Python list of dictionaries search

Question

Assume I have this       name    Tom    age   10     name    Mark    age   5     name    Pam    age   7      and by searching  Pam  as name  I want to retrieve the related dictionary   name   Pam   age  7   How to achieve this

User · Answer

You can use a list comprehension   def search name  people       return  element for element in people if element  name      name

User · Answer

dicts     name    Tom    age   10     name    Mark    age   5     name    Pam    age   7     from collections import defaultdict dicts by name defaultdict list  for d in dicts      dicts by name d  name    d  print dicts by name  Tom     output   gt  gt  gt     age   10   name    Tom

User · Answer

Have you ever tried out the pandas package  It s perfect for this kind of search task and optimized too   import pandas as pd  listOfDicts       name    Tom    age   10     name    Mark    age   5     name    Pam    age   7       Create a data frame  keys are used as column headers    Dict items with the same key are entered into the same respective column  df   pd DataFrame listOfDicts     The pandas dataframe allows you to pick out specific values like so   df2   df   df  name       Pam    amp   df  age      7       Alternate syntax  same thing  df2   df   df name     Pam    amp   df age    7      I ve added a little bit of benchmarking below to illustrate pandas  faster runtimes on a larger scale i e  100k  entries   setup large    dicts         dicts extend     name    Tom    age   10      name    Mark    age   5        name    Pam    age   7      name    Dick    age   12     for   in range 25000     from operator import itemgetter import pandas as pd   df   pd DataFrame dicts     setup small    dicts        dicts extend     name    Tom    age   10      name    Mark    age   5        name    Pam    age   7      name    Dick    age   12       from operator import itemgetter import pandas as pd   df   pd DataFrame dicts     method1     item for item in dicts if item  name       Pam    method2    df df  name       Pam     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 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 Pandas      str t timeit 100      Small Method LC  0 000191926956177  Small Method Pandas  0 044392824173  Large Method LC  1 98827004433  Large Method Pandas  0 324505090714

User · Answer

This looks to me the most pythonic way   people       name    Tom    age   10     name    Mark    age   5     name    Pam    age   7     filter lambda person  person  name       Pam   people    result  returned as a list in Python 2       age   7   name    Pam      Note  In Python 3  a filter object is returned  So the python3 solution would be   list filter lambda person  person  name       Pam   people

User · Answer

Here is a comparison using iterating throuhg list  using filter lambda or refactoring if needed or valid to your case  your code to dict of dicts rather than list of dicts  import time    Build list of dicts list of dicts   list   for i in range 100000       list of dicts append   id   i   name    Tom       Build dict of dicts dict of dicts   dict   for i in range 100000       dict of dicts i      name    Tom       Find the one with ID of 99    1  iterate through the list lod ts   time time   for elem in list of dicts      if elem  id      99999          break lod tf   time time   lod td   lod tf - lod ts    2  Use filter f ts   time time   x   filter lambda k  k  id      99999  list of dicts  f tf   time time   f td   f tf- f ts    3  find it in dict of dicts dod ts   time time   x   dict of dicts 99999  dod tf   time time   dod td   dod tf - dod ts   print  List of Dictionries took   s    lod td print  Using filter took   s    f td print  Dict of Dicts took   s    dod td   And the output is this   List of Dictionries took  0 0099310874939 Using filter took  0 0121960639954 Dict of Dicts took  4 05311584473e-06   Conclusion  Clearly having a dictionary of dicts is the most efficient way to be able to search in those cases  where you know say you will be searching by id s only  interestingly using filter is the slowest solution

