关于：
numpy.bincount(arr，weights = None，min_len = 0)：在+ ve个整数数组中，它计算每个元素的出现。每个bin值是其索引的出现。您也可以相应地设置箱子大小。
参数：
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
返回：
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.
代码1：在NumPy中运行bincount()
# Python Program explaining
# working of numpy.bincount() method
import numpy as geek
# 1D array with +ve integers
array1 = [1, 6, 1, 1, 1, 2, 2]
bin = geek.bincount(array1)
print("Bincount output :\n ", bin)
print("size of bin:", len(bin), "\n")
array2 = [1, 5, 5, 5, 4, 5, 5, 2, 2, 2]
bin = geek.bincount(array2)
print("Bincount output :\n ", bin)
print("size of bin:", len(bin), "\n")
# using min_length attribute
length = 10
bin1 = geek.bincount(array2, None, length)
print("Bincount output :\n ", bin1)
print("size of bin:", len(bin1), "\n")输出：
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
代码2：我们可以按权重bincount()的元素执行加法
# Python Program explaining
# working of numpy.bincount() 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:\n", bin)
#index 0:0
#index 1:10 + 4 + 2 = 16
#index 2:1 + 9 = 10
#index 3:11 + 6 = 17输出：
Summation element-wise: [ 0. 16. 10. 17.]
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注：本文由纯净天空筛选整理自 numpy.bincount() in Python。非经特殊声明，原始代码版权归原作者所有，本译文未经允许或授权，请勿转载或复制。