Map list to bin numbersquantilization (if that is a word)Map efficiently over duplicates in listGenerating a list of cubefree numbersIs there an equivalent to MATLAB's linspace?Convert a list of hexadecimal numbers to decimalTaking one list Mod a second listlist of items and group of alternative itemsHow find numbers in this list of inequalities?Selecting list entries with a True False index list of similar lengthReplace element in array by checking condition in another listAttempting to fill a table with the number of elements in each bin and make a table with the elements in the bins?

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Map list to bin numbers


quantilization (if that is a word)Map efficiently over duplicates in listGenerating a list of cubefree numbersIs there an equivalent to MATLAB's linspace?Convert a list of hexadecimal numbers to decimalTaking one list Mod a second listlist of items and group of alternative itemsHow find numbers in this list of inequalities?Selecting list entries with a True False index list of similar lengthReplace element in array by checking condition in another listAttempting to fill a table with the number of elements in each bin and make a table with the elements in the bins?













3












$begingroup$


Does WL have the equivalent of Matlab's discretize or NumPy's digitize? I.e., a function that takes a length-N list and a list of bin edges and returns a length-N list of bin numbers, mapping each list item to its bin number?










share|improve this question











$endgroup$











  • $begingroup$
    HistogramList seems similar. This could also be done efficiently with GroupBy and some easy little Compile-d selection determiner. Or maybe hit it first with Sort then write something that only checks the next bin up. Again, can be easily Compile-d.
    $endgroup$
    – b3m2a1
    7 hours ago











  • $begingroup$
    I need it to work like a map (in terms of the order of the items in the resulting list). Of course it is possible to write something ...
    $endgroup$
    – Alan
    6 hours ago











  • $begingroup$
    Related: 140577
    $endgroup$
    – Carl Woll
    2 hours ago







  • 1




    $begingroup$
    Did you try BinCounts? I guess it is what you need.
    $endgroup$
    – Rom38
    1 hour ago










  • $begingroup$
    @Rom38 You probably meant BinLists, right?
    $endgroup$
    – Henrik Schumacher
    49 mins ago















3












$begingroup$


Does WL have the equivalent of Matlab's discretize or NumPy's digitize? I.e., a function that takes a length-N list and a list of bin edges and returns a length-N list of bin numbers, mapping each list item to its bin number?










share|improve this question











$endgroup$











  • $begingroup$
    HistogramList seems similar. This could also be done efficiently with GroupBy and some easy little Compile-d selection determiner. Or maybe hit it first with Sort then write something that only checks the next bin up. Again, can be easily Compile-d.
    $endgroup$
    – b3m2a1
    7 hours ago











  • $begingroup$
    I need it to work like a map (in terms of the order of the items in the resulting list). Of course it is possible to write something ...
    $endgroup$
    – Alan
    6 hours ago











  • $begingroup$
    Related: 140577
    $endgroup$
    – Carl Woll
    2 hours ago







  • 1




    $begingroup$
    Did you try BinCounts? I guess it is what you need.
    $endgroup$
    – Rom38
    1 hour ago










  • $begingroup$
    @Rom38 You probably meant BinLists, right?
    $endgroup$
    – Henrik Schumacher
    49 mins ago













3












3








3





$begingroup$


Does WL have the equivalent of Matlab's discretize or NumPy's digitize? I.e., a function that takes a length-N list and a list of bin edges and returns a length-N list of bin numbers, mapping each list item to its bin number?










share|improve this question











$endgroup$




Does WL have the equivalent of Matlab's discretize or NumPy's digitize? I.e., a function that takes a length-N list and a list of bin edges and returns a length-N list of bin numbers, mapping each list item to its bin number?







