How to count the frequency of the elements in an unordered list

Question

I need to find the frequency of elements in an unordered list  a    1 1 1 1 2 2 2 2 3 3 4 5 5    output-   b    4 4 2 1 2    Also I want to remove the duplicates from a  a    1 2 3 4 5

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Python 2 7  introduces Dictionary Comprehension  Building the dictionary from the list will get you the count as well as get rid of duplicates    gt  gt  gt  a    1 1 1 1 2 2 2 2 3 3 4 5 5   gt  gt  gt  d    x a count x  for x in a   gt  gt  gt  d  1  4  2  4  3  2  4  1  5  2   gt  gt  gt  a  b   d keys    d values    gt  gt  gt  a  1  2  3  4  5   gt  gt  gt  b  4  4  2  1  2

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from collections import Counter a   E   D   C   G   B   A   B   F   D   D   C   A   G   A   C   B   F   C   B    counter Counter a   kk  list counter keys    list counter values      pd DataFrame np array kk  T  columns   Letter   Count

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usr bin python def frq words       freq          for w in words              if w in freq                      freq w    freq get w  1             else                      freq w   1     return freq  fp   open  poem   r   list   fp read   fp close   input   list split   print input d   frq input  print  frequency of input n    print d fp1   open  output txt   w    for k v in d items    fp1 write str k      str v    n   fp1 close

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Simple solution using a dictionary   def frequency l        d           for i in l          if i in d keys               d i     1         else             d i    1       for k  v in d iteritems            if v   max  d values                return k d keys    print frequency  10 10 10 10 20 20 20 20 40 40 50 50 30

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This approach can be tried if you don t want to use any library and keep it simple and short   a    1 1 1 1 2 2 2 2 3 3 4 5 5  marked      b     a count i   marked append i   0  for i in a if i not in marked  print b    o p   4  4  2  1  2

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seta   set a  b    a count el  for el in seta  a   list seta   Only if you really want it

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I would simply use scipy stats itemfreq in the following manner   from scipy stats import itemfreq  a    1 1 1 1 2 2 2 2 3 3 4 5 5   freq   itemfreq a   a   freq   0  b   freq   1    you may check the documentation here  http   docs scipy org doc scipy-0 16 0 reference generated scipy stats itemfreq html

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To count the number of appearances   from collections import defaultdict  appearances   defaultdict int   for curr in a      appearances curr     1   To remove duplicates   a   set a

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Another approach of doing this  albeit by using a heavier but powerful library - NLTK   import nltk  fdist   nltk FreqDist a  fdist values   fdist most common

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a    1 1 1 1 2 2 2 2 3 3 4 5 5     1  Get counts and store in another list output      for i in set a       output append a count i   print output     2  Remove duplicates using set constructor a   list set a   print a     Set collection does not allow duplicates  passing a list to the set   constructor will give an iterable of totally unique objects  count   function returns an integer count when an object that is in a list is passed  With that the unique objects are counted and each count value is stored by appending to an empty list output list   constructor is used to convert the set a  into list and referred by the same variable a    Output  D  MLrec venv Scripts python exe D  MLrec listgroup py  4  4  2  1  2   1  2  3  4  5

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Yet another solution with another algorithm without using collections   def countFreq A      n len A     count  0  n                       Create a new list initialized with  0     for i in range n         count A i     1                increase occurrence for value A i     return  x for x in count if x     return non-zero count

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str1  the cat sat on the hat hat  list1 str1 split    list2 str1 split     count 0  m      for i in range len list1        t list1 pop 0       print t     for j in range len list2            if t  list2 j                count count 1              print count     m append count      print m     count 0   print m

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Data  Let s say we have a list  fruits     banana    banana    apple    banana    Solution  Then we can find out how many of each fruit we have in the list by doing this  import numpy as np      unique  counts    np unique fruits  return counts True   x y for x y in zip unique  counts    Output    banana   3   apple   1

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def frequencyDistribution data       return  i  data count i  for i in data      print frequencyDistribution  1 2 3 4            1  1  2  1  3  1  4  1      originalNumber  count

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i m using Counter to generate a freq  dict from text file words in 1 line of code  def  fileIndex fh       create a dict using Counter of a flat list of words  re findall re compile r  a-zA-Z      lines   in  lines in file- gt for lines in fh      return Counter       wrd lower   for wrdList in       words for words in        re findall re compile r  a-zA-Z      lines  for lines in fh        for wrd in wrdList

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Here s another succint alternative using itertools groupby which also works for unordered input   from itertools import groupby  items    5  1  1  2  2  1  1  2  2  3  4  3  5   results    value  len list freq   for value  freq in groupby sorted items      results   1  4  2  4  3  2  4  1  5  2

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You can use the in-built function provided in python  l count l i       d      for i in range len l            if l i  not in d               d append l i                print l count l i     The above code automatically removes duplicates in a list and also prints the frequency of each element in original list and the list without duplicates   Two birds for one shot   X D

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a  1 2 3 4 5 1 2 3  b  0 0 0 0 0 0 0  for i in range 0 len a        b a i    1

