How to sort two lists which reference each other in the exact same way

Question

Say I have two lists   list1    3  2  4  1  1  list2     three    two    four    one    one2     If I run list1 sort    it ll sort it to  1 1 2 3 4  but is there a way to get list2 in sync as well  so I can say item 4 belongs to  three     So  the expected output would be   list1    1  1  2  3  4  list2     one    one2    two    three    four     My problem is I have a pretty complex program that is working fine with lists but I sort of need to start referencing some data  I know this is a perfect situation for dictionaries but I m trying to avoid dictionaries in my processing because I do need to sort the key values  if I must use dictionaries I know how to use them    Basically the nature of this program is  the data comes in a random order  like above   I need to sort it  process it and then send out the results  order doesn t matter but users need to know which result belongs to which key   I thought about putting it in a dictionary first  then sorting list one but I would have no way of differentiating of items in the with the same value if order is not maintained  it may have an impact when communicating the results to users    So ideally  once I get the lists I would rather figure out a way to sort both lists together   Is this possible

User · Answer

I would like to suggest a solution if you need to sort more than 2 lists in sync  def SortAndSyncList Multi ListToSort   ListsToSync       y   sorted zip ListToSort  zip  ListsToSync        w    n for n in zip  y       return list w 0    tuple list a  for a in zip  w 1

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I have used the answer given by senderle for a long time until I discovered np argsort  Here is how it works     idx works on np array and not lists  list1   np array  3 2 4 1   list2   np array   three   two   four   one    idx     np argsort list1   list1   np array list1  idx  list2   np array list2  idx    I find this solution more intuitive  and it works really well  The perfomance   def sorting l1  l2         l1 and l2 has to be numpy arrays     idx   np argsort l1      return l1 idx   l2 idx     list1 and list2 are np arrays here     timeit sorting list1  list2  100000 loops  best of 3  3 53 us per loop    This works best when the lists are NOT np array  timeit zip  sorted zip list1  list2    100000 loops  best of 3  2 41 us per loop    0 01us better for np array  I think this is negligible   timeit tups   zip list1  list2   tups sort    zip  tups  100000 loops  best for 3 loops  1 96 us per loop   Even though np argsort isn t the fastest one  I find it easier to use

User · Answer

You can use the zip   and sort   functions to accomplish this   Python 2 6 5  r265 79063  Jun 12 2010  17 07 01   GCC 4 3 4 20090804  release  1  on cygwin  gt  gt  gt  list1    3 2 4 1 1   gt  gt  gt  list2     three    two    four    one    one2    gt  gt  gt  zipped   zip list1  list2   gt  gt  gt  zipped sort    gt  gt  gt  slist1    i for  i  s  in zipped   gt  gt  gt  slist1  1  1  2  3  4   gt  gt  gt  slist2    s for  i  s  in zipped   gt  gt  gt  slist2   one    one2    two    three    four     Hope this helps

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You can sort indexes using values as keys   indexes   range len list1   indexes sort key list1   getitem      To get sorted lists given sorted indexes   sorted list1   map list1   getitem    indexes  sorted list2   map list2   getitem    indexes    In your case you shouldn t have list1  list2 but rather a single list of pairs   data     3   three     2   two     4   four     1   one     1   one2      It is easy to create  it is easy to sort in Python   data sort     sort using a pair as a key   Sort by the first value only   data sort key lambda pair  pair 0

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You can use the key argument in sorted   method unless you have two same values in list2   The code is given below   sorted list2  key   lambda x  list1 list2 index x       It sorts list2 according to corresponding values in list1  but make sure that while using this  no two values in list2 evaluate to be equal because list index   function give the first value

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an algorithmic solution   list1    3 2 4 1  1  list2     three    two    four    one    one2     lis     list1 i   list2 i   for i in range len list1    list1 sort   list2    x 1  for i in range len list1   for x in lis if x 0     i    Outputs  - gt  Output speed  0 2s   gt  gt  gt list1  gt  gt  gt  1  1  2  3  4   gt  gt  gt list2  gt  gt  gt   one    one2    two    three    four

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Schwartzian transform  The built-in Python sorting is stable  so the two 1s don t cause a problem    gt  gt  gt  l1    3  2  4  1  1   gt  gt  gt  l2     three    two    four    one    second one    gt  gt  gt  zip  sorted zip l1  l2      1  1  2  3  4     one    second one    two    three    four

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newsource    newtarget    for valueT in targetFiles      for valueS in sourceFiles              l1 len valueS  l2 len valueT               j 0             while  j lt  l1                       if  str valueT     valueS j l1                                 newsource append valueS                              newtarget append valueT                      j  1