list-manipulation data






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









user64494

3,58411122




3,58411122










asked 7 hours ago









AlanAlan

6,6431125




6,6431125











  • $begingroup$
    HistogramList seems similar. This could also be done efficiently with GroupBy and some easy little Compile-d selection determiner. Or maybe hit it first with Sort then write something that only checks the next bin up. Again, can be easily Compile-d.
    $endgroup$
    – b3m2a1
    7 hours ago











  • $begingroup$
    I need it to work like a map (in terms of the order of the items in the resulting list). Of course it is possible to write something ...
    $endgroup$
    – Alan
    6 hours ago











  • $begingroup$
    Related: 140577
    $endgroup$
    – Carl Woll
    2 hours ago







  • 1




    $begingroup$
    Did you try BinCounts? I guess it is what you need.
    $endgroup$
    – Rom38
    1 hour ago










  • $begingroup$
    @Rom38 You probably meant BinLists, right?
    $endgroup$
    – Henrik Schumacher
    49 mins ago
















  • $begingroup$
    HistogramList seems similar. This could also be done efficiently with GroupBy and some easy little Compile-d selection determiner. Or maybe hit it first with Sort then write something that only checks the next bin up. Again, can be easily Compile-d.
    $endgroup$
    – b3m2a1
    7 hours ago











  • $begingroup$
    I need it to work like a map (in terms of the order of the items in the resulting list). Of course it is possible to write something ...
    $endgroup$
    – Alan
    6 hours ago











  • $begingroup$
    Related: 140577
    $endgroup$
    – Carl Woll
    2 hours ago







  • 1




    $begingroup$
    Did you try BinCounts? I guess it is what you need.
    $endgroup$
    – Rom38
    1 hour ago










  • $begingroup$
    @Rom38 You probably meant BinLists, right?
    $endgroup$
    – Henrik Schumacher
    49 mins ago















$begingroup$
HistogramList seems similar. This could also be done efficiently with GroupBy and some easy little Compile-d selection determiner. Or maybe hit it first with Sort then write something that only checks the next bin up. Again, can be easily Compile-d.
$endgroup$
– b3m2a1
7 hours ago





$begingroup$
HistogramList seems similar. This could also be done efficiently with GroupBy and some easy little Compile-d selection determiner. Or maybe hit it first with Sort then write something that only checks the next bin up. Again, can be easily Compile-d.
$endgroup$
– b3m2a1
7 hours ago













$begingroup$
I need it to work like a map (in terms of the order of the items in the resulting list). Of course it is possible to write something ...
$endgroup$
– Alan
6 hours ago





$begingroup$
I need it to work like a map (in terms of the order of the items in the resulting list). Of course it is possible to write something ...
$endgroup$
– Alan
6 hours ago













$begingroup$
Related: 140577
$endgroup$
– Carl Woll
2 hours ago





$begingroup$
Related: 140577
$endgroup$
– Carl Woll
2 hours ago





1




1




$begingroup$
Did you try BinCounts? I guess it is what you need.
$endgroup$
– Rom38
1 hour ago




$begingroup$
Did you try BinCounts? I guess it is what you need.
$endgroup$
– Rom38
1 hour ago












$begingroup$
@Rom38 You probably meant BinLists, right?
$endgroup$
– Henrik Schumacher
49 mins ago




$begingroup$
@Rom38 You probably meant BinLists, right?
$endgroup$
– Henrik Schumacher
49 mins ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

This is a very quick-n-dirty, but may serve as a simple example.



This creates a piecewise function following the first definition in Matlab's discretize documentation, then applies that to the data.



disc[data_, edges_] := Module[e = Partition[edges, 2, 1], p, l,
l = Length@e;
Table[Piecewise[
Append[Table[i, e[[i, 1]] <= x < e[[i, 2]], i, l - 1]
, l,e[[l, 1]] <= x <= e[[l, 2]]]
, "NaN"]
, x, data]];


From the first example in the above referenced documentation:



data=1, 1, 2, 3, 6, 5, 8, 10, 4, 4;
edges=2, 4, 6, 8, 10;

disc[data,edges]



