Re: Appending a list to a list and getting one list.
Re: Appending a list to a list and getting one list.
- Subject: Re: Appending a list to a list and getting one list.
- From: Lachlan Deck <email@hidden>
- Date: Tue, 14 May 2002 11:14:02 +1000
Hi there,
From: Paul Skinner <email@hidden>
set bigListRef to a reference to bigList
set numItems to 10000
set t to (time of (current date)) --Start timing operations.
repeat with n from 1 to numItems
copy n to the end of bigListRef
end repeat
set total to (time of (current date)) - t --End timing.
total --result: 1 second
--But...
set bigList to {}
set bigListRef to a reference to bigList
set numItems to 10000
set t to (time of (current date)) --Start timing operations.
repeat with n from 1 to numItems
set bigListRef to bigListRef & n --SuperSlowDowner!
end repeat
set total to (time of (current date)) - t --End timing.
total --result: 109 seconds
The example made me think that they are doing a similar thing
(concatenating) in the 10K iterations as they show prior to it. Not so.
If you want to concatenate a list to a list then you have to wait.
References aren't a bit of help.
Can anyone tell me how to append a list to a list and get just one
list?
With text lists under 4K text items combined I can...
set l to {"red", "orange", "yellow"}
set l2 to {"green", "blue", "indigo"}
set AppleScript's text item delimiters to "o#?o#?"--Sentinel high ascii
string
set l to every text item of ((l as text) & AppleScript's text item
delimiters & l2 as text)
-->{"red", "orange", "yellow", "green", "blue", "indigo"}
But I am hoping someone can tell me there's a better way.
on cat_lists(main_list, the_lists)
repeat with i from 1 to the number of items in the_lists
set this_list to item i of the_lists
repeat with j from 1 to the number of items in this_list
set the end of main_list to (item j of this_list)
end repeat
end repeat
return main_list
end cat_lists
set a to {"A", "big", "list"}
set b to {"red", "orange", "yellow"}
set c to {"green", "blue", "indigo"}
set a to cat_lists(a, {b, c})
Obviously the order you pass in the lists determines what order they'll
come back as...
--
Paul Skinner
with regards,
--
Lachlan Deck
email@hidden
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