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This answer is more explicit  a    1 1 1 1 2 2 2 2 3 3 3 4 4   d      for item in a      if item in d          d item    d get item  1     else          d item    1  for k v in d items        print str k      str v      output  1 4  2 4  3 3  4 2   remove dups d   set a  print d    1  2  3  4

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For an unordered list you should use   a count el  for el in set a    The output is  4  4  2  1  2

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To find unique elements in the list  a    1 1 1 1 2 2 2 2 3 3 4 5 5  a   list set a    To find the count of unique elements in a sorted array using dictionary  def CountFrequency my list      Creating an empty dictionary   freq       for item in my list       if  item in freq            freq item     1     else           freq item    1  for key  value in freq items         print   quot   d     d quot   key  value      Driver function  if   name       quot   main   quot     my list   1  1  1  5  5  3  1  3  3  1  4  4  4  2  2  2  2    CountFrequency my list   Reference  GeeksforGeeks

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Counting the frequency of elements is probably best done with a dictionary   b      for item in a      b item    b get item  0    1   To remove the duplicates  use a set   a   list set a

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In Python 2 7   you could use collections Counter to count items   gt  gt  gt  a    1 1 1 1 2 2 2 2 3 3 4 5 5   gt  gt  gt   gt  gt  gt  from collections import Counter  gt  gt  gt  c Counter a   gt  gt  gt   gt  gt  gt  c values    4  4  2  1  2   gt  gt  gt   gt  gt  gt  c keys    1  2  3  4  5

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For your first question  iterate the list and use a dictionary to keep track of an elements existsence    For your second question  just use the set operator

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num  3 2 3 5 5 3 7 6 4 6 7 2  print    nelements are  t  num  count dict    for elements in num      count dict elements  num count elements  print    nfrequency  t  count dict

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In Python 2 7  or newer   you can use collections Counter  import collections a    1 1 1 1 2 2 2 2 3 3 4 5 5  counter collections Counter a  print counter    Counter  1  4  2  4  3  2  5  2  4  1   print counter values       4  4  2  1  2  print counter keys       1  2  3  4  5  print counter most common 3       1  4    2  4    3  2    If you are using Python 2 6 or older  you can download it here

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You can do this   import numpy as np a    1 1 1 1 2 2 2 2 3 3 4 5 5  np unique a  return counts True    Output    array  1  2  3  4  5    array  4  4  2  1  2   dtype int64     The first array is values  and the second array is the number of elements with these values   So If you want to get just array with the numbers you should use this   np unique a  return counts True  1

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One more way is to use a dictionary and the list count  below a naive way to do it   dicio   dict    a    1 1 1 1 2 2 2 2 3 3 4 5 5   b   list    c   list    for i in a      if i in dicio  continue      else         dicio i    a count i         b append a count i          c append i   print  b   print  c

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For the record  a functional answer    gt  gt  gt  L    1 1 1 1 2 2 2 2 3 3 4 5 5   gt  gt  gt  import functools  gt  gt  gt   gt  gt  gt  functools reduce lambda acc  e   v  i  e  for i  v in enumerate acc 1   if e lt  len acc  else acc  0 for   in range e-len acc -1    1   L       4  4  2  1  2    It s cleaner if you count zeroes too    gt  gt  gt  functools reduce lambda acc  e   v  i  e  for i  v in enumerate acc   if e lt len acc  else acc  0 for   in range e-len acc     1   L       0  4  4  2  1  2    An explanation    we start with an empty acc list  if the next element e of L is lower than the size of acc  we just update this element  v  i  e  means v 1 if the index i of acc is the current element e  otherwise the previous value v  if the next element e of L is greater or equals to the size of acc  we have to expand acc to host the new 1    The elements do not have to be sorted  itertools groupby   You ll get weird results if you have negative numbers

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I am quite late  but this will also work  and will help others    a    1 1 1 1 2 2 2 2 3 3 4 5 5  freq list      a l   list set a    for x in a l      freq list append a count x     print  Freq  freq list print  number  a l   will produce this    Freq   4  4  2  1  2  number 1  2  3  4  5

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Note  You should sort the list before using groupby  You can use groupby from itertools package if the list is an ordered list  a    1 1 1 1 2 2 2 2 3 3 4 5 5  from itertools import groupby  len list group   for key  group in groupby a    Output   4  4  2  1  2   update  Note that sorting takes O n log n   time

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from collections import OrderedDict a    1 1 1 1 2 2 2 2 3 3 4 5 5  def get count lists       dictionary   OrderedDict       for val in lists          dictionary setdefault val     append 1      return  sum val  for val in dictionary values    print get count a    gt  gt  gt  4  4  2  1  2    To remove duplicates and Maintain order   list dict fromkeys get count a     gt  gt  gt  4  2  1

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Found another way of doing this  using sets    ar is the list of elements  convert ar to set to get unique elements sock set   set ar    create dictionary of frequency of socks sock dict       for sock in sock set      sock dict sock    ar count sock

[python] How to count the frequency of the elements in an unordered list?

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