Return first N key value pairs from dict

Question

Consider the following dictionary  d   d     a   3   b   2   c   3   d   4   e   5    I want to return the first N key value pairs from d  N  lt   4 in this case   What is the most efficient method of doing this

User · Answer

A very efficient way to retrieve anything is to combine list or dictionary comprehensions with slicing. If you don't need to order the items (you just want n random pairs), you can use a dictionary comprehension like this:

# Python 2
first2pairs = {k: mydict[k] for k in mydict.keys()[:2]}
# Python 3
first2pairs = {k: mydict[k] for k in list(mydict)[:2]}

Generally a comprehension like this is always faster to run than the equivalent "for x in y" loop. Also, by using .keys() to make a list of the dictionary keys and slicing that list you avoid 'touching' any unnecessary keys when you build the new dictionary.

If you don't need the keys (only the values) you can use a list comprehension:

first2vals = [v for v in mydict.values()[:2]]

If you need the values sorted based on their keys, it's not much more trouble:

first2vals = [mydict[k] for k in sorted(mydict.keys())[:2]]

or if you need the keys as well:

first2pairs = {k: mydict[k] for k in sorted(mydict.keys())[:2]}

User · Answer

See PEP 0265 on sorting dictionaries   Then use the aforementioned iterable code   If you need more efficiency in the sorted key-value pairs   Use a different data structure   That is  one that maintains sorted order and the key-value associations   E g   import bisect  kvlist      a   1     b   2     c   3     e   5   bisect insort left kvlist    d   4    print kvlist      a   1     b   2     c   3     d   4     e   5

User · Answer

I have tried a few of the answers above and note that some of them are version dependent and do not work in version 3 7   I also note that since 3 6 all dictionaries are ordered by the sequence in which items are inserted   Despite dictionaries being ordered since 3 6 some of the statements you expect to work with ordered structures don t seem to work   The answer to the OP question that worked best for me   itr   iter dic items    lst    next itr  for i in range 3

User · Answer

For Python 3 and above To select first n Pairs  n 4 firstNpairs    k  Diction k  for k in list Diction keys     n

User · Answer

Were d is your dictionary and n is the printing number  for idx   k  v  in enumerate d     if idx    n  break   print  k  v    Casting your dictionary to list can be slow  Your dictionary may be too large and you don t need to cast all of it just for printing a few of the first

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Dictionary maintains no order   so before picking top N key value pairs lets make it sorted   import operator d     a   3   b   2   c   3   d   4  d dict sorted d items   key operator itemgetter 1  reverse True    itemgetter 0  sort by keys  itemgetter 1  sort by values   Now we can do the retrieval of top  N  elements   using the method structure like this   def return top elements dictionary element          Takes the dictionary and the  N  elements needed in return             topers        for h i in enumerate dictionary element           if h lt elements              topers update  i dictionary element i        return topers   to get the top 2 elements then simply use this structure   d     a   3   b   2   c   3   d   4  d dict sorted d items   key operator itemgetter 1  reverse True   d return top 2 d  print d

User · Answer

Did not see it on here  Will not be ordered but the simplest syntactically if you need to just take some elements from a dictionary   n   2  key value for key value in d items   0 n

User · Answer

Python s dicts are not ordered  so it s meaningless to ask for the  first N  keys   The collections OrderedDict class is available if that s what you need  You could efficiently get its first four elements as  import itertools import collections  d   collections OrderedDict    foo    bar     1   a     2   b     3   c     4   d     x   itertools islice d items    0  4   for key  value in x      print key  value   itertools islice allows you to lazily take a slice of elements from any iterator  If you want the result to be reusable you d need to convert it to a list or something  like so   x   list itertools islice d items    0  4

User · Answer

You can get dictionary items by calling  items   on the dictionary  then convert that to a list and from there get first N items as you would on any list  below code prints first 3 items of the dictionary object e g  d     a   3   b   2   c   3   d   4   e   5   first three items   list d items     3   print first three items   Outputs     a   3     b   2     c   3

User · Answer

There s no such thing a the  first n  keys because a dict doesn t remember which keys were inserted first   You can get any n key-value pairs though   n items   take n  d iteritems      This uses the implementation of take from the itertools recipes   from itertools import islice  def take n  iterable        Return first n items of the iterable as a list      return list islice iterable  n     See it working online  ideone   Update for Python 3 6  n items   take n  d items

User · Answer

You can approach this a number of ways   If order is important you can do this   for key in sorted d keys       item   d pop key    If order isn t a concern you can do this   for i in range 4     item   d popitem

User · Answer

This might not be very elegant  but works for me   d     a   3   b   2   c   3   d   4   e   5   x  0 for key  val in d items        if x    2          break     else          x    1           Do something with the first two key-value pairs

User · Answer

consider a dict   d     a   3   b   2   c   3   d   4   e   5   from itertools import islice n   3 list islice d items   n     islice will do the trick     hope it helps

User · Answer

This depends on what is  most efficient  in your case   If you just want a semi-random sample of a huge dictionary foo  use foo iteritems   and take as many values from it as you need  it s a lazy operation that avoids creation of an explicit list of keys or items    If you need to sort keys first  there s no way around using something like keys   foo keys    keys sort   or sorted foo iterkeys     you ll have to build an explicit list of keys  Then slice or iterate through first N keys   BTW why do you care about the  efficient  way  Did you profile your program  If you did not  use the obvious and easy to understand way first  Chances are it will do pretty well without becoming a bottleneck

User · Answer

just add an answer using zip    k  d k  for k    in zip d  range n

User · Answer

For Python 3 8 the correct answer should be  import more itertools  d     a   3   b   2   c   3   d   4   e   5   first n   more itertools take 3  d items    print len first n   print first n   Whose output is  3    a   3     b   2     c   3    After pip install more-itertools of course

User · Answer

in py3  this will do the trick   A N for  A N  in  x for x in d items     4       a   3   b   2   c   3   d   4

User · Answer

To get the top N elements from your python dictionary one can use the following line of code   list dictionaryName items     N    In your case you can change it to   list d items     4

User · Answer

def GetNFirstItems self       self dict    f Item i   1    round uniform 20 40  50 50   2  for i in range 10   Example Dict     self get items   int input        for self index self item in zip range len self dict   self dict items             if self index  self get items            break         else              print self item  quot   quot  end  quot  quot    Unusual approach  as it gives out intense O N  time complexity

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foo     a  1   b  2   c  3   d  4   e  5   f  6  iterator   iter foo items    for i in range 3       print next iterator     Basically  turn the view  dict items  into an iterator  and then iterate it with next

[python] Return first N key:value pairs from dict

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