擅长:python、mysql、java
<p>你可以自己写:</p>
<pre><code>def numpy_gcd(a, b):
a, b = np.broadcast_arrays(a, b)
a = a.copy()
b = b.copy()
pos = np.nonzero(b)[0]
while len(pos) > 0:
b2 = b[pos]
a[pos], b[pos] = b2, a[pos] % b2
pos = pos[b[pos]!=0]
return a
</code></pre>
<p>下面是测试结果和速度的代码:</p>
<pre><code>In [181]:
n = 2000
a = np.random.randint(100, 1000, n)
b = np.random.randint(1, 100, n)
al = a.tolist()
bl = b.tolist()
cl = zip(al, bl)
from fractions import gcd
g1 = numpy_gcd(a, b)
g2 = [gcd(x, y) for x, y in cl]
print np.all(g1 == g2)
True
In [182]:
%timeit numpy_gcd(a, b)
1000 loops, best of 3: 721 us per loop
In [183]:
%timeit [gcd(x, y) for x, y in cl]
1000 loops, best of 3: 1.64 ms per loop
</code></pre>