4
$\begingroup$

I would like to reverse the effect of Counts, i.e., starting with

x = <|a -> 4, b -> 2, c -> 1|>

I would like to reconstruct a sorted list

y = {a, a, a, a, b, b, c}

such that Counts[y] == x. I have come up with

y = Join @@ ConstantArray @@@ Normal @ x

but I'd like to know if there's anything more elegant than this, with less intermediary expressions.

Improve this question
$\endgroup$
3
  • $\begingroup$ Essentially a duplicate on Stack Overflow: (763915) $\endgroup$
    – Mr.Wizard
    Apr 26, 2017 at 16:27
  • $\begingroup$ Wow, I would never find that question in search for relateds, @Mr.Wizard. But what I take from your answer there is that the method I'm suggesting is about as good as it gets, right? $\endgroup$
    – The Vee
    Apr 26, 2017 at 17:31
  • $\begingroup$ Well, it's what I would use, still; you can take that for what it's worth. :-) $\endgroup$
    – Mr.Wizard
    Apr 26, 2017 at 19:17

3 Answers 3

Reset to default
5
$\begingroup$ 
Catenate @ KeyValueMap[ConstantArray] @ x

Related topics:

Reverse DeleteDuplicates using Information from Tally

Generating repeated elements in a list

Replicate sublist in new list

List creation/manipulation

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ Catenate @ KeyValueMap[Table] @ x ought to work as well. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Apr 19, 2017 at 11:16
  • 1
    $\begingroup$ @J.M. yep, I think ConstantArray has better performance but I don't remember where to find benchmarks, I just use it by default. p.s. any idea why only the first link is automatically formatted? $\endgroup$
    – Kuba ♦
    Apr 19, 2017 at 11:18
  • $\begingroup$ I changed your links to use https://. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Apr 19, 2017 at 11:25
  • 1
    $\begingroup$ @Kuba A little and I think we even can ignored it. $\endgroup$
    – yode
    Apr 26, 2017 at 11:28
  • 1
    $\begingroup$ @Kuba I believe the key difference is that if it is numeric input, ConstantArray generates packed arrays. But, Catenate doesn't respect packed arrays, so you instead have to use Join @@ KeyValueMap[ConstantArray]@<|1 -> 4, 2 -> 2, 3 -> 1|>, instead. $\endgroup$
    – rcollyer
    Apr 26, 2017 at 16:23
2
$\begingroup$ 
KeyValueMap[Table /* Apply[Sequence], x]

{a, a, a, a, b, b, c}

Improve this answer
$\endgroup$
1
$\begingroup$

Multiplication of Matrices

In:

xs = <|a -> 4, b -> 2, c -> 1|>;
A = Table[1, #] & /@ Values[xs]
B = Keys[xs]
A B // Flatten

Out:

{{1, 1, 1, 1}, {1, 1}, {1}}
{a, b, c}
{a, a, a, a, b, b, c}
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.