python - 对数组的两个值进行更快的循环操作

Question

考虑以下函数：

def dostuff(n, f):
    array = numpy.arange(0, n)
    for i in range(1, n):                    # Line 1                      
        array[i] = f(array[i-1], array[i])   # Line 2
    return numpy.sum(array)

如何重写第 1 行/第 2 行以使循环在 python 3 中更快（不使用 cython）？

Answer 1

我鼓励你检查这个关于SO NumPy/SciPy 中广义累积函数的问题？，因为你想要一个generalized cumulative function.

还可以在此处查看函数的 scipy 文档frompyfunc

func = np.frompyfunc(f , 2 , 1)

def dostuff(n,f):
    final_array = func.accumulate(np.arange(0,n), dtype=np.object).astype(np.int)
    return np.sum(final_array)

例子

---

In [86]:
def f(num1 , num2):
    return num1 + num2

In [87]:
func = np.frompyfunc(f , 2 , 1)

In [88]:
def dostuff(n,f):
    final_array = func.accumulate(np.arange(0,n), dtype=np.object).astype(np.int)
    return np.sum(final_array)

In [108]:
dostuff(15,f)
Out[108]:
560

In [109]:
dostuff(10,f)
Out[109]:
165

基准

---

def dostuff1(n, f):
    array = np.arange(0, n)
    for i in range(1, n):                    # Line 1                      
        array[i] = f(array[i-1], array[i])   # Line 2
    return np.sum(array)

def dostuff2(n,f):
    final_array = func.accumulate(np.arange(0,n), dtype=np.object).astype(np.int)
    return np.sum(final_array)

In [126]:
%timeit dostuff1(100,f)
10000 loops, best of 3: 40.6 µs per loop

In [127]:
%timeit dostuff2(100,f)
The slowest run took 4.98 times longer than the fastest. This could mean that an intermediate result is being cached 
10000 loops, best of 3: 23.8 µs per loop