Question

Python，Numpy

是否有更紧凑的方式对数组元素进行操作，而不必使用标准for循环。？

例如，请考虑以下功能：

filterData(A):
    B = numpy.zeros(len(A));
    B[0] = (A[0] + A[1])/2.0;
    for i in range(1, len(A)): 
        B[i] = (A[i]-A[i-1])/2.0;
    return B;

Answer 1

Numpy有一个diff operator，适用于numpy数组和Python本机数组。您可以将代码重写为：

def filterData(A):
    B = numpy.zeros(len(A));
    B[1:] = np.diff( A )/2.0
    B[0] = (A[0] + A[1])/2.0;
    return B

Answer 2

还有numpy.ediff1d，它允许您使用to_end和to_begin参数明确地前置或附加到差异，例如：

>>> import numpy as np
>>> a = np.arange(10.)
>>> diff = np.ediff1d(a,to_begin = a[:2].sum()) / 2.
>>> diff
array([ 0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5,  0.5])
>>> diff.size
10

python：for循环紧凑表示

2 个答案: