# Convert frequency counts to long notation in Mathematica

I have a series of frequencies of values, e.g. 5 times 1, 10 times 2 and 5 times 3, as in

list={{1,5},{2,10},{3,5}}


and I would like to convert this to long notation as in

list2={1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3}


what is the most elegant way to do this in Mathematica?

I added timings - 3rd from the bottom is fastest. I am sure there are faster versions. If speed is important you can parallelize or come up with a Compile-ed solution.

In:= list = RandomInteger[{3, 12}, {10^7, 2}];

In:= list // DeveloperPackedArrayQ
Out= True

In:= Table[#1, {#2}] & @@@ list // Flatten; // AbsoluteTiming
Out= {22.015290, Null}

In:= Join @@ (Table[#1, {#2}] & @@@ list); // AbsoluteTiming
Out= {18.528328, Null}

In:= Join @@ ConstantArray @@@ list; // AbsoluteTiming
Out= {18.261945, Null}

In:= ConstantArray[#1, #2] & @@@ list // Flatten; // AbsoluteTiming
Out= {43.177745, Null}

In:= NestList[# &, #1, #2 - 1] & @@@ list // Flatten; // AbsoluteTiming
Out= {30.278883, Null}

Out= {15.465663, Null}

Out= {40.184748, Null}

Out= {18.716637, Null}

In:= Inner[ConstantArray, Sequence @@ Transpose@list, Join]; // AbsoluteTiming
Out= {16.525300, Null}

• Thx millions for the many possible solutions!! Apr 12, 2014 at 10:25
• @TomWenseleers I added some timings ;-) Apr 12, 2014 at 10:30
• Great - many thx! Apr 12, 2014 at 10:39
• I am getting better timings with Join @@ ConstantArray @@@ list: about 50 percent of the timings for ConstantArray[#1, #2] & @@@ list // Flatten and about as good as Join @@MapThread[ConstantArray, Thread[list]]. (+1)
– kglr
Apr 12, 2014 at 11:00
• ... Similarly for Inner[ConstantArray, Sequence @@ Transpose@list, Join] :)
– kglr
Apr 12, 2014 at 11:56

## InternalRepetitionFromMultiplicity

list = {{1, 5}, {2, 10}, {3, 5}};
InternalRepetitionFromMultiplicity @ list


{1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3}

This is faster than the fastest method in @Vitaliy's post.

list = RandomInteger[{3, 12}, {10^7, 2}];


8.5771

res2 = InternalRepetitionFromMultiplicity[list]; // AbsoluteTiming // First


6.6958

res1 == res2


True

• Where did you dig that one out? Jan 14, 2018 at 9:42
• @HenrikSchumacher, some time back, searching (I think) for *Repeated*, I mistyped ?? **Repet* and this was one in a short list of results. As the name suggests what it does and the syntax, my first or second guess worked:)
– kglr
Jan 14, 2018 at 10:17
• Great intuition! =D Jan 14, 2018 at 10:20

kglr made a very interesting find with InternalRepetitionFromMultiplicity. However, InternalRepetitionFromMultiplicity produces unpacked arrays and that tells me that it is not as efficient as it could be.

Here is an attempt to produce a compiled version that also allows for parallelization:

getRepetitionFromMultiplicity =
Compile[{{list, _Integer, 2}, {start, _Integer}, {stop, _Integer}},
Block[{a, x, y, c = 0},
a = Table[0, {i, 1, Total[list[[start ;; stop, 2]]]}];
Do[
x = CompileGetElement[list, i, 1];
y = CompileGetElement[list, i, 2];
Do[c++; a[[c]] = x, {i, 1, y}],
{i, start, stop}
];
a
],
CompilationTarget -> "C",
RuntimeAttributes -> {Listable},
Parallelization -> True,
RuntimeOptions -> "Speed"
];

repetitionFromMultiplicity[list_?MatrixQ, jobs_: 1000] :=
Module[{len, starts, stops},
If[jobs <= Length[list],
len = Floor[Length[list]/jobs];
starts = len Range[0, jobs - 1] + 1;
stops = len Range[1, jobs];
stops[[-1]] = Length[list];
Join @@ getRepetitionFromMultiplicity[list, starts, stops]
,
getRepetitionFromMultiplicity[list, 1, Length[list]]
]
]


These are the timings (on a quad core machine):

list = RandomInteger[{3, 12}, {10^7 + 1, 2}];
res2 = InternalRepetitionFromMultiplicity[list]; // AbsoluteTiming // First
res3 = repetitionFromMultiplicity[list]; // AbsoluteTiming // First
DeveloperToPackedArray@res2 == res3
`

4.85631

0.586881

True