How do I sort a list of dictionaries by a value of the dictionary

Question

I have a list of dictionaries and want each item to be sorted by a specific value  Take into consideration the list     name   Homer    age  39     name   Bart    age  10    When sorted by name  it should become     name   Bart    age  10     name   Homer    age  39

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You have to implement your own comparison function that will compare the dictionaries by values of name keys  See Sorting Mini-HOW TO from PythonInfo Wiki

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my list      name   Homer    age  39     name   Bart    age  10    my list sort lambda x y   cmp x  name    y  name      my list will now be what you want  Or better  Since Python 2 4  there s a key argument is both more efficient and neater  my list   sorted my list  key lambda k  k  name        the lambda is  IMO  easier to understand than operator itemgetter  but your mileage may vary

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my list      name   Homer    age  39     name   Bart    age  10    my list sort lambda x y   cmp x  name    y  name      my list will now be what you want  Or better  Since Python 2 4  there s a key argument is both more efficient and neater  my list   sorted my list  key lambda k  k  name        the lambda is  IMO  easier to understand than operator itemgetter  but your mileage may vary

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If you want to sort the list by multiple keys  you can do the following  my list      name   Homer    age  39     name   Milhouse    age  10     name   Bart    age  10    sortedlist   sorted my list   key lambda elem   quot  02d  s quot     elem  age    elem  name      It is rather hackish  since it relies on converting the values into a single string representation for comparison  but it works as expected for numbers including negative ones  although you will need to format your string appropriately with zero paddings if you are using numbers

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my list      name   Homer    age  39     name   Bart    age  10    my list sort lambda x y   cmp x  name    y  name      my list will now be what you want  Or better  Since Python 2 4  there s a key argument is both more efficient and neater  my list   sorted my list  key lambda k  k  name        the lambda is  IMO  easier to understand than operator itemgetter  but your mileage may vary

User · Answer

If you do not need the original list of dictionaries  you could modify it in-place with sort   method using a custom key function   Key function   def get name d           Return the value of a key in a dictionary           return d  name     The list to be sorted   data one      name    Homer    age   39     name    Bart    age   10     Sorting it in-place   data one sort key get name    If you need the original list  call the sorted   function passing it the list and the key function  then assign the returned sorted list to a new variable   data two      name    Homer    age   39     name    Bart    age   10   new data   sorted data two  key get name    Printing data one and new data    gt  gt  gt  print data one     name    Bart    age   10     name    Homer    age   39    gt  gt  gt  print new data     name    Bart    age   10     name    Homer    age   39

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If performance is a concern  I would use operator itemgetter instead of lambda as built-in functions perform faster than hand-crafted functions  The itemgetter function seems to perform approximately 20  faster than lambda based on my testing  From https   wiki python org moin PythonSpeed   Likewise  the builtin functions run faster than hand-built equivalents  For example  map operator add  v1  v2  is faster than map lambda x y  x y  v1  v2    Here is a comparison of sorting speed using lambda vs itemgetter  import random import operator    Create a list of 100 dicts with random 8-letter names and random ages from 0 to 100  l      name      join random choices string ascii lowercase  k 8     age   random randint 0  100   for i in range 100      Test the performance with a lambda function sorting on name  timeit sorted l  key lambda x  x  name    13   s    388 ns per loop  mean    std  dev  of 7 runs  100000 loops each     Test the performance with itemgetter sorting on name  timeit sorted l  key operator itemgetter  name    10 7   s    38 1 ns per loop  mean    std  dev  of 7 runs  100000 loops each     Check that each technique produces the same sort order sorted l  key lambda x  x  name       sorted l  key operator itemgetter  name    True  Both techniques sort the list in the same order  verified by execution of the final statement in the code block   but the first one is a little faster

User · Answer

If you want to sort the list by multiple keys  you can do the following  my list      name   Homer    age  39     name   Milhouse    age  10     name   Bart    age  10    sortedlist   sorted my list   key lambda elem   quot  02d  s quot     elem  age    elem  name      It is rather hackish  since it relies on converting the values into a single string representation for comparison  but it works as expected for numbers including negative ones  although you will need to format your string appropriately with zero paddings if you are using numbers

User · Answer

I guess you ve meant      name   Homer    age  39     name   Bart    age  10     This would be sorted like this   sorted l cmp lambda x y  cmp x  name   y  name