NaN,NaN,1,1,3,2,4,4,2,2




I'm sure there are more efficient/elegant solutions, and will revisit as time permits.






share|improve this answer









$endgroup$




















    2












    $begingroup$

    Here's a version based on Nearest:



    digitize[edges_] := DigitizeFunction[edges, Nearest[edges -> "Index"]]
    digitize[data_, edges_] := digitize[edges][data]

    DigitizeFunction[edges_, nf_NearestFunction][data_] := With[init = nf[data][[All, 1]],
    init + UnitStep[data - edges[[init]]] - 1
    ]


    For example:



    SeedRandom[1]
    data = RandomReal[10, 10]
    digitize[data, 2, 4, 5, 7, 8]



    8.17389, 1.1142, 7.89526, 1.87803, 2.41361, 0.657388, 5.42247, 2.31155, 3.96006, 7.00474



    5, 0, 4, 0, 1, 0, 3, 1, 1, 4




    Note that I broke up the definition of digitize into two pieces, so that if you do this for multiple data sets with the same edges list, you only need to compute the nearest function once.






    share|improve this answer











    $endgroup$




















      1












      $begingroup$

      You may also use Interpolation with InterpolationOrder -> 0. However, employing Nearest as Carl Woll did will usually be much faster.



      First, we prepare the interplating function.



      m = 20;
      binboundaries = Join[-1., Sort[RandomReal[-1, 1, m - 1]], 1.];

      f = Interpolation[Transpose[binboundaries, Range[0, m]], InterpolationOrder -> 0];


      Now you can apply it to lists of values as follows:



      vals = RandomReal[-1, 1, 1000]; 
      Round[f[vals]]





      share|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        This is a very quick-n-dirty, but may serve as a simple example.



        This creates a piecewise function following the first definition in Matlab's discretize documentation, then applies that to the data.



        disc[data_, edges_] := Module[e = Partition[edges, 2, 1], p, l,
        l = Length@e;
        Table[Piecewise[
        Append[Table[i, e[[i, 1]] <= x < e[[i, 2]], i, l - 1]
        , l,e[[l, 1]] <= x <= e[[l, 2]]]
        , "NaN"]
        , x, data]];


        From the first example in the above referenced documentation:



        data=1, 1, 2, 3, 6, 5, 8, 10, 4, 4;
        edges=2, 4, 6, 8, 10;

        disc[data,edges]



        NaN,NaN,1,1,3,2,4,4,2,2




        I'm sure there are more efficient/elegant solutions, and will revisit as time permits.






        share|improve this answer









        $endgroup$

















          5












          $begingroup$

          This is a very quick-n-dirty, but may serve as a simple example.



          This creates a piecewise function following the first definition in Matlab's discretize documentation, then applies that to the data.



          disc[data_, edges_] := Module[e = Partition[edges, 2, 1], p, l,
          l = Length@e;
          Table[Piecewise[
          Append[Table[i, e[[i, 1]] <= x < e[[i, 2]], i, l - 1]
          , l,e[[l, 1]] <= x <= e[[l, 2]]]
          , "NaN"]
          , x, data]];


          From the first example in the above referenced documentation:



          data=1, 1, 2, 3, 6, 5, 8, 10, 4, 4;
          edges=2, 4, 6, 8, 10;

          disc[data,edges]



          NaN,NaN,1,1,3,2,4,4,2,2




          I'm sure there are more efficient/elegant solutions, and will revisit as time permits.






          share|improve this answer









          $endgroup$















            5












            5








            5





            $begingroup$

            This is a very quick-n-dirty, but may serve as a simple example.



