擅长:python、mysql、java
<p>多亏了jasonharper,我意识到我的实现效率低得离谱,可以简单得多。你知道吗</p>
<p>他的方法实现如下:</p>
<pre><code>def BestApprox_fast(value):
mindiff = 1000000000
for Dp in np.arange(0, 32, 1):
M = round(value*2**Dp)
if abs(M) < 1000:
diff = Diff([M, Dp], value)
if diff < mindiff:
mindiff = diff
M_best = M
Dp_best = Dp
return M_best, Dp_best
</code></pre>
<p>大约快200倍。你知道吗</p>