User · Answer

import operator   To sort the list of dictionaries by key  name    list of dicts sort key operator itemgetter  name      To sort the list of dictionaries by key  age    list of dicts sort key operator itemgetter  age

User · Answer

Here is the alternative general solution - it sorts elements of a dict by keys and values  The advantage of it - no need to specify keys  and it would still work if some keys are missing in some of dictionaries  def sort key func item        quot  quot  quot  Helper function used to sort list of dicts       param item  dict      return  sorted list of tuples  k  v       quot  quot  quot      pairs          for k  v in item items            pairs append  k  v       return sorted pairs  sorted A  key sort key func

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You could use a custom comparison function  or you could pass in a function that calculates a custom sort key  That s usually more efficient as the key is only calculated once per item  while the comparison function would be called many more times   You could do it this way   def mykey adict   return adict  name   x      name    Homer    age   39     name    Bart    age  10   sorted x  key mykey    But the standard library contains a generic routine for getting items of arbitrary objects  itemgetter  So try this instead   from operator import itemgetter x      name    Homer    age   39     name    Bart    age  10   sorted x  key itemgetter  name

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Using the Schwartzian transform from Perl  py      name   Homer    age  39     name   Bart    age  10    do sort on    quot name quot  decorated     dict  sort on   dict   for dict  in py  decorated sort   result    dict  for  key  dict   in decorated   gives  gt  gt  gt  result    age   10   name    Bart      age   39   name    Homer     More on the Perl Schwartzian transform   In computer science  the Schwartzian transform is a Perl programming idiom used to improve the efficiency of sorting a list of items  This idiom is appropriate for comparison-based sorting when the ordering is actually based on the ordering of a certain property  the key  of the elements  where computing that property is an intensive operation that should be performed a minimal number of times  The Schwartzian Transform is notable in that it does not use named temporary arrays

User · Answer

import operator a list of dicts sort key operator itemgetter  name       key  is used to sort by an arbitrary value and  itemgetter  sets that value to each item s  name  attribute

User · Answer

It may look cleaner using a key instead a cmp   newlist   sorted list to be sorted  key lambda k  k  name       or as J F Sebastian and others suggested   from operator import itemgetter newlist   sorted list to be sorted  key itemgetter  name       For completeness  as pointed out in comments by fitzgeraldsteele   add reverse True to sort descending  newlist   sorted l  key itemgetter  name    reverse True

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import operator   To sort the list of dictionaries by key  name    list of dicts sort key operator itemgetter  name      To sort the list of dictionaries by key  age    list of dicts sort key operator itemgetter  age

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import operator   To sort the list of dictionaries by key  name    list of dicts sort key operator itemgetter  name      To sort the list of dictionaries by key  age    list of dicts sort key operator itemgetter  age

User · Answer

Sometimes we need to use lower    For example  lists      name   Homer    age  39       name   Bart    age  10       name   abby    age  9    lists   sorted lists  key lambda k  k  name    print lists       name   Bart    age  10     name   Homer    age  39     name   abby    age  9    lists   sorted lists  key lambda k  k  name   lower    print lists        name   abby    age  9     name   Bart    age  10     name   Homer    age  39

User · Answer

import operator a list of dicts sort key operator itemgetter  name       key  is used to sort by an arbitrary value and  itemgetter  sets that value to each item s  name  attribute

User · Answer

It may look cleaner using a key instead a cmp   newlist   sorted list to be sorted  key lambda k  k  name       or as J F Sebastian and others suggested   from operator import itemgetter newlist   sorted list to be sorted  key itemgetter  name       For completeness  as pointed out in comments by fitzgeraldsteele   add reverse True to sort descending  newlist   sorted l  key itemgetter  name    reverse True

