the one-but-last list [4]. After running out of the n- items avalaible it should continue with the second item of list 0 and so on for all items on all the lists.
Intput, six lists of 30 items each
[0] (n=30)
[1] (n=30)
[2] (n=30)
[3] (n=30)
[4] (n=30)
[5] (n=30)
Output, 18 lists of 10 items each
[0],i=0;[5],i=4; [4],i=7;...
[0],i=1;[5],i=5; [4],i=8;...
...
[5],i=1;[4],i=5; [3],i=0;...
I thought perhaps the weave component or the relative tree item component but didn't manage to figure out how to compose the mask. I couldn't find much on how to use these. I guess it should wrap the lists, but not the items.
Any help would be greatly appreciated.…
Added by Thorsten Lang at 2:27am on January 24, 2011
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013