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Say I have a list called A and I want to divide it into two separate list B and C by looping through A and picking out one element to add to B and one to add to A throughout the length of A.

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    $\begingroup$ Welcome to Mma.SE. Start by taking the tour now and learning about asking and what's on-topic. Always edit if improvable, show due diligence, give brief context, include minimal working example of code and data in formatted form. By doing all this you help us to help you and likely you will inspire great answers. You need to provide us with: the data, the criteria to Select elements, which Cases define how to Split the list, and most importantly: the code of what you have tried so far. $\endgroup$
    – rhermans
    Nov 1 '18 at 18:25
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    $\begingroup$ A hint: the best solution will not use a For loop, but probably something in the lines of Select, Cases or Split. Your question may be put on-hold because its unanswerable for lack of details. We can't help unless you edit your question to improve it and make it specific, with all the details one would need to reproduce your problem exactly. Please don't be discouraged by that cleaning-up policy. Your questions are and will be most welcomed. Learn about good questions here. $\endgroup$
    – rhermans
    Nov 1 '18 at 18:28
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    $\begingroup$ A For loop is a very awkward and inefficient way to accomplish what you seek. By what criterion do you place elements into $B$ or $C$? Simple alternates? $\endgroup$
    – David G. Stork
    Nov 1 '18 at 18:30
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    $\begingroup$ Yes, Simple alternatives or even random separation $\endgroup$
    – Bryan
    Nov 1 '18 at 18:34
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Don't use For. Really. Never.

There are much more elegant ways, for example

a = Range[20];
{b, c} = Transpose[Partition[a, 2]]

{{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}, {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}}

or (should be the fastest; look up Span to understand it)

b = a[[1;; ;;2]]
c = a[[2;; ;;2]]

{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

or

{b, c} = GatherBy[a, OddQ]

{{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}, {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}}

or

b = Select[a, OddQ]
c = Select[a, EvenQ]

{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

In the last two examples, you may replace OddQ and EvenQ by any other function that returns True or False in order to realize other decision rules.

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  • $\begingroup$ Only the middle approach is robust. +1 for it. $\endgroup$
    – Bob Hanlon
    Nov 2 '18 at 0:25
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As it turns out, there is a special function for taking every other element called DownSample:

list = Range[20];
a = Downsample[list, 2]
b = Downsample[list, 2, 2]

a has all the odd values, b has all the evens.

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list = Range@20;

here are alternatives 

TakeDrop[list, {1, -1, 2}]

{{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}, {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}}

and here is random separation 

{d = RandomSample[s, Length@list/2], Complement[list, d]}

{{4, 12, 5, 19, 3, 11, 17, 16, 6, 10}, {1, 2, 7, 8, 9, 13, 14, 15, 18, 20}}

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