尝试用 numpy 向量化迭代计算 [英] Trying to vectorize iterative calculation with numpy

问题描述

我试图通过在 numpy 中使用矢量化形式来使某些代码更高效.让我给你举个例子，让你明白我的意思.

I am trying to make some piece of code more efficient by using the vectorized form in numpy. Let me show you an example so you know what I mean.

给定以下代码:

a = np.zeros([4,4])
a[0] = [1., 2., 3., 4.]
for i in range(len(a)-1):
    a[i+1] = 2*a[i]
print a

输出

[[  1.   2.   3.   4.]
 [  2.   4.   6.   8.]
 [  4.   8.  12.  16.]
 [  8.  16.  24.  32.]]

当我现在尝试像这样矢量化代码时:

When I now try to vectorize the code like this:

a = np.zeros([4,4])
a[0] = [1., 2., 3., 4.]
a[1:] = 2*a[0:-1]
print a

我刚刚得到正确的第一次迭代:

I just get the first iteration correct:

[[ 1.  2.  3.  4.]
 [ 2.  4.  6.  8.]
 [ 0.  0.  0.  0.]
 [ 0.  0.  0.  0.]]

是否可以以矢量化形式(其中下一次迭代始终访问前一次迭代)有效地编写上述代码，还是必须保持 for 循环?

Is it possible to write the code above efficiently in a vectorized form (where the next iteration always accesses the previous iteration) or do I have to keep the for loop?

推荐答案

可以使用 scipy.signal.lfilter:

In [19]: from scipy.signal import lfilter

In [20]: num = np.array([1.0])

In [21]: alpha = 2.0

In [22]: den = np.array([1.0, -alpha])

In [23]: a = np.zeros((4,4))

In [24]: a[0,:] = [1,2,3,4]

In [25]: lfilter(num, den, a, axis=0)
Out[25]: 
array([[  1.,   2.,   3.,   4.],
       [  2.,   4.,   6.,   8.],
       [  4.,   8.,  12.,  16.],
       [  8.,  16.,  24.,  32.]])

有关更多详细信息，请参阅以下内容:python 递归向量化与时间序列, Pandas 中的递归定义

See the following for more details: python recursive vectorization with timeseries, Recursive definitions in Pandas

请注意，使用 lfilter 只有在解决非齐次问题时才有意义，例如 x[i+1] = alpha*x[i] + u[i]，其中 u 是给定的输入数组.对于简单的递归a[i+1] = alpha*a[i]，可以使用精确解a[i] = a[0]*alpha**i.可以使用广播对多个初始值的解决方案进行矢量化.例如，

Note that using lfilter really only makes sense if you are solving a nonhomogeneous problem such as x[i+1] = alpha*x[i] + u[i], where u is a given input array. For the simple recurrence a[i+1] = alpha*a[i], you can use the exact solution a[i] = a[0]*alpha**i. The solution for multiple initial values can be vectorized using broadcasting. For example,

In [271]: alpha = 2.0

In [272]: a0 = np.array([1, 2, 3, 4])

In [273]: n = 5

In [274]: a0 * (alpha**np.arange(n).reshape(-1, 1))
Out[274]: 
array([[  1.,   2.,   3.,   4.],
       [  2.,   4.,   6.,   8.],
       [  4.,   8.,  12.,  16.],
       [  8.,  16.,  24.,  32.],
       [ 16.,  32.,  48.,  64.]])

这篇关于尝试用 numpy 向量化迭代计算的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！