User · Answer

I found this thread when I was searching for an answer to the same question  While I realize that it s a late answer  I thought I d contribute it in case it s useful to anyone else   def find dict in list dicts  default None    kwargs          Find first matching  obj  dict  in  obj  list         param list dicts  List of dictionaries       param dict default  Optional  Default dictionary to return          Defaults to  None        param   kwargs   key value  pairs to match in  obj  dict         returns  First matching  obj  dict  from  dicts        rtype  dict               rval   default     for d in dicts          is found   False            Search for keys in dict          for k  v in kwargs items                if d get k  None     v                  is found   True              else                  is found   False                 break          if is found              rval   d             break      return rval   if   name         main           Tests     dicts          keys    spam eggs shrubbery knight  split        start   0     for   in range 4           dct    k  v for k  v in zip keys  range start  start 4            dicts append dct          start    4        Find each dict based on  spam  key only        for x in range len dicts            spam   x 4         assert find dict in list dicts  spam spam     dicts x         Find each dict based on  spam  and  shrubbery  keys      for x in range len dicts            spam   x 4         assert find dict in list dicts  spam spam  shrubbery spam 2     dicts x         Search for one correct key  one incorrect key      for x in range len dicts            spam   x 4         assert find dict in list dicts  spam spam  shrubbery spam 1  is None        Search for non-existent dict      for x in range len dicts            spam   x 100         assert find dict in list dicts  spam spam  is None

User · Answer

One simple way using list comprehensions is   if  l is the list  l       name    Tom    age   10     name    Mark    age   5     name    Pam    age   7      then    d  age   for d in l if d  name     Tom

User · Answer

Simply using list comprehension    i for i in dct if i  name       Pam   0    Sample code   dct           name    Tom    age   10         name    Mark    age   5         name    Pam    age   7     print  i for i in dct if i  name       Pam   0     gt    age   7   name    Pam

User · Answer

names      name   Tom    age   10     name    Mark    age   5     name    Pam    age   7   resultlist    d    for d in names     if d get  name           Pam   first result   resultlist 0    This is one way

User · Answer

def dsearch lod    kw       return filter lambda i  all  i k     v for  k  v  in kw items      lod   lod    a  33   b   test2    c   a ing333           a  22   b   ihaha    c   fbgval           a  33   b   TEst1    c   s ing123           a  22   b   ihaha    c   dfdvbfjkv       list dsearch lod  a 22       a   22   b    ihaha    c    fbgval       a   22   b    ihaha    c    dfdvbfjkv       list dsearch lod  a 22  b  ihaha        a   22   b    ihaha    c    fbgval       a   22   b    ihaha    c    dfdvbfjkv      list dsearch lod  a 22  c  fbgval        a   22   b    ihaha    c    fbgval

User · Answer

I tested various methods to go through a list of dictionaries and return the dictionaries where key x has a certain value     Results       Speed  list comprehension   generator expression    normal list iteration     filter      All scale linear with the number of dicts in the list  10x list size -  10x time     The keys per dictionary does not affect speed significantly for large amounts  thousands  of keys  Please see this graph I calculated  https   imgur com a quQzv  method names see below       All tests done with Python 3 6 4  W7x64   from random import randint from timeit import timeit   list dicts      for   in range 1000         number of dicts in the list     dict tmp          for i in range 10       number of keys for each dict         dict tmp f key i      randint 0 50      list dicts append  dict tmp      def a          normal iteration over all elements     for dict  in list dicts          if dict   key3      20              pass  def b          use  generator      for dict  in  x for x in list dicts if x  key3      20           pass  def c          use  list      for dict  in  x for x in list dicts if x  key3      20           pass  def d          use  filter      for dict  in filter lambda x  x  key3      20  list dicts           pass   Results   1 7303   normal list iteration  1 3849   generator expression  1 3158   list comprehension  7 7848   filter

User · Answer

You can use a generator expression    gt  gt  gt  dicts                name    Tom    age   10               name    Mark    age   5               name    Pam    age   7               name    Dick    age   12           gt  gt  gt  next item for item in dicts if item  name       Pam     age   7   name    Pam     If you need to handle the item not being there  then you can do what user Matt suggested in his comment and provide a default using a slightly different API   next  item for item in dicts if item  name       Pam    None    And to find the index of the item  rather than the item itself  you can enumerate   the list   next  i for i  item in enumerate dicts  if item  name       Pam    None

