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Topic: List help (Read 3404 times)
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trevmeister66
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«
on:
November 16, 2007, 01:44:00 am »
What would be the fastest way to move every value in a list up one position in the list (thus deleting the last number in the list, and adding a new one at the beginning)?
Example:
{1,2,3,4,5->L1
//What would I do here?
#->L1(1
L1 = {#,1,2,3,4
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JonimusPrime
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Reply #1 on:
November 16, 2007, 02:06:00 am »
c1-->
CODE
ec1
Agument({#},L1->
dim(L1)-1->L1c2
ec2
that should do it but ti think I spelled that comand wrong.
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Reply #2 on:
November 16, 2007, 04:00:00 am »
Augment*?
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trevmeister66
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Reply #3 on:
November 16, 2007, 09:46:00 am »
Sweet it worked
Thanks.
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Speler
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Reply #4 on:
November 16, 2007, 02:15:00 pm »
No problem (See! I can be helpful sometimes!)
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JonimusPrime
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Reply #5 on:
November 17, 2007, 05:32:00 am »
QuoteBegin-Super Speler+16 Nov, 2007, 20:15-->
QUOTE
(Super Speler @ 16 Nov, 2007, 20:15)
No problem (See! I can be helpful sometimes!)
Wait I posted the code Not you Speler? :
:
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Speler
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Reply #6 on:
November 17, 2007, 06:56:00 am »
I was posting in the wrong topic... sorry.
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JoostinOnline
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Reply #7 on:
November 17, 2007, 09:16:00 am »
QuoteBegin-Super Speler+17 Nov, 2007, 12:56-->
QUOTE
(Super Speler @ 17 Nov, 2007, 12:56)
I was posting in the wrong topic... sorry.
Of course you were
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Harrierfalcon
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«
Reply #8 on:
November 17, 2007, 09:40:00 am »
c1-->
CODE
ec1?List(cumSum(L1→L1c2
ec2(A lot faster during 100 repetitions) and smaller by 10 bytes.
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simplethinker
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snjwffl
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«
Reply #9 on:
November 17, 2007, 10:26:00 am »
That cuts out the first element and doesn't add on a new one.
@trevmeister66, if you wanted to move everything down (left in your list) and cut off the first element and add a new number to the end(which is the exact opposite of what you're doing now) you can use c1-->
CODE
ec1?List(cumsum(augment(L1,{Nc2
ec2
which is 7 shorter. All you would have to do is switch your list around.
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