=1), {1,1}(N=1)...{1,15}(N=1). While Y is made of {0}(N=6), {1}(N=15).
What I need to do, is to cycle through all the branched items of the X lists with every single item of the corresponding list of Y.
So:
{0,0}(N=1) with 1st item of Y{0},
{0,1}(N=1) with 1st item of Y{0}
...
{0,15}(N=1) with 1st item of Y{0} (and this is the first cycle).
Then (cycle 2)
{0,0}(N=1) with 2nd item of Y{0},
{0,1}(N=1) with 2nd item of Y{0}
...
{0,15}(N=1) with 2nd item of Y{0}
Then do the same thing with the X list {1,0} .. {1,15} with the corresponding Y{1}
As a result I need the following list structure:
{0,0}N=6, {0,1}N=6, {0,2}N=6 ... {0,15}N=6,
{1,0}N=15, {1,1}N=15, {1,2}N=15 ... {0,15}N=15.
If I use the component you suggested "shift path" it will match the items per item-order instead of cycling through all the items of the first sublist...
Hope this makes sense.
Any suggestion?…
e aquí!
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you want each "element" to be a single Item or a single item for ALL elements. See Below
0. 20
1. 30
2. 59
3. 60
4. {9,45,29}
5. 0.0
6. 3.0
7. 6.0
Or
0. 20 30 59 60 {9,45,29} 0.0 3.0 6.0
…
Added by Danny Boyes at 3:13am on October 29, 2013