如何顺序更新Numpy数组而无循环 [英] How to update a Numpy array sequentially without loop

问题描述

我有一个Numpy数组 v ，我想使用该数组当前元素上的函数来更新每个元素:

I have a Numpy array v and I want to update each element using a function on the current element of the array :

v[i] = f(v, i)

执行此操作的基本方法是使用循环

A basic way to do this is to use a loop

for i in xrange(2, len(v)):
    v[i] = f(v, i)

因此，用于更新 v [i] 的值就是更新后的数组 v .有没有办法进行这些更新而没有循环?

Hence the value used to update v[i] is the updated array v. Is there a way to do these updates without a loop ?

例如

v = [f(v, i) for i in xrange(len(v))]

不起作用，因为在综合列表中使用 v [i-1] 时不会对其进行更新.

does not work since the v[i-1] is not updated when it is used in the comprehensive list.

I函数 f 可以取决于列表中的几个元素，索引应小于i的那些元素应被更新，而索引大于i的那些元素尚未被更新，如以下示例所示:

IThe function f can depend on several elements on the list, those with index lower than i should be updated and those with an index greater than i are not yet updated, as in the following example :

v = [1, 2, 3, 4, 5]
f = lambda v, i: (v[i-1] + v[i]) / v[i+1]  # for i = [1,3]
f = lambda v, i: v[i]                      # for i = {0,4}

它应该返回

v = [1, (1+2)/3, (1+4)/4, ((5/4)+4)/5, 5]

推荐答案

为此提供了一个功能:

import numpy

v = numpy.array([1, 2, 3, 4, 5])

numpy.add.accumulate(v)
#>>> array([ 1,  3,  6, 10, 15])

这适用于许多不同类型的ufunc:

This works on many different types of ufunc:

numpy.multiply.accumulate(v)
#>>> array([  1,   2,   6,  24, 120])

对于执行这种累加的任意函数，您可以创建自己的ufunc，尽管这会慢很多:

For an arbitrary function doing this kind of accumulation, you can make your own ufunc, although this will be much slower:

myfunc = numpy.frompyfunc(lambda x, y: x + y, 2, 1)
myfunc.accumulate([1, 2, 3], dtype=object)
#>>> array([1, 3, 6], dtype=object)

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