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I would like to expand open jfs s answer  which worked great for my problem  sorting two lists by a third  decorated list  We can create our decorated list in any way  but in this case we will create it from the elements of one of the two original lists  that we want to sort    say we have the following list and we want to sort both by the algorithms name     if we were to sort by the string list  it would sort by the numerical    value in the strings  string list     quot 0 123 Algo  XYZ quot    quot 0 345 Algo  BCD quot    quot 0 987 Algo  ABC quot   dict list      quot dict xyz quot    quot XYZ quot      quot dict bcd quot    quot BCD quot      quot dict abc quot    quot ABC quot       thus we need to create the decorator list  which we can now use to sort decorated    text 6   for text in string list      decorated list to sort  gt  gt  gt  decorated   Algo  XYZ    Algo  BCD    Algo  ABC    Now we can apply jfs s solution to sort our two lists by the third   create and sort the list of indices sorted indices   list range len string list    sorted indices sort key decorated   getitem       map sorted indices to the two  original lists sorted stringList   list map string list   getitem    sorted indices   sorted dictList   list map dict list   getitem    sorted indices      output  gt  gt  gt  sorted stringList   0 987 Algo  ABC    0 345 Algo  BCD    0 123 Algo  XYZ    gt  gt  gt  sorted dictList    dict abc    ABC      dict bcd    BCD      dict xyz    XYZ

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What about   list1    3 2 4 1  1  list2     three    two    four    one    one2    sortedRes   sorted zip list1  list2   key lambda x  x 0     use 0 or 1 depending on what you want to sort  gt  gt  gt    1   one     1   one2     2   two     3   three     4   four

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One classic approach to this problem is to use the  decorate  sort  undecorate  idiom  which is especially simple using python s built-in zip function    gt  gt  gt  list1    3 2 4 1  1   gt  gt  gt  list2     three    two    four    one    one2    gt  gt  gt  list1  list2   zip  sorted zip list1  list2     gt  gt  gt  list1  1  1  2  3  4   gt  gt  gt  list2    one    one2    two    three    four     These of course are no longer lists  but that s easily remedied  if it matters    gt  gt  gt  list1  list2    list t  for t in zip  sorted zip list1  list2      gt  gt  gt  list1  1  1  2  3  4   gt  gt  gt  list2   one    one2    two    three    four     It s worth noting that the above may sacrifice speed for terseness  the in-place version  which takes up 3 lines  is a tad faster on my machine for small lists    gt  gt  gt   timeit zip  sorted zip list1  list2    100000 loops  best of 3  3 3 us per loop  gt  gt  gt   timeit tups   zip list1  list2   tups sort    zip  tups  100000 loops  best of 3  2 84 us per loop   On the other hand  for larger lists  the one-line version could be faster    gt  gt  gt   timeit zip  sorted zip list1  list2    100 loops  best of 3  8 09 ms per loop  gt  gt  gt   timeit tups   zip list1  list2   tups sort    zip  tups  100 loops  best of 3  8 51 ms per loop   As Quantum7 points out  JSF s suggestion is a bit faster still  but it will probably only ever be a little bit faster  because Python uses the very same DSU idiom internally for all key-based sorts  It s just happening a little closer to the bare metal   This shows just how well optimized the zip routines are    I think the zip-based approach is more flexible and is a little more readable  so I prefer it

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Another approach to retaining the order of a string list when sorting against another list is as follows   list1    3 2 4 1  1  list2     three    two    four    one    one2      sort on list1 while retaining order of string list sorted list1    y for   y in sorted zip list1 list2  key lambda x  x 0    sorted list2   sorted list1   print sorted list1  print sorted list2    output    one    one2    two    three    four    1  1  2  3  4

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If you are using numpy you can use np argsort to get the sorted indices and apply those indices to the list  This works for any number of list that you would want to sort   import numpy as np  arr1   np array  4 3 1 32 21   arr2   arr1   10 sorted idxs   np argsort arr1   print sorted idxs   gt  gt  gt  array  2  1  0  4  3    print arr1 sorted idxs    gt  gt  gt  array   1   3   4  21  32    print arr2 sorted idxs    gt  gt  gt  array   10   30   40  210  320

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One way is to track where each index goes to by sorting the identity  0 1 2   n   This works for any number of lists   Then move each item to its position  Using splices is best   list1    3 2 4 1  1  list2     three    two    four    one    one2    index   list range len list1    print index    0  1  2  3  4    index sort key   list1   getitem    print index    3  4  1  0  2    list1       list1 i  for i in index  list2       list2 i  for i in index   print list1  print list2    1  1  2  3  4      one    one2    two    three    four      Note we could have iterated the lists without even sorting them   list1 iter    list1 i  for i in index

[python] How to sort two lists (which reference each other) in the exact same way

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