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Using the Pandas package is another method  though its runtime at large scale is much slower than the more traditional methods proposed by others  import pandas as pd  listOfDicts      name   Homer    age  39     name   Bart    age  10   df   pd DataFrame listOfDicts  df   df sort values  name   sorted listOfDicts   df T to dict   values    Here are some benchmark values for a tiny list and a large  100k   list of dicts  setup large    quot listOfDicts         listOfDicts extend    name   Homer    age  39     name   Bart    age  10    for   in range 50000     from operator import itemgetter import pandas as pd   df   pd DataFrame listOfDicts   quot   setup small    quot listOfDicts        listOfDicts extend    name   Homer    age  39     name   Bart    age  10      from operator import itemgetter import pandas as pd   df   pd DataFrame listOfDicts   quot   method1    quot newlist   sorted listOfDicts  key lambda k  k  name    quot  method2    quot newlist   sorted listOfDicts  key itemgetter  name     quot  method3    quot df   df sort values  name     sorted listOfDicts   df T to dict   values   quot   import timeit t   timeit Timer method1  setup small  print  Small Method LC      str t timeit 100    t   timeit Timer method2  setup small  print  Small Method LC2      str t timeit 100    t   timeit Timer method3  setup small  print  Small Method Pandas      str t timeit 100     t   timeit Timer method1  setup large  print  Large Method LC      str t timeit 100    t   timeit Timer method2  setup large  print  Large Method LC2      str t timeit 100    t   timeit Timer method3  setup large  print  Large Method Pandas      str t timeit 1      Small Method LC  0 000163078308105  Small Method LC2  0 000134944915771  Small Method Pandas  0 0712950229645  Large Method LC  0 0321750640869  Large Method LC2  0 0206089019775  Large Method Pandas  5 81405615807

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Using the Pandas package is another method  though its runtime at large scale is much slower than the more traditional methods proposed by others  import pandas as pd  listOfDicts      name   Homer    age  39     name   Bart    age  10   df   pd DataFrame listOfDicts  df   df sort values  name   sorted listOfDicts   df T to dict   values    Here are some benchmark values for a tiny list and a large  100k   list of dicts  setup large    quot listOfDicts         listOfDicts extend    name   Homer    age  39     name   Bart    age  10    for   in range 50000     from operator import itemgetter import pandas as pd   df   pd DataFrame listOfDicts   quot   setup small    quot listOfDicts        listOfDicts extend    name   Homer    age  39     name   Bart    age  10      from operator import itemgetter import pandas as pd   df   pd DataFrame listOfDicts   quot   method1    quot newlist   sorted listOfDicts  key lambda k  k  name    quot  method2    quot newlist   sorted listOfDicts  key itemgetter  name     quot  method3    quot df   df sort values  name     sorted listOfDicts   df T to dict   values   quot   import timeit t   timeit Timer method1  setup small  print  Small Method LC      str t timeit 100    t   timeit Timer method2  setup small  print  Small Method LC2      str t timeit 100    t   timeit Timer method3  setup small  print  Small Method Pandas      str t timeit 100     t   timeit Timer method1  setup large  print  Large Method LC      str t timeit 100    t   timeit Timer method2  setup large  print  Large Method LC2      str t timeit 100    t   timeit Timer method3  setup large  print  Large Method Pandas      str t timeit 1      Small Method LC  0 000163078308105  Small Method LC2  0 000134944915771  Small Method Pandas  0 0712950229645  Large Method LC  0 0321750640869  Large Method LC2  0 0206089019775  Large Method Pandas  5 81405615807

User · Answer

You could use a custom comparison function  or you could pass in a function that calculates a custom sort key  That s usually more efficient as the key is only calculated once per item  while the comparison function would be called many more times   You could do it this way   def mykey adict   return adict  name   x      name    Homer    age   39     name    Bart    age  10   sorted x  key mykey    But the standard library contains a generic routine for getting items of arbitrary objects  itemgetter  So try this instead   from operator import itemgetter x      name    Homer    age   39     name    Bart    age  10   sorted x  key itemgetter  name

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a      name   Homer    age  39           This changes the list a a sort key lambda k   k  name       This returns a new list  a is not modified  sorted a  key lambda k   k  name

User · Answer

I guess you ve meant      name   Homer    age  39     name   Bart    age  10     This would be sorted like this   sorted l cmp lambda x y  cmp x  name   y  name

User · Answer

It may look cleaner using a key instead a cmp   newlist   sorted list to be sorted  key lambda k  k  name       or as J F Sebastian and others suggested   from operator import itemgetter newlist   sorted list to be sorted  key itemgetter  name       For completeness  as pointed out in comments by fitzgeraldsteele   add reverse True to sort descending  newlist   sorted l  key itemgetter  name    reverse True

User · Answer

You could use a custom comparison function  or you could pass in a function that calculates a custom sort key  That s usually more efficient as the key is only calculated once per item  while the comparison function would be called many more times   You could do it this way   def mykey adict   return adict  name   x      name    Homer    age   39     name    Bart    age  10   sorted x  key mykey    But the standard library contains a generic routine for getting items of arbitrary objects  itemgetter  So try this instead   from operator import itemgetter x      name    Homer    age   39     name    Bart    age  10   sorted x  key itemgetter  name

