re: the new Where?

hogi — Thu, 27 Jun 2024 18:58:14 GMT

The first one was meant as working example :)

Print inside Where:
to show that xs indeed refers to each individual element during the loop.
[wow! - but explains the easy syntax]

"refers to each individual element"

... actually like in the second example.

Which makes me wonder about the second result - xs is [10 20], but Where doesn't find the match.

Re: the new Where?

jthi — Fri, 28 Jun 2024 07:15:39 GMT

If you were to add one more item to your list, Where will break. I cannot really understand the error message

Not all sequences are the same size in access or evaluation of 'Where' , Bad Argument(
	Print(xs);
	xs == [10 20];
), Where/*###*/(Print(xs) ; xs == [10 20])

Names Default To Here(1);

xs = {[10 20], [30 50], [10 20]};
Where(
	Print(xs);
	xs == [10 20];
);

Other example in the Scripting Index seems to be closest to this, but using index 1 doesn't really work in this case.

@EvanMcCorkle

If you make a simple comparison like

Names Default To Here(1);

xs = {[10 20], [30 50], [10 20]};
mask = Repeat(Eval List({[10 20]}), N Items(xs));
xs[1] == mask[1];

you won't get back 1, you will get back [1 1] (so one for each match). You should be able to convert this to 1 using All, but Where doesn't like that either (it seems to only compare first item in the matrix?)

Names Default To Here(1);

xs = {[10 20], [30 20]};

Where(Show(xs == [10 20])); // [1 0] and [0 1]

If you try with xs[1], it will work until you add third item (or remove one) so most likely it is comparing wrong thing

Names Default To Here(1);

xs = {[10 20], [30 20]};

r = Where(
	Show(xs[1] == [10 20]);
	xs[1] == [10 20];
);

Finally with a lot of trial, error and guess work this worked at least from time to time (not really sure if you want to create large matrices for a mask). This isn't what you would expect from Where() to be for this type of comparisons unless it is documented properly

Names Default To Here(1);

xs = {[10 20], [30 20], [10 20]};
mask = Repeat(Eval List({[10 20]}), N Items(xs));

r = Where(
	//show(xs, mask, xs == mask, All(xs == mask));
	All(xs == mask);
);

Also doing it this way seems to be slower than just using Loc() but my test might be flawed

Names Default To Here(1);
testitems = 300;
//xs = {[10 20], [30 20], [10 20]};
xs = Repeat({[10 20]}, testitems / 3);
xs = xs || Repeat({[30 20]}, testitems / 3);
xs = xs || Repeat({[1 2]}, testitems / 3);
xs = xs[Random Shuffle(1::testitems)];


testcount = 10000;

where_speed = Repeat({.}, testcount);
loc_speed = Repeat({.}, testcount);
matches = Repeat({.}, testcount);
category = Repeat({""}, testcount);


expr1 = Expr(
	wait(0);	
	a = HP Time();
	mask = Repeat(Eval List({[10 20]}), N Items(xs));
	r1 = Where(All(xs == mask));
	wait(0);
	b = HP Time();
);

expr2 = Expr(
	wait(0);
	
	c = HP Time();
	r2 = Loc(xs, [10 20]);
	wait(0);
	d = HP Time();	
);

For(i = 1, i <= testcount, i++,
	wait(0);
	If(Random Integer(1, 2) == 2,
		expr2;
		expr1;
		cur_cat = "Loc";
	,
		expr1;
		expr2;
		cur_cat = "Where";
	);
	
	where_speed[i] = b - a;
	loc_speed[i] = d - c;
	category[i] = "Loc";
	matches[i] = Try(All(r1 == r2), 0);
	
	wait(0);
);

dt = New Table("Test",
	Add Rows(testcount),
	New Column("Row", Numeric, Ordinal, Formula(Row())),
	New Column("Locs", Numeric, Continuous, Values(loc_speed)),
	New Column("Where", Numeric, Continuous, Values(where_speed)),
	New Column("Matches", Numeric, Nominal, Values(matches)),
	New Column("Category", Character, Nominal, Values(category))
);

tab = dt << Tabulate(
	Show Control Panel(0),
	Add Table(
		Column Table(Analysis Columns(:Locs, :Where)),
		Row Table(
			Grouping Columns(:Category),
			Statistics(Mean, Median, Std Dev, Interquartile Range)
		)
	)
);

Re: the new Where?

hogi — Fri, 18 Jul 2025 20:19:05 GMT

Thanks for the insights and the trick via Mask - much faster than

Where( xs[0,1] ==10 & xs[0,2] ==20 );

... but no chance to beat Loc.

kind of surprising ...

Re: the new Where?

simopod — Fri, 18 Jul 2025 20:07:42 GMT

Ran into the same error:

Not all sequences are the same size in access or evaluation of 'Where' , Bad Argument

This time when using the "Contains" function. The same argument works as expected when used inside the "Get Rows Where" function.

Here's a minimal example:

Names Default To Here( 1 );
dt = Open( "$SAMPLE_DATA/Big Class.jmp" );
names = {"KATIE", "JAMES"};
r1 = dt << Get Rows Where( Contains(names, :name) );
r2 = Where( Contains(names, :name) );
Show( r1, r2 );

r1 works, r2 will throw the error.

topic Re: the new Where? in Discussions

the new Where?

Re: the new Where?

Re: the new Where?

Re: the new Where?

Re: the new Where?

Re: the new Where?