numpy.vectorize:为什么这么慢? [英] numpy.vectorize: Why so slow?
问题描述
scipy.special中的expit函数是向量化的S型函数.它计算1/(1 + e ^(-x))很复杂，可能涉及泰勒级数.
The expit function, in scipy.special, is a vectorized sigmoid function. It computes 1 / (1+e^(-x)), which is complicated, probably involving a Taylor series.
我了解了快速S形"，1/(1 + abs(x))，它应该快得多-但是内置的expit函数的性能大大超过了它，即使当我将其作为lambda表达式传递给numpy.vectorize.
I learned about the "fast sigmoid", 1 / (1 + abs(x)), which ought to be much faster -- but the builtin expit function massively outperforms it, even when I hand it as a lambda expression to numpy.vectorize.
这里是测试它们的一种方法:
Here's one way to test them:
from scipy.special import expit
data = np.random.rand(1000000)
内置的，复杂的Sigmoid速度很快:
The built-in, complicated sigmoid is fast:
%prun expit(data)
3 function calls in 0.064 seconds
Ordered by: internal time
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.064 0.064 0.064 0.064 <string>:1(<module>)
1 0.000 0.000 0.064 0.064 {built-in method builtins.exec}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
更简单的乙状结肠慢了约20倍:
The simpler sigmoid is around 20 times slower:
%prun np.vectorize( lambda x: (x / (1 + abs(x)) + 1) / 2 )(data)
2000023 function calls in 1.992 seconds
Ordered by: internal time
ncalls tottime percall cumtime percall filename:lineno(function)
1000001 1.123 0.000 1.294 0.000 <string>:1(<lambda>)
1 0.558 0.558 1.950 1.950 function_base.py:2276(_vectorize_call)
1000001 0.170 0.000 0.170 0.000 {built-in method builtins.abs}
4 0.098 0.025 0.098 0.025 {built-in method numpy.core.multiarray.array}
1 0.041 0.041 1.991 1.991 function_base.py:2190(__call__)
1 0.000 0.000 0.068 0.068 function_base.py:2284(<listcomp>)
1 0.000 0.000 1.992 1.992 {built-in method builtins.exec}
1 0.000 0.000 1.991 1.991 <string>:1(<module>)
1 0.000 0.000 0.000 0.000 function_base.py:2220(_get_ufunc_and_otypes)
1 0.000 0.000 0.000 0.000 function_base.py:2162(__init__)
1 0.000 0.000 0.000 0.000 function_base.py:2242(<listcomp>)
2 0.000 0.000 0.000 0.000 numeric.py:414(asarray)
1 0.000 0.000 0.000 0.000 {built-in method numpy.core.umath.frompyfunc}
1 0.000 0.000 0.000 0.000 function_base.py:2266(<listcomp>)
2 0.000 0.000 0.000 0.000 {built-in method builtins.isinstance}
1 0.000 0.000 0.000 0.000 {built-in method builtins.len}
1 0.000 0.000 0.000 0.000 {method 'join' of 'str' objects}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
推荐答案
I'll just quote the the vectorize docstring: "The vectorize function is provided primarily for convenience, not for performance. The implementation is essentially a for loop."
您想使 1/(1 + abs(x))快速.numpy具有称为 numpy.abs 的功能(也称为
You want to make 1/(1 + abs(x)) fast. numpy has a function called numpy.abs (also called numpy.absolute--they are different names for the same object). It computes the absolute value of each element of its argument, and it does this in C code, so it is fast. Also, the Python built-in function abs knows how to dispatch arguments to objects that have the method __abs__, which numpy arrays do, so you can also use Python's abs() to compute the element-wise absolute value of a numpy array. In the following, though, I'll use np.abs.
以下是使用 np.abs 的示例:
In [25]: x = np.array([-2, -1.5, 0, 5, 10])
In [26]: np.abs(x)
Out[26]: array([ 2. , 1.5, 0. , 5. , 10. ])
这里是大型阵列 x的: scipy.special.expit 和 1.0/(1 + np.abs(x))性能的比较
Here's a comparision of the performance of scipy.special.expit and 1.0/(1 + np.abs(x)) for a large array x:
In [34]: from scipy.special import expit
In [35]: x = np.random.randn(100000)
In [36]: %timeit expit(x)
1000 loops, best of 3: 771 µs per loop
In [37]: %timeit 1.0/(1 + np.abs(x))
1000 loops, best of 3: 279 µs per loop
所以 1.0/(1 + np.abs(x))比 expit(x)快很多.
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