Suppose I had the following data
a={1,2,3,4,5}
How would I turn this into
b={{1,2,3}->4,{2,3,4}->5}
That is, transforming the data into rules where each class has a definition as the three before it.
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4 Answers
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a = {1, 2, 3, 4, 5};
(Most[#] -> Last[#]) & /@ Partition[a, 4, 1]
{{1, 2, 3} -> 4, {2, 3, 4} -> 5}
answered Apr 27, 2015 at 18:17
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$\begingroup$ Yes, the 2nd argument to
Partitionis n+1; however, n must have a numeric (positive integer) value. $\endgroup$ Commented Apr 27, 2015 at 18:29
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Try this solution:
ruleTransformer[from___, to_] := {from} -> to
ruleTransformer @@@ Partition[Range[5], 4, 1]
(* {{1, 2, 3} -> 4, {2, 3, 4} -> 5} *)
Partition breaks the data into groups of 4, with start of each group shifted by 1 from the previous. We then apply the helper function ruleTransformer, which takes a list of arguments from___ to be put into the beginning of the Rule, and a single argument to_ to be put at the end of the rule.
We can accomplish the same thing (with a slight modification) with an anonymous
Function:
Function[# -> Reverse[{##2}]] @@@ Reverse /@ Partition[Range[5], 4, 1]
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Developer`PartitionMap[Most@# -> Last@# &, a, 4, 1]
(* {{1, 2, 3} -> 4, {2, 3, 4} -> 5} *)
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♃ = {#, #[[0]] @@ (# /. #[[0]] -> (1 + {##} &))} &@(#[[;; 3]] -> #[[4]] &@#) &;
♃ @ {1, 2, 3, 4, 5}
(* {{1, 2, 3} -> 4, {2, 3, 4} -> 5} *)
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$\begingroup$ ... posted this separately since it belongs to a different ... well, species. In case some/most of your alphabetic keys are not functioning replace
♃with$. $\endgroup$– kglrCommented Apr 28, 2015 at 6:04