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Is it possible to split a list into two lists at a specific position? The main list is for example:


I want to split this list into two new lists:




at a specific y-value at position i. I was not successful in using Part. Maybe there is another possibility?

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up vote 8 down vote accepted

This can be accomplished easily using Part ([[ ]]) and Span (;;), as follows:

data = {{x1, y1}, {x2, y2}, {x3, y3}, {x4, y4}, {xz, yz}};
data[[;; 3]]
data[[4 ;;]]

(* ->
{{x1, y1}, {x2, y2}, {x3, y3}}
{{x4, y4}, {xz, yz}}
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You beat me to it, so I added my answer to yours. – rcollyer May 5 '12 at 19:34
If you don't want Part[], there's always Take[]... – J. M. May 5 '12 at 19:43

Following up on J.M.'s suggestion,

m = 3;
data1 = Take[data, m]
data2 = Take[data, -(Length[data] - m)]

You might also obtain data2 as follows:

data2 = Complement[data, data1]

I'm uncertain whether the second approach would maintain order invariant if there are identical sublists.

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Or, perhaps, {Take[#, 3], Drop[ #, 3]} &@data – user 106 May 5 '12 at 22:58
Surprisingly this is faster than Part as recommended by belisarius, and considerably faster than Take/Drop as recommended by Tom. It could also be written: {Take[data, m], Take[data, {m + 1, -1}]} with about the same performance. – Mr.Wizard May 6 '12 at 0:01
I'm surprised it's faster than Take...Drop. – DavidC May 6 '12 at 0:32

Starting from 10.2 you can use TakeDrop

data = {{x1, y1}, {x2, y2}, {x3, y3}, {x4, y4}, {xz, yz}};

{data1, data2} = TakeDrop[data, 3]
(* {{{x1, y1}, {x2, y2}, {x3, y3}}, {{x4, y4}, {xz, yz}}} *)
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