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Finding positions of minimum values in list

List manipulation: position & max value combinationHow to efficiently find positions of duplicates?Finding elements from a sparse matrixInsert at specific resulting positions?Insert at specific resulting positions in multidimensional list?How can I get a list of the positions of the running minima in a list?Add values from a list, at positions from a list, to a matrixPositions of elements in a listAssemble rule replacement from pairs of positionsList positions without bracketsFinding the minimum value in a column and extracting the row where the minimum value placed there

3

1

$begingroup$

I would like to find the positions of multiple occurrences of a minimum value in a list. For instance, for the list {1, 2, 1, 3}, I want to obtain the two positions for the occurrences of the number 1.

MinimalBy[{1, 2, 1, 3}, id (x)] gives me these occurrences, but not the positions.

list-manipulation

share|improve this question

asked 3 hours ago

MikeMike

926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  MinimalBy[Range@Length@list, list[[#]] &]?
  $endgroup$
  – Michael E2
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or modifying this answer: minpos[a_] := SparseArray[UnitStep[# - a]]["AdjacencyLists"] &@Min@a and then minpos[list].
  $endgroup$
  – Michael E2
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate: mathematica.stackexchange.com/questions/40231/…
  $endgroup$
  – Michael E2
  3 hours ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

I would like to find the positions of multiple occurrences of a minimum value in a list. For instance, for the list {1, 2, 1, 3}, I want to obtain the two positions for the occurrences of the number 1.

MinimalBy[{1, 2, 1, 3}, id (x)] gives me these occurrences, but not the positions.

list-manipulation

share|improve this question

asked 3 hours ago

MikeMike

926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  MinimalBy[Range@Length@list, list[[#]] &]?
  $endgroup$
  – Michael E2
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or modifying this answer: minpos[a_] := SparseArray[UnitStep[# - a]]["AdjacencyLists"] &@Min@a and then minpos[list].
  $endgroup$
  – Michael E2
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate: mathematica.stackexchange.com/questions/40231/…
  $endgroup$
  – Michael E2
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I would like to find the positions of multiple occurrences of a minimum value in a list. For instance, for the list {1, 2, 1, 3}, I want to obtain the two positions for the occurrences of the number 1.

MinimalBy[{1, 2, 1, 3}, id (x)] gives me these occurrences, but not the positions.

list-manipulation

share|improve this question

asked 3 hours ago

MikeMike

926

$endgroup$

share|improve this question

asked 3 hours ago

MikeMike

926

share|improve this question

asked 3 hours ago

MikeMike

926

asked 3 hours ago

MikeMike

926

asked 3 hours ago

MikeMike

926

1 Answer
1

active

oldest

votes

3

$begingroup$

lst = {1, 2, 1, 3};

Random`Private`PositionsOf[lst, Min @ lst]

{1, 3}

Also

 Flatten@Position[lst, Min @ lst]

{1, 3}

share|improve this answer

edited 2 hours ago

answered 2 hours ago

kglrkglr

190k10209427

$endgroup$

add a comment | 

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3

$begingroup$

lst = {1, 2, 1, 3};

Random`Private`PositionsOf[lst, Min @ lst]

{1, 3}

Also

 Flatten@Position[lst, Min @ lst]

{1, 3}

share|improve this answer

edited 2 hours ago

answered 2 hours ago

kglrkglr

190k10209427

$endgroup$

add a comment | 

3

$begingroup$

lst = {1, 2, 1, 3};

Random`Private`PositionsOf[lst, Min @ lst]

{1, 3}

Also

 Flatten@Position[lst, Min @ lst]

{1, 3}

share|improve this answer

edited 2 hours ago

answered 2 hours ago

kglrkglr

190k10209427

$endgroup$

add a comment | 

$begingroup$

lst = {1, 2, 1, 3};

Random`Private`PositionsOf[lst, Min @ lst]

{1, 3}

Also

 Flatten@Position[lst, Min @ lst]

{1, 3}

share|improve this answer

edited 2 hours ago

answered 2 hours ago

kglrkglr

190k10209427

$endgroup$

lst = {1, 2, 1, 3};

Random`Private`PositionsOf[lst, Min @ lst]

 Flatten@Position[lst, Min @ lst]

share|improve this answer

edited 2 hours ago

answered 2 hours ago

kglrkglr

190k10209427

answered 2 hours ago

kglrkglr

190k10209427

answered 2 hours ago

kglrkglr

190k10209427