Item frequency count in Python

Question

Assume I have a list of words  and I want to find the number of times each word appears in that list   An obvious way to do this is   words    apple banana apple strawberry banana lemon  uniques   set words split    freqs     item  words split   count item   for item in uniques  print freqs    But I find this code not very good  because the program runs through the word list twice  once to build the set  and a second time to count the number of appearances   Of course  I could write a function to run through the list and do the counting  but that wouldn t be so Pythonic  So  is there a more efficient and Pythonic way

User · Answer

Without defaultdict   words    apple banana apple strawberry banana lemon  my count      for word in words split        try  my count word     1     except KeyError  my count word    1

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words    apple banana apple strawberry banana lemon  w words split   e list set w          word freqs      for i in e      word freqs i  w count i  print word freqs       Hope this helps

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The Counter class in the collections module is purpose built to solve this type of problem   from collections import Counter words    apple banana apple strawberry banana lemon  Counter words split      Counter   apple   2   banana   2   strawberry   1   lemon   1

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The answer below takes some extra cycles  but it is another method  def func tup       return tup -1    def print words filename       f   open  small txt   r       whole content    f read    lower       print whole content     list content   whole content split       dict          for one word in list content          dict one word    0     for one word in list content          dict one word     1     print dict items       print sorted dict items   key func

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Use reduce   to convert the list to a single dict   words    apple banana apple strawberry banana lemon  reduce  lambda d  c  d update   c  d get c 0  1    or d  words split          returns    strawberry   1   lemon   1   apple   2   banana   2

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If you don t want to use the standard dictionary method  looping through the list incrementing the proper dict  key   you can try this    gt  gt  gt  from itertools import groupby  gt  gt  gt  myList   words split       apple    banana    apple    strawberry    banana    lemon    gt  gt  gt    k  len list g    for k  g in groupby sorted myList       apple   2     banana   2     lemon   1     strawberry   1     It runs in O n log n  time

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freqs      for word in words      freqs word    freqs get word  0    1   fetch and increment OR initialize   I think this results to the same as Triptych s solution  but without importing collections  Also a bit like Selinap s solution  but more readable imho  Almost identical to Thomas Weigel s solution  but without using Exceptions   This could be slower than using defaultdict   from the collections library however  Since the value is fetched  incremented and then assigned again  Instead of just incremented  However using    might do just the same internally

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Standard approach   from collections import defaultdict  words    apple banana apple strawberry banana lemon  words   words split   result   defaultdict int  for word in words      result word     1  print result   Groupby oneliner   from itertools import groupby  words    apple banana apple strawberry banana lemon  words   words split    result   dict  key  len list group    for key  group in groupby sorted words    print result

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Can t you just use count   words    the quick brown fox jumps over the lazy gray dog  words count  z    output  1

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I happened to work on some Spark exercise  here is my solution   tokens     quick    brown    fox    jumps    lazy    dog    print  n  float tokens count n   float len tokens   for n in tokens       output of the above       brown   0 16666666666666666   lazy   0 16666666666666666   jumps   0 16666666666666666   fox   0 16666666666666666   dog   0 16666666666666666   quick   0 16666666666666666

User · Answer

defaultdict to the rescue   from collections import defaultdict  words    apple banana apple strawberry banana lemon   d   defaultdict int  for word in words split        d word     1   This runs in O n

[python] Item frequency count in Python

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