 # numpy.bincount () in Python

Around:
numpy.bincount (arr, weights = None, min_len = 0): in the + ve array of integers, it counts the occurrence of each element. Each bin value is an entry for its index. You can also set the size of the basket accordingly.
Parameters:

`  arr:  [array_like, 1D] Input array, having positive numbers  weights:  [ array_like, optional] same shape as that of arr  min_len:  Minimum number of bins we want in the output array `

Return:

` Output array with no. of occurrence of index value of bin in input - arr. Output array, by default is of the length max element of arr + 1. `

Code 1: Working bincount () in NumPy

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``` # Python program explaining # working method numpy.bincount ()   import numpy as geek   # 1D array with + ve integers array1 = [ 1 , 6 , 1 , 1 , 1 , < code class = "value"> 2 , 2 ] bin = geek.bincount (array1) print ( "Bincount output :" , bin ) print ( "size of bin:" , len ( bin ), " " )    array2 = [ 1 , 5 , 5 , 5 , 4 , 5 , 5 , 2 , 2 , 2 ] bin = geek.bincount (array2) print ( " Bincount output : " , bin ) print ( "size of bin:" , len ( bin ), "" )   # using min_length attribute length = 10 bin1 = geek.bincount (array2, None , length) print ( "Bincount output :" , bin1)   print ( "size of bin:" , len (bin1), "" ) ```

Exit :

` Bincount output: [0 4 2 0 0 0 1] size of bin: 7 Bincount output: [0 1 3 0 1 5] size of bin: 6 Bincount output: [0 1 3 0 1 5 0 0 0 0] size of bin: 10 `

Code 2: we can add according to the element using bincount ()

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``` # Python program explaining # numpy.bincount () working method   import numpy as geek   # 1D array with + ve integers array2 = [ 10  , 11 , 4 , 6 , 2 , 1 , 9 ] array1 = [ 1 , 3 , 1 , 3 , 1 , 2 , 2 ]   # array2: weight bin = geek.bincount (array1, array2) print ( "Summation element-wise:" , bin )   # Index 0: 0 # Index 1: 10 + 4 + 2 = 16 # Index 2: 1 + 9 = 10 # Index 3: 11 + 6 = 17 ```

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Output:

` Summation element-wise: [0. 16. 10. 17.] `