numpy矢量化方法来计算整数数组中的非零位 [英] numpy vectorized way to count non-zero bits in array of integers

问题描述

我有一个整数数组:

[int1, int2, ..., intn]

我想计算这些整数的二进制表示中有多少非零位.

I want to count how many non-zero bits are in the binary representation of these integers.

例如:

bin(123) -> 0b1111011, there are 6 non-zero bits

当然我可以遍历整数列表，使用 bin() 和 count('1') 函数，但我正在寻找向量化的方法

Of course I can loop over list of integers, use bin() and count('1') functions, but I'm looking for vectorized way to do it.

推荐答案

假设你的数组是 a，你可以简单地做:

Assuming your array is a, you can simply do:

np.unpackbits(a.view('uint8')).sum()

示例:

a = np.array([123, 44], dtype=np.uint8)
#bin(a) is [0b1111011, 0b101100]
np.unpackbits(a.view('uint8')).sum()
#9

---

比较使用benchit:

#@Ehsan's solution
def m1(a):
  return np.unpackbits(a.view('uint8')).sum()

#@Valdi_Bo's solution
def m2(a):
  return sum([ bin(n).count('1') for n in a ])

in_ = [np.random.randint(100000,size=(n)) for n in [10,100,1000,10000,100000]]

m1 明显更快.

这篇关于numpy矢量化方法来计算整数数组中的非零位的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！