how to return index of a sorted list

Question

I need  to sort a list and then return a list with the index of the sorted items in the list  For example  if the list I want to sort is  2 3 1 4 5   I need  2 0 1 3 4  to be returned    This question was posted on bytes  but I thought I would repost it here  http   bytes com topic python answers 44513-sorting-list-then-return-index-sorted-item  My specific need to sort a list of objects based on a property of the objects  I then need to re-order a corresponding list to match the order of the newly sorted list    Is there a good way to do this

User · Answer

you can use numpy argsort  or you can do   test     2 3 1 4 5  idxs   list zip  sorted   val  i  for i  val in enumerate test      1    zip will rearange the list so that the first element is test and the second is the idxs

User · Answer

If you need both the sorted list and the list of indices  you could do   L    2 3 1 4 5  from operator import itemgetter indices  L sorted   zip  sorted enumerate L   key itemgetter 1    list L sorted   gt  gt  gt   1  2  3  4  5  list indices   gt  gt  gt   2  0  1  3  4    Or  for Python  lt 2 4  no itemgetter or sorted    temp     v i  for i v in enumerate L   temp sort indices  L sorted   zip  temp    p s  The zip  iterable  idiom reverses the zip process  unzip      Update   To deal with your specific requirements       my specific need to sort a list of objects based on a property of the objects  i then need to re-order a corresponding list to match the order of the newly sorted list      That s a long-winded way of doing it  You can achieve that with a single sort by zipping both lists together then sort using the object property as your sort key  and unzipping after    combined   zip obj list  secondary list  zipped sorted   sorted combined  key lambda x  x 0  some obj attribute  obj list  secondary list   map list  zip  zipped sorted     Here s a simple example  using strings to represent your object  Here we use the length of the string as the key for sorting    str list     banana    apple    nom    Eeeeeeeeeeek   sec list    0 123423  9 231  23  10 11001  temp   sorted zip str list  sec list   key lambda x  len x 0    str list  sec list   map list  zip  temp   str list  gt  gt  gt    nom    apple    banana    Eeeeeeeeeeek   sec list  gt  gt  gt   23  9 231  0 123423  10 11001

User · Answer

You can use the python sorting functions  key parameter to sort the index array instead   gt  gt  gt  s    2  3  1  4  5  3   gt  gt  gt  sorted range len s    key lambda k  s k    2  0  1  5  3  4   gt  gt  gt

User · Answer

You can do this with numpy s argsort method if you have numpy available    gt  gt  gt  import numpy  gt  gt  gt  vals   numpy array  2 3 1 4 5    gt  gt  gt  vals array  2  3  1  4  5    gt  gt  gt  sort index   numpy argsort vals   gt  gt  gt  sort index array  2  0  1  3  4     If not available  taken from this question  this is the fastest method    gt  gt  gt  vals    2 3 1 4 5   gt  gt  gt  sorted range len vals    key vals   getitem     2  0  1  3  4

User · Answer

How about  l1    2 3 1 4 5  l2    l1 index x  for x in sorted l1

User · Answer

Straight out of the documentation for collections OrderedDict    gt  gt  gt    dictionary sorted by value  gt  gt  gt  OrderedDict sorted d items    key lambda t  t 1    OrderedDict    pear   1     orange   2     banana   3     apple   4      Adapted to the example in the original post    gt  gt  gt  l  2 3 1 4 5   gt  gt  gt  OrderedDict sorted enumerate l   key lambda x  x 1    keys    2  0  1  3  4    See http   docs python org library collections html collections OrderedDict for details

User · Answer

What I would do  looking at your specific need   Say you have list a with some values  and your keys are in the attribute x of the objects stored in list b  keys    i j x for i j in zip a  b   a sort key keys   get item      With this method you get your list ordered without having to construct the intermediate permutation list you were asking for

[python] how to return index of a sorted list?

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