            This creates a piecewise function following the first definition in Matlab's discretize documentation, then applies that to the data.



            disc[data_, edges_] := Module[e = Partition[edges, 2, 1], p, l,
            l = Length@e;
            Table[Piecewise[
            Append[Table[i, e[[i, 1]] <= x < e[[i, 2]], i, l - 1]
            , l,e[[l, 1]] <= x <= e[[l, 2]]]
            , "NaN"]
            , x, data]];


            From the first example in the above referenced documentation:



            data=1, 1, 2, 3, 6, 5, 8, 10, 4, 4;
            edges=2, 4, 6, 8, 10;

            disc[data,edges]



            NaN,NaN,1,1,3,2,4,4,2,2




            I'm sure there are more efficient/elegant solutions, and will revisit as time permits.






            share|improve this answer









            $endgroup$



            This is a very quick-n-dirty, but may serve as a simple example.



            This creates a piecewise function following the first definition in Matlab's discretize documentation, then applies that to the data.



            disc[data_, edges_] := Module[e = Partition[edges, 2, 1], p, l,
            l = Length@e;
            Table[Piecewise[
            Append[Table[i, e[[i, 1]] <= x < e[[i, 2]], i, l - 1]
            , l,e[[l, 1]] <= x <= e[[l, 2]]]
            , "NaN"]
            , x, data]];


            From the first example in the above referenced documentation:



            data=1, 1, 2, 3, 6, 5, 8, 10, 4, 4;
            edges=2, 4, 6, 8, 10;

            disc[data,edges]



            NaN,NaN,1,1,3,2,4,4,2,2




            I'm sure there are more efficient/elegant solutions, and will revisit as time permits.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 6 hours ago









            ciaociao

            17.4k138109




            17.4k138109





















                2












                $begingroup$

                Here's a version based on Nearest:



                digitize[edges_] := DigitizeFunction[edges, Nearest[edges -> "Index"]]
                digitize[data_, edges_] := digitize[edges][data]

                DigitizeFunction[edges_, nf_NearestFunction][data_] := With[init = nf[data][[All, 1]],
                init + UnitStep[data - edges[[init]]] - 1
                ]


                For example:



                SeedRandom[1]
                data = RandomReal[10, 10]
                digitize[data, 2, 4, 5, 7, 8]



                8.17389, 1.1142, 7.89526, 1.87803, 2.41361, 0.657388, 5.42247, 2.31155, 3.96006, 7.00474



                5, 0, 4, 0, 1, 0, 3, 1, 1, 4




                Note that I broke up the definition of digitize into two pieces, so that if you do this for multiple data sets with the same edges list, you only need to compute the nearest function once.






                share|improve this answer











                $endgroup$

















                  2












                  $begingroup$

                  Here's a version based on Nearest:



                  digitize[edges_] := DigitizeFunction[edges, Nearest[edges -> "Index"]]
                  digitize[data_, edges_] := digitize[edges][data]

                  DigitizeFunction[edges_, nf_NearestFunction][data_] := With[init = nf[data][[All, 1]],
                  init + UnitStep[data - edges[[init]]] - 1
                  ]


                  For example:



                  SeedRandom[1]
                  data = RandomReal[10, 10]
                  digitize[data, 2, 4, 5, 7, 8]



                  8.17389, 1.1142, 7.89526, 1.87803, 2.41361, 0.657388, 5.42247, 2.31155, 3.96006, 7.00474



                  5, 0, 4, 0, 1, 0, 3, 1, 1, 4




                  Note that I broke up the definition of digitize into two pieces, so that if you do this for multiple data sets with the same edges list, you only need to compute the nearest function once.






                  share|improve this answer











                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    Here's a version based on Nearest:



                    digitize[edges_] := DigitizeFunction[edges, Nearest[edges -> "Index"]]
                    digitize[data_, edges_] := digitize[edges][data]

                    DigitizeFunction[edges_, nf_NearestFunction][data_] := With[init = nf[data][[All, 1]],
                    init + UnitStep[data - edges[[init]]] - 1
                    ]


                    For example:



                    SeedRandom[1]
                    data = RandomReal[10, 10]
                    digitize[data, 2, 4, 5, 7, 8]