User · Answer

I have been a big fan of a filter with lambda  However  it is not best option if you consider time complexity  First option sorted list   sorted list to sort  key  lambda x  x  name      Returns list of values  Second option list to sort sort key operator itemgetter  name      Edits the list  and does not return a new list  Fast comparison of execution times   First option python3 6 -m timeit -s  quot list to sort      name   Homer    age  39     name   Bart    age  10     name   Faaa    age  57     name   Errr    age  20   quot  -s  quot sorted l    quot   quot sorted l   sorted list to sort  key lambda e  e  name    quot    1000000 loops  best of 3  0 736   sec per loop    Second option python3 6 -m timeit -s  quot list to sort      name   Homer    age  39     name   Bart    age  10     name   Faaa    age  57     name   Errr    age  20   quot  -s  quot sorted l    quot  -s  quot import operator quot   quot list to sort sort key operator itemgetter  name    quot    1000000 loops  best of 3  0 438   sec per loop

User · Answer

You have to implement your own comparison function that will compare the dictionaries by values of name keys  See Sorting Mini-HOW TO from PythonInfo Wiki

User · Answer

a      name   Homer    age  39           This changes the list a a sort key lambda k   k  name       This returns a new list  a is not modified  sorted a  key lambda k   k  name

User · Answer

Let s say I have a dictionary D with the elements below  To sort  just use the key argument in sorted to pass a custom function as below  D     eggs   3   ham   1   spam   2  def get count tuple       return tuple 1   sorted D items    key   get count  reverse True    Or sorted D items    key   lambda x  x 1   reverse True     Avoiding get count function call  Check this out

User · Answer

Using the Schwartzian transform from Perl  py      name   Homer    age  39     name   Bart    age  10    do sort on    quot name quot  decorated     dict  sort on   dict   for dict  in py  decorated sort   result    dict  for  key  dict   in decorated   gives  gt  gt  gt  result    age   10   name    Bart      age   39   name    Homer     More on the Perl Schwartzian transform   In computer science  the Schwartzian transform is a Perl programming idiom used to improve the efficiency of sorting a list of items  This idiom is appropriate for comparison-based sorting when the ordering is actually based on the ordering of a certain property  the key  of the elements  where computing that property is an intensive operation that should be performed a minimal number of times  The Schwartzian Transform is notable in that it does not use named temporary arrays

User · Answer

You have to implement your own comparison function that will compare the dictionaries by values of name keys  See Sorting Mini-HOW TO from PythonInfo Wiki

User · Answer

Sometimes we need to use lower    For example  lists      name   Homer    age  39       name   Bart    age  10       name   abby    age  9    lists   sorted lists  key lambda k  k  name    print lists       name   Bart    age  10     name   Homer    age  39     name   abby    age  9    lists   sorted lists  key lambda k  k  name   lower    print lists        name   abby    age  9     name   Bart    age  10     name   Homer    age  39

User · Answer

If performance is a concern  I would use operator itemgetter instead of lambda as built-in functions perform faster than hand-crafted functions  The itemgetter function seems to perform approximately 20  faster than lambda based on my testing  From https   wiki python org moin PythonSpeed   Likewise  the builtin functions run faster than hand-built equivalents  For example  map operator add  v1  v2  is faster than map lambda x y  x y  v1  v2    Here is a comparison of sorting speed using lambda vs itemgetter  import random import operator    Create a list of 100 dicts with random 8-letter names and random ages from 0 to 100  l      name      join random choices string ascii lowercase  k 8     age   random randint 0  100   for i in range 100      Test the performance with a lambda function sorting on name  timeit sorted l  key lambda x  x  name    13   s    388 ns per loop  mean    std  dev  of 7 runs  100000 loops each     Test the performance with itemgetter sorting on name  timeit sorted l  key operator itemgetter  name    10 7   s    38 1 ns per loop  mean    std  dev  of 7 runs  100000 loops each     Check that each technique produces the same sort order sorted l  key lambda x  x  name       sorted l  key operator itemgetter  name    True  Both techniques sort the list in the same order  verified by execution of the final statement in the code block   but the first one is a little faster

User · Answer

You could use a custom comparison function  or you could pass in a function that calculates a custom sort key  That s usually more efficient as the key is only calculated once per item  while the comparison function would be called many more times   You could do it this way   def mykey adict   return adict  name   x      name    Homer    age   39     name    Bart    age  10   sorted x  key mykey    But the standard library contains a generic routine for getting items of arbitrary objects  itemgetter  So try this instead   from operator import itemgetter x      name    Homer    age   39     name    Bart    age  10   sorted x  key itemgetter  name

