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<p>我刚刚注意到,我的脚本执行时间几乎减半,只需将乘法改为除法。在</p>
<p>为了调查这一点,我写了一个小例子:</p>
<pre><code>import numpy as np
import timeit
# uint8 array
arr1 = np.random.randint(0, high=256, size=(100, 100), dtype=np.uint8)
# float32 array
arr2 = np.random.rand(100, 100).astype(np.float32)
arr2 *= 255.0
def arrmult(a):
"""
mult, read-write iterator
"""
b = a.copy()
for item in np.nditer(b, op_flags=["readwrite"]):
item[...] = (item + 5) * 0.5
def arrmult2(a):
"""
mult, index iterator
"""
b = a.copy()
for i, j in np.ndindex(b.shape):
b[i, j] = (b[i, j] + 5) * 0.5
def arrmult3(a):
"""
mult, vectorized
"""
b = a.copy()
b = (b + 5) * 0.5
def arrdiv(a):
"""
div, read-write iterator
"""
b = a.copy()
for item in np.nditer(b, op_flags=["readwrite"]):
item[...] = (item + 5) / 2
def arrdiv2(a):
"""
div, index iterator
"""
b = a.copy()
for i, j in np.ndindex(b.shape):
b[i, j] = (b[i, j] + 5) / 2
def arrdiv3(a):
"""
div, vectorized
"""
b = a.copy()
b = (b + 5) / 2
def print_time(name, t):
print("{: <10}: {: >6.4f}s".format(name, t))
timeit_iterations = 100
print("uint8 arrays")
print_time("arrmult", timeit.timeit("arrmult(arr1)", "from __main__ import arrmult, arr1", number=timeit_iterations))
print_time("arrmult2", timeit.timeit("arrmult2(arr1)", "from __main__ import arrmult2, arr1", number=timeit_iterations))
print_time("arrmult3", timeit.timeit("arrmult3(arr1)", "from __main__ import arrmult3, arr1", number=timeit_iterations))
print_time("arrdiv", timeit.timeit("arrdiv(arr1)", "from __main__ import arrdiv, arr1", number=timeit_iterations))
print_time("arrdiv2", timeit.timeit("arrdiv2(arr1)", "from __main__ import arrdiv2, arr1", number=timeit_iterations))
print_time("arrdiv3", timeit.timeit("arrdiv3(arr1)", "from __main__ import arrdiv3, arr1", number=timeit_iterations))
print("\nfloat32 arrays")
print_time("arrmult", timeit.timeit("arrmult(arr2)", "from __main__ import arrmult, arr2", number=timeit_iterations))
print_time("arrmult2", timeit.timeit("arrmult2(arr2)", "from __main__ import arrmult2, arr2", number=timeit_iterations))
print_time("arrmult3", timeit.timeit("arrmult3(arr2)", "from __main__ import arrmult3, arr2", number=timeit_iterations))
print_time("arrdiv", timeit.timeit("arrdiv(arr2)", "from __main__ import arrdiv, arr2", number=timeit_iterations))
print_time("arrdiv2", timeit.timeit("arrdiv2(arr2)", "from __main__ import arrdiv2, arr2", number=timeit_iterations))
print_time("arrdiv3", timeit.timeit("arrdiv3(arr2)", "from __main__ import arrdiv3, arr2", number=timeit_iterations))
</code></pre>
<p>这将打印以下计时:</p>
^{pr2}$
<p>我一直认为乘法在计算上比除法便宜。然而,对于<code>uint8</code>来说,除法的效果几乎是前者的两倍。这是否与<code>* 0.5</code>必须计算浮点乘法,然后将结果转换回整数?在</p>
<p>至少对于float,乘法似乎比除法快。这通常是真的吗?在</p>
<p>为什么<code>uint8</code>中的乘法比<code>float32</code>中的乘法更膨胀?我以为8位无符号整数应该比32位浮点运算快得多?!在</p>
<p>有人能“解开”这个谜团吗?在</p>
<p><strong>编辑</strong>:为了获得更多的数据,我还包括了向量化函数(如建议的),并添加了索引迭代器。矢量化的函数要快得多,因此不具有真正的可比性。然而,如果向量化函数的<code>timeit_iterations</code>设置得更高,则<code>uint8</code>和{<cd4>}的乘法速度更快。我想这更让人困惑?!在</p>
<p>也许乘法实际上总是比除法快,但是for循环的主要性能漏洞不是算术运算,而是循环本身。尽管这并不能解释为什么循环在不同的操作中表现不同。在</p>
<p><strong>EDIT2</strong>:如@jotasi所述,我们正在寻找<code>division</code>与{<cd9>}和{<cd10>}(或{<cd1>})与{<cd12>}(或{<cd4>})的完整解释。此外,解释矢量化方法和迭代器的不同趋势也很有趣,因为在矢量化的情况下,除法似乎比较慢,而在迭代器的情况下则更快。在</p>