                    8.17389, 1.1142, 7.89526, 1.87803, 2.41361, 0.657388, 5.42247, 2.31155, 3.96006, 7.00474



                    5, 0, 4, 0, 1, 0, 3, 1, 1, 4




                    Note that I broke up the definition of digitize into two pieces, so that if you do this for multiple data sets with the same edges list, you only need to compute the nearest function once.






                    share|improve this answer











                    $endgroup$



                    Here's a version based on Nearest:



                    digitize[edges_] := DigitizeFunction[edges, Nearest[edges -> "Index"]]
                    digitize[data_, edges_] := digitize[edges][data]

                    DigitizeFunction[edges_, nf_NearestFunction][data_] := With[init = nf[data][[All, 1]],
                    init + UnitStep[data - edges[[init]]] - 1
                    ]


                    For example:



                    SeedRandom[1]
                    data = RandomReal[10, 10]
                    digitize[data, 2, 4, 5, 7, 8]



                    8.17389, 1.1142, 7.89526, 1.87803, 2.41361, 0.657388, 5.42247, 2.31155, 3.96006, 7.00474



                    5, 0, 4, 0, 1, 0, 3, 1, 1, 4




                    Note that I broke up the definition of digitize into two pieces, so that if you do this for multiple data sets with the same edges list, you only need to compute the nearest function once.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 2 hours ago

























                    answered 3 hours ago









                    Carl WollCarl Woll

                    73k396189




                    73k396189





















                        1












                        $begingroup$

                        You may also use Interpolation with InterpolationOrder -> 0. However, employing Nearest as Carl Woll did will usually be much faster.



                        First, we prepare the interplating function.



                        m = 20;
                        binboundaries = Join[-1., Sort[RandomReal[-1, 1, m - 1]], 1.];

                        f = Interpolation[Transpose[binboundaries, Range[0, m]], InterpolationOrder -> 0];


                        Now you can apply it to lists of values as follows:



                        vals = RandomReal[-1, 1, 1000]; 
                        Round[f[vals]]





                        share|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          You may also use Interpolation with InterpolationOrder -> 0. However, employing Nearest as Carl Woll did will usually be much faster.



                          First, we prepare the interplating function.



                          m = 20;
                          binboundaries = Join[-1., Sort[RandomReal[-1, 1, m - 1]], 1.];

                          f = Interpolation[Transpose[binboundaries, Range[0, m]], InterpolationOrder -> 0];


                          Now you can apply it to lists of values as follows:



                          vals = RandomReal[-1, 1, 1000]; 
                          Round[f[vals]]





                          share|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            You may also use Interpolation with InterpolationOrder -> 0. However, employing Nearest as Carl Woll did will usually be much faster.



                            First, we prepare the interplating function.



                            m = 20;
                            binboundaries = Join[-1., Sort[RandomReal[-1, 1, m - 1]], 1.];

                            f = Interpolation[Transpose[binboundaries, Range[0, m]], InterpolationOrder -> 0];


                            Now you can apply it to lists of values as follows:



                            vals = RandomReal[-1, 1, 1000]; 
                            Round[f[vals]]





                            share|improve this answer









                            $endgroup$



                            You may also use Interpolation with InterpolationOrder -> 0. However, employing Nearest as Carl Woll did will usually be much faster.



                            First, we prepare the interplating function.



                            m = 20;
                            binboundaries = Join[-1., Sort[RandomReal[-1, 1, m - 1]], 1.];

                            f = Interpolation[Transpose[binboundaries, Range[0, m]], InterpolationOrder -> 0];


                            Now you can apply it to lists of values as follows:



                            vals = RandomReal[-1, 1, 1000]; 
                            Round[f[vals]]






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 39 mins ago









                            Henrik SchumacherHenrik Schumacher

                            59.5k582166




                            59.5k582166



























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