User · Answer

I guess you ve meant      name   Homer    age  39     name   Bart    age  10     This would be sorted like this   sorted l cmp lambda x y  cmp x  name   y  name

User · Answer

Here is the alternative general solution - it sorts elements of a dict by keys and values  The advantage of it - no need to specify keys  and it would still work if some keys are missing in some of dictionaries  def sort key func item        quot  quot  quot  Helper function used to sort list of dicts       param item  dict      return  sorted list of tuples  k  v       quot  quot  quot      pairs          for k  v in item items            pairs append  k  v       return sorted pairs  sorted A  key sort key func

User · Answer

You have to implement your own comparison function that will compare the dictionaries by values of name keys  See Sorting Mini-HOW TO from PythonInfo Wiki

User · Answer

import operator a list of dicts sort key operator itemgetter  name       key  is used to sort by an arbitrary value and  itemgetter  sets that value to each item s  name  attribute

User · Answer

It may look cleaner using a key instead a cmp   newlist   sorted list to be sorted  key lambda k  k  name       or as J F Sebastian and others suggested   from operator import itemgetter newlist   sorted list to be sorted  key itemgetter  name       For completeness  as pointed out in comments by fitzgeraldsteele   add reverse True to sort descending  newlist   sorted l  key itemgetter  name    reverse True

User · Answer

import operator a list of dicts sort key operator itemgetter  name       key  is used to sort by an arbitrary value and  itemgetter  sets that value to each item s  name  attribute

User · Answer

my list      name   Homer    age  39     name   Bart    age  10    my list sort lambda x y   cmp x  name    y  name      my list will now be what you want  Or better  Since Python 2 4  there s a key argument is both more efficient and neater  my list   sorted my list  key lambda k  k  name        the lambda is  IMO  easier to understand than operator itemgetter  but your mileage may vary

User · Answer

I have been a big fan of a filter with lambda  However  it is not best option if you consider time complexity  First option sorted list   sorted list to sort  key  lambda x  x  name      Returns list of values  Second option list to sort sort key operator itemgetter  name      Edits the list  and does not return a new list  Fast comparison of execution times   First option python3 6 -m timeit -s  quot list to sort      name   Homer    age  39     name   Bart    age  10     name   Faaa    age  57     name   Errr    age  20   quot  -s  quot sorted l    quot   quot sorted l   sorted list to sort  key lambda e  e  name    quot    1000000 loops  best of 3  0 736   sec per loop    Second option python3 6 -m timeit -s  quot list to sort      name   Homer    age  39     name   Bart    age  10     name   Faaa    age  57     name   Errr    age  20   quot  -s  quot sorted l    quot  -s  quot import operator quot   quot list to sort sort key operator itemgetter  name    quot    1000000 loops  best of 3  0 438   sec per loop

User · Answer

I guess you ve meant      name   Homer    age  39     name   Bart    age  10     This would be sorted like this   sorted l cmp lambda x y  cmp x  name   y  name

User · Answer

If you do not need the original list of dictionaries  you could modify it in-place with sort   method using a custom key function   Key function   def get name d           Return the value of a key in a dictionary           return d  name     The list to be sorted   data one      name    Homer    age   39     name    Bart    age   10     Sorting it in-place   data one sort key get name    If you need the original list  call the sorted   function passing it the list and the key function  then assign the returned sorted list to a new variable   data two      name    Homer    age   39     name    Bart    age   10   new data   sorted data two  key get name    Printing data one and new data    gt  gt  gt  print data one     name    Bart    age   10     name    Homer    age   39    gt  gt  gt  print new data     name    Bart    age   10     name    Homer    age   39

User · Answer

import operator   To sort the list of dictionaries by key  name    list of dicts sort key operator itemgetter  name      To sort the list of dictionaries by key  age    list of dicts sort key operator itemgetter  age

User · Answer

Let s say I have a dictionary D with the elements below  To sort  just use the key argument in sorted to pass a custom function as below  D     eggs   3   ham   1   spam   2  def get count tuple       return tuple 1   sorted D items    key   get count  reverse True    Or sorted D items    key   lambda x  x 1   reverse True     Avoiding get count function call  Check this out

[python] How do I sort a list of dictionaries by a value of the dictionary?

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