User · Answer

people       name    Tom    age   10     name    Mark    age   5     name    Pam    age   7     def search name       for p in people          if p  name      name              return p  search  Pam

User · Answer

I would create a dict of dicts like so  names     quot Tom quot    quot Mark quot    quot Pam quot   ages    10  5  7  my d       for i  j in zip names  ages       my d i      quot name quot   i   quot age quot   j   or  using exactly the same info as in the posted question  info list      quot name quot    quot Tom quot    quot age quot   10     quot name quot    quot Mark quot    quot age quot   5     quot name quot    quot Pam quot    quot age quot   7   my d       for d in info list      my d d  quot name quot      d  Then you could do my d  quot Pam quot   and get   quot name quot    quot Pam quot    quot age quot   7

User · Answer

You can achieve this with the usage of filter and next methods in Python   filter method filters the given sequence and returns an iterator  next method accepts an iterator and returns the next element in the list   So you can find the element by   my dict           name    Tom    age   10         name    Mark    age   5         name    Pam    age   7     next filter lambda obj  obj get  name       Pam   my dict   None    and the output is     name    Pam    age   7    Note  The above code will return None incase if the name we are searching is not found

User · Answer

You have to go through all elements of the list  There is not a shortcut   Unless somewhere else you keep a dictionary of the names pointing to the items of the list  but then you have to take care of the consequences of popping an element from your list

User · Answer

Most  if not all  implementations proposed here have two flaws    They assume only one key to be passed for searching  while it may be interesting to have more for complex dict They assume all keys passed for searching exist in the dicts  hence they don t deal correctly with KeyError occuring when it is not    An updated proposition   def find first in list objects    kwargs       return next  obj for obj in objects if                  len set obj keys    intersection kwargs keys      gt  0 and                  all  obj k     v for k  v in kwargs items   if k in obj keys                       None    Maybe not the most pythonic  but at least a bit more failsafe   Usage     gt  gt  gt  obj1   find first in list list of dict  name  Pam   age 7   gt  gt  gt  obj2   find first in list list of dict  name  Pam   age 27   gt  gt  gt  obj3   find first in list list of dict  name  Pam   address  nowhere    gt  gt  gt    gt  gt  gt  print obj1  obj2  obj3    name    Pam    age   7   None    name    Pam    age   7    The gist

User · Answer

My first thought would be that you might want to consider creating a dictionary of these dictionaries     if  for example  you were going to be searching it more a than small number of times   However that might be a premature optimization   What would be wrong with   def get records key  store dict            Return a list of all records containing name  key from our store             assert key is not None     return  d for d in store if d  name    key

User · Answer

To add just a tiny bit to  Fr  d  ricHamidi   In case you are not sure a key is in the the list of dicts  something like this would help   next  item for item in dicts if item get  name   and item  name       Pam    None

User · Answer

You can try this       lst  list of dictionaries     lst      name    Tom    age   10     name    Mark    age   5     name    Pam    age   7    search   raw input  What name      Input name that needs to be searched  say  Pam    print   lst i  for i in range len lst   if lst i   name    search    0   Output  gt  gt  gt    age   7   name    Pam

User · Answer

This is a general way of searching a value in a list of dictionaries   def search dictionaries key  value  list of dictionaries       return  element for element in list of dictionaries if element key     value

User · Answer

Fr  d  ric Hamidi s answer is great   In Python 3 x the syntax for  next   changed slightly   Thus a slight modification    gt  gt  gt  dicts             name    Tom    age   10            name    Mark    age   5            name    Pam    age   7            name    Dick    age   12       gt  gt  gt  next item for item in dicts if item  name       Pam     age   7   name    Pam     As mentioned in the comments by  Matt  you can add a default value as such    gt  gt  gt  next  item for item in dicts if item  name       Pam    False    name    Pam    age   7   gt  gt  gt  next  item for item in dicts if item  name       Sam    False  False  gt  gt  gt

[python] Python list of dictionaries search

Examples related to python

Examples related to search

Examples related to dictionary