Python Numpy项目Euler 1广义优化

我已经开始学习Numpy，我正在寻找写这篇文章的方法。我写了Euler 1的一个推广。它需要一个除数列表和一个数字，例如Euler 1中的[3,5]和1000

在纯python中天真地：

def subn1(divisors, n):
    return sum(i for i in range(1,n) if not all(i % d for d in divisors))

对于范围（2,20），1000000，这大约需要2.5秒

我迄今为止第一次也是最好的尝试如下：

def subn2(divisors, n):
    a = np.arange(1,n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[b]) 

def subn4(divisors, n):
    a = np.arange(np.min(divisors),n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[np.where(b)])

def subn4_(divisors, n):
    a = np.arange(1,n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[np.where(b)]) 

对于范围（2,20）和1000000，运行时间约为0.45秒

我的第三次尝试是删除foor循环并使用纯numpy，但是它在速度方面损失了一小部分，并且使用了更多内存

def subn3(divisors, n):
    nums = np.arange(1,n)     
    divs = np.array([divisors]).T
    return np.sum(nums[np.logical_or.reduce(np.mod(nums, divs) == 0, axis=0)])

对于射程（2,20）和100000，这大约需要0.5秒

有没有一种更快的方法可以用“纯”numpy来编写它，还是foor循环不必回避

---

注意：我知道您可以通过减少除数列表来优化它，因此无需对此进行评论：）

一种具有-

运行时测试和验证-

In [14]: # Inputs
    ...: n = 1000
    ...: divisors = range(2,20)
    ...: 

In [15]: print subn1(divisors, n)
    ...: print subn2(divisors, n)
    ...: print subn3(divisors, n)
    ...: print subn_broadcasting(divisors, n)
    ...: 
416056
416056
416056
416056

In [16]: %timeit subn1(divisors, n)
    ...: %timeit subn2(divisors, n)
    ...: %timeit subn3(divisors, n)
    ...: %timeit subn_broadcasting(divisors, n)
    ...: 
1000 loops, best of 3: 1.39 ms per loop
1000 loops, best of 3: 480 µs per loop
1000 loops, best of 3: 434 µs per loop
1000 loops, best of 3: 428 µs per loop

---

嗯，与

n2

和

n3

版本相比，似乎没有多大改进。

-

运行时测试和验证-

In [14]: # Inputs
    ...: n = 1000
    ...: divisors = range(2,20)
    ...: 

In [15]: print subn1(divisors, n)
    ...: print subn2(divisors, n)
    ...: print subn3(divisors, n)
    ...: print subn_broadcasting(divisors, n)
    ...: 
416056
416056
416056
416056

In [16]: %timeit subn1(divisors, n)
    ...: %timeit subn2(divisors, n)
    ...: %timeit subn3(divisors, n)
    ...: %timeit subn_broadcasting(divisors, n)
    ...: 
1000 loops, best of 3: 1.39 ms per loop
1000 loops, best of 3: 480 µs per loop
1000 loops, best of 3: 434 µs per loop
1000 loops, best of 3: 428 µs per loop

嗯，与

n2

和

n3

版本相比，似乎没有多大改进。

您可以使用，如下所示：

def subn2(divisors, n):
    a = np.arange(1,n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[b]) 

def subn4(divisors, n):
    a = np.arange(np.min(divisors),n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[np.where(b)])

def subn4_(divisors, n):
    a = np.arange(1,n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[np.where(b)]) 

这些测试，如前所述：

%timeit subn1(divisors, n)
%timeit subn2(divisors, n)
%timeit subn3(divisors, n)
%timeit subn_broadcasting(divisors, n)
%timeit subn4(divisors, n)
%timeit subn4_(divisors, n)


1 loops, best of 3: 596 ms per loop
10 loops, best of 3: 30.1 ms per loop
10 loops, best of 3: 32 ms per loop
10 loops, best of 3: 31.9 ms per loop
10 loops, best of 3: 28.2 ms per loop
10 loops, best of 3: 27.4 ms per loop

您可以这样使用：

def subn2(divisors, n):
    a = np.arange(1,n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[b]) 

def subn4(divisors, n):
    a = np.arange(np.min(divisors),n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[np.where(b)])

def subn4_(divisors, n):
    a = np.arange(1,n) 
    b = np.zeros(a.shape[0], dtype=bool) 
    for d in divisors:
        b += a % d == 0
    return np.sum(a[np.where(b)]) 

这些测试，如前所述：

%timeit subn1(divisors, n)
%timeit subn2(divisors, n)
%timeit subn3(divisors, n)
%timeit subn_broadcasting(divisors, n)
%timeit subn4(divisors, n)
%timeit subn4_(divisors, n)


1 loops, best of 3: 596 ms per loop
10 loops, best of 3: 30.1 ms per loop
10 loops, best of 3: 32 ms per loop
10 loops, best of 3: 31.9 ms per loop
10 loops, best of 3: 28.2 ms per loop
10 loops, best of 3: 27.4 ms per loop