Use the
delete (1 v) of [list v]block
Last edited by topazdragonlord (2013-04-10 15:41:07)
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perhaps you could try....
delete (1 v) of [list v]//just change the 1 to a customised number add [number] to [Replica List v]and then if you mean compare them maybe you could use:
say (List) for (4) secs say (Replica List) for (4) secsIs that what you meant?
Last edited by topazdragonlord (2013-04-10 16:07:52)
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No... .
Try this:
set [a v] to [1] delete (all v) of [templist v] repeat (20) if <not <[templist v] contains (item (a) of [yourlist v])>> add (item (a) of [yourlist v]) to [templist v] end change [a v] by (1) endUsing yourlist as the place the numbers are, the output will be in templist and contain no duplicates.
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Hmm... well there's a couple way you could do this. You could scan the list for each number checking for duplicate instances. For example,
set (i) to [0] repeat until <not <(i) < (length of [list])>> set (temp) to item (i) of list set (i2) to 0 repeat until <not <(i2) < (length of [list])>> if <(item (i2) of [list]) = (temp)> delete item (i2) of [list] end end endOf course, that's horribly inefficient. If your list is sorted you could make one that's more efficient by searching only the indexes adjacent to each item.
Last edited by Harakou (2013-04-10 23:55:23)
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