How to count the occurrence of certain item in an ndarray

Question

In Python  I have an ndarray y that is printed as array  0  0  0  1  0  1  1  0  0  0  0  1    I m trying to count how many 0s and how many 1s are there in this array    But when I type y count 0  or y count 1   it says      numpy ndarray object has no attribute count   What should I do

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Convert your array y to list l and then do l count 1  and l count 0    gt  gt  gt  y   numpy array  0  0  0  1  0  1  1  0  0  0  0  1    gt  gt  gt  l   list y   gt  gt  gt  l count 1  4  gt  gt  gt  l count 0  8

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Try this   a   np array  0  0  0  1  0  1  1  0  0  0  0  1   list a  count 1

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y tolist   count val   with val 0 or 1  Since a python list has a native function count  converting to list before using that function is a simple solution

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y   np array  0  0  0  1  0  1  1  0  0  0  0  1     If you know that they are just 0 and 1   np sum y    gives you the number of ones   np sum 1-y  gives the zeroes   For slight generality  if you want to count 0 and not zero  but possibly 2 or 3    np count nonzero y    gives the number of nonzero   But if you need something more complicated  I don t think numpy will provide a nice count option   In that case  go to collections   import collections collections Counter y   gt  Counter  0  8  1  4     This behaves like a dict  collections Counter y  0   gt  8

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It involves one more step  but a more flexible solution which would also work for 2d arrays and more complicated filters is to create a boolean mask and then use  sum   on the mask    gt  gt  gt  gt y   np array  0  0  0  1  0  1  1  0  0  0  0  1    gt  gt  gt  gt mask   y    0  gt  gt  gt  gt mask sum   8

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dict zip  numpy unique y  return counts True      Just copied  Seppo Enarvi s comment here which deserves to be a proper answer

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I d use np where   how many 0   len np where a  0   0   how many 1   len np where a  1   0

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here I have something  through which you can count the number of occurrence of a particular number  according to your code  count of zero list y y  0   count 0    print count of zero      according to the match there will be boolean values and according to True value the number 0 will be return

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Numpy has a module for this  Just a small hack  Put your input array as bins    numpy histogram y  bins y    The output are 2 arrays  One with the values itself  other with the corresponding frequencies

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Yet another simple solution might be to use numpy count nonzero     import numpy as np y   np array  0  0  0  1  0  1  1  0  0  0  0  1   y nonzero num   np count nonzero y  1  y zero num   np count nonzero y  0  y nonzero num 4 y zero num 8   Don t let the name mislead you  if you use it with the boolean just like in the example  it will do the trick

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Since your ndarray contains only 0 and 1   you can use sum   to get the occurrence of 1s  and len  -sum   to get the occurrence of 0s   num of ones   sum array  num of zeros   len array -sum array

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To count the number of occurrences  you can use np unique array  return counts True    In  75   boo   np array  0  0  0  1  0  1  1  0  0  0  0  1      use bool value  True  or equivalently  1  In  77   uniq  cnts   np unique boo  return counts 1  In  81   uniq Out 81   array  0  1      unique elements in input array are  0  1  In  82   cnts Out 82   array  8  4       0 occurs 8 times  1 occurs 4 times

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take advantage of the methods offered by a Series    gt  gt  gt  import pandas as pd  gt  gt  gt  y    0  0  0  1  0  1  1  0  0  0  0  1   gt  gt  gt  pd Series y  value counts   0    8 1    4 dtype  int64

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If you know exactly which number you re looking for  you can use the following   lst   np array  1 1 2 3 3 6 6 6 3 2 1    lst    2  sum     returns how many times 2 is occurred in your array

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What about len y y  0   and len y y  1

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using numpy count    a    0  0  0  1  0  1  1  0  0  0  0  1     np count a  1

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This funktion returns the number of occurences of a variable in an array  def count array variable       number   0     for i in range array shape 0            for j in range array shape 1                if array i j     variable                  number    1     return number

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For your case you could also look into numpy bincount  In  56   a   np array  0  0  0  1  0  1  1  0  0  0  0  1    In  57   np bincount a  Out 57   array  8  4     count of zeros is at index 0   8                          count of ones is at index 1   4

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a   numpy array  0  3  0  1  0  1  2  1  0  0  0  0  1  3  4   unique  counts   numpy unique a  return counts True  dict zip unique  counts       0  7  1  4  2  1  3  2  4  1   Non-numpy way  Use collections Counter  import collections  numpy a   numpy array  0  3  0  1  0  1  2  1  0  0  0  0  1  3  4   collections Counter a     Counter  0  7  1  4  3  2  2  1  4  1

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This can be done easily in the following method  y   np array  0  0  0  1  0  1  1  0  0  0  0  1   y tolist   count 1

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You can use dictionary comprehension to create a neat one-liner  More about dictionary comprehension can be found here    gt  gt  gt counts    int value   list y  count value  for value in set y    gt  gt  gt print counts   0  8  1  4    This will create a dictionary with the values in your ndarray as keys  and the counts of the values as the values for the keys respectively   This will work whenever you want to count occurences of a value in arrays of this format

