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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?

list-manipulation data

share|improve this question

edited 56 mins ago

user64494

3,58411122

asked 7 hours ago

AlanAlan

6,6431125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  27 mins ago
  

  
  
  
  

add a comment | 

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?

list-manipulation data

share|improve this question

edited 56 mins ago

user64494

3,58411122

asked 7 hours ago

AlanAlan

6,6431125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  27 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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?

list-manipulation data

share|improve this question

edited 56 mins ago

user64494

3,58411122

asked 7 hours ago

AlanAlan

6,6431125

$endgroup$

share|improve this question

edited 56 mins ago

user64494

3,58411122

asked 7 hours ago

AlanAlan

6,6431125

share|improve this question

edited 56 mins ago

user64494

3,58411122

asked 7 hours ago

AlanAlan

6,6431125

edited 56 mins ago

user64494

3,58411122

edited 56 mins ago

user64494

3,58411122

asked 7 hours ago

AlanAlan

6,6431125

asked 7 hours ago

AlanAlan

6,6431125

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

answered 5 hours ago

ciaociao

17.4k138109

$endgroup$

add a comment | 

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

edited 2 hours ago

answered 2 hours ago

Carl WollCarl Woll

73k396189

$endgroup$

add a comment | 

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

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

add a comment | 

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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

answered 5 hours ago

ciaociao

17.4k138109

$endgroup$

add a comment | 

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

answered 5 hours ago

ciaociao

17.4k138109

$endgroup$

add a comment | 

$begingroup$

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

answered 5 hours ago

ciaociao

17.4k138109

$endgroup$

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]];

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

disc[data,edges]

share|improve this answer

answered 5 hours ago

ciaociao

17.4k138109

answered 5 hours ago

ciaociao

17.4k138109

answered 5 hours ago

ciaociao

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

edited 2 hours ago

answered 2 hours ago

Carl WollCarl Woll

73k396189

$endgroup$

add a comment | 

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

edited 2 hours ago

answered 2 hours ago

Carl WollCarl Woll

73k396189

$endgroup$

add a comment | 

$begingroup$

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

edited 2 hours ago

answered 2 hours ago

Carl WollCarl Woll

73k396189

$endgroup$

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
]

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

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Carl WollCarl Woll

73k396189

answered 2 hours ago

Carl WollCarl Woll

73k396189

answered 2 hours ago

Carl WollCarl Woll

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

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

add a comment | 

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

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

add a comment | 

$begingroup$

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

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

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

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

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

share|improve this answer

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166

answered 17 mins ago

Henrik SchumacherHenrik Schumacher

59.5k582166