List comparison questions

Scripting Forums AppleScript | Mac OS X
#1

I came across this issue last week, and am hoping someone can offer an explanation or some guidance.

Given a long list of 2 numbers, I need to find any matching instances, but this will not work:

set a to {{1, 2}, {2, 3}, {3, 4}}
set b to {2, 3}
a contains b
-->false

Neither will this:

set a to {{1, 2}, {2, 3}, {3, 4}}
set b to {2, 3}
set the_count to 0
repeat with c in a
	if c is equal to b then
		set the_count to the_count + 1
	end if
end repeat
the_count
-->0

But, if you test a single item of a list to see if it contains another list, it works:

set a to {{1, 2}, {2, 3}, {3, 4}}
set b to {2, 3}
set the_count to 0
repeat with c in a
	if c contains b then
		set the_count to the_count + 1
	end if
end repeat
the_count
-->1

Is there some kind of rule that needs to be remembered when trying to find out whether or not a list is already present in another list?

#2

Your second one works with this change:

set a to {{1, 2}, {2, 3}, {1, 2}}
set b to {1, 2}
set the_count to 0
repeat with c in a
	if contents of c is equal to contents of b then
		set the_count to the_count + 1
	end if
end repeat
the_count
--> 2

With several variations, I cannot get the first to work, but cannot explain why.

#3

Hi,

Here’s an article:

http://www.applescriptsourcebook.com/tips/gotchas/listoflists.html

Basically, AppleScript requires list of lists on both sides of the ‘contained by’ operator.

set a to {{1, 2}, {2, 3}, {3, 4}}
set b to {{2, 3}}
a contains b

gl,

#4

kel:

Thanks loads; good article! That speeded things up nicely.