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Personally  I d go for   y    0  sum   and  y    1  sum    E g   import numpy as np y   np array  0  0  0  1  0  1  1  0  0  0  0  1   num zeros    y    0  sum   num ones    y    1  sum

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Honestly I find it easiest to convert to a pandas Series or DataFrame   import pandas as pd import numpy as np  df   pd DataFrame   data  np array  0  0  0  1  0  1  1  0  0  0  0  1     print df  data   value counts     Or this nice one-liner suggested by Robert Muil   pd Series  0  0  0  1  0  1  1  0  0  0  0  1   value counts

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You have a special array with only 1 and 0 here  So a trick is to use   np mean x    which gives you the percentage of 1s in your array  Alternatively  use  np sum x  np sum 1-x    will give you the absolute number of 1 and 0 in your array

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If you are interested in the fastest execution  you know in advance which value s  to look for  and your array is 1D  or you are otherwise interested in the result on the flattened array  in which case the input of the function should be np flatten arr  rather than just arr   then Numba is your friend   import numba as nb    nb jit def count nb arr  value       result   0     for x in arr          if x    value              result    1     return result   or  for very large arrays where parallelization may be beneficial    nb jit parallel True  def count nbp arr  value       result   0     for i in nb prange arr size           if arr i     value              result    1     return result   Benchmarking these against np count nonzero    which also has a problem of creating a temporary array which may be avoided  and np unique  -based solution  import numpy as np   def count np arr  value       return np count nonzero arr    value    import numpy as np   def count np2 arr  value       uniques  counts   np unique a  return counts True      counter   dict zip uniques  counts       return counter value  if value in counter else 0    for input generated with   def gen input n  a 0  b 100       return np random randint a  b  n    the following plots are obtained  the second row of plots is a zoom on the faster approach       Showing that Numba-based solution are noticeably faster than the NumPy counterparts  and  for very large inputs  the parallel approach is faster than the naive one     Full code available here

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if you are dealing with very large arrays using generators could be an option  The nice thing here it that this approach works fine for both arrays and lists and you dont need any additional package  Additionally  you are not using that much memory    my array   np array  0  0  0  1  0  1  1  0  0  0  0  1   sum 1 for val in my array if val  0  Out  8

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A general and simple answer would be   numpy sum MyArray  x      sum of a binary list of the occurence of x   0 or 1  in MyArray   which would result into this full code as exemple  import numpy MyArray numpy array  0  0  0  1  0  1  1  0  0  0  0  1      array we want to search in x 0     the value I want to count  can be iterator  in a list  etc   numpy sum MyArray  0      sum of a binary list of the occurence of x in MyArray   Now if MyArray is in multiple dimensions and you want to count the occurence of a distribution of values in line    pattern hereafter   MyArray numpy array   6  1   4  5   0  7   5  1   2  5   1  2   3  2   0  2   2  5   5  1   3  0    x numpy array  5 1       the value I want to count  can be iterator  in a list  etc   temp   numpy ascontiguousarray MyArray  view numpy dtype  numpy void  MyArray dtype itemsize   MyArray shape 1        convert the 2d-array into an array of analyzable patterns xt numpy ascontiguousarray x  view numpy dtype  numpy void  x dtype itemsize   x shape 0        convert what you search into one analyzable pattern numpy sum temp  xt     count of the searched pattern in the list of patterns

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What about using numpy count nonzero  something like    gt  gt  gt  import numpy as np  gt  gt  gt  y   np array  1  2  2  2  2  0  2  3  3  3  0  0  2  2  0     gt  gt  gt  np count nonzero y    1  1  gt  gt  gt  np count nonzero y    2  7  gt  gt  gt  np count nonzero y    3  3

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For generic entries   x   np array  11  2  3  5  3  2  16  10  10  3  11  4  5  16  3  11  4   n    i len  j for j in np where x  i  0    for i in set x   ix    i  j for j in np where x  i  0   for i in set x     Will output a count    2  2  3  4  4  2  5  2  10  2  11  3  16  2    And indices    2   1  5   3   2  4  9  14   4   11  16   5   3  12   10   7  8   11   0  10  15   16   6  13

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No one suggested to use numpy bincount input  minlength  with minlength   np size input   but it seems to be a good solution  and definitely the fastest   In  1   choices   np random randint 0  100  10000   In  2    timeit   np sum choices    k  for k in range min choices   max choices  1    100 loops  best of 3  2 67 ms per loop  In  3    timeit np unique choices  return counts True  1000 loops  best of 3  388   s per loop  In  4    timeit np bincount choices  minlength np size choices   100000 loops  best of 3  16 3   s per loop   That s a crazy speedup between numpy unique x  return counts True  and numpy bincount x  minlength np max x

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If you don t want to use numpy or a collections module you can use a dictionary   d   dict   a    0  0  0  1  0  1  1  0  0  0  0  1  for item in a      try          d item   1     except KeyError          d item  1   result    gt  gt  gt d  0  8  1  4    Of course you can also use an if else statement   I think the Counter function does almost the same thing but this is more transparant

[python] How to count the occurrence of certain item in an ndarray?

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