的Python：在数组元素条件 [英] Python: Conditional elements in array

问题描述

这是一个完整的Python新手问题。

A question from a complete Python novice.

我有一个列数组，我需要根据应用到另一个阵列条件语句来迫使某些值为零。我已经发现两种解决方案，这两者都提供了正确的答案。但它们消耗了较大的阵列我通常需要（> 1E6元素）都相当的时间 - 我也怀疑这是编程技术较差。这两个版本是：

I have a column array where I need to force certain values to zero depending on a conditional statement applied to another array. I have found two solutions, which both provide the correct answer. But they are both quite time consuming for the larger arrays I typically need (>1E6 elements) - also I suspect that it is poor programming technique. The two versions are:

from numpy import zeros,abs,multiply,array,reshape

def testA(y, f, FC1, FC2):
    c = zeros((len(f),1))
    for n in xrange(len(f)):
        if abs(f[n,0]) >= FC1 and abs(f[n,0]) <= FC2:
            c[n,0] = 1.
    w = multiply(c,y)
    return w

def testB(y, f, FC1, FC2):
    z = [(abs(f[n,0])>=FC1 and abs(f[n,0])<=FC2) for n in xrange(len(f))]
    z = multiply(array(z,dtype=float).reshape(len(f),1), y)
    return z

输入数组的数组列，因为这匹配的后处理工作要做。测试可以像来完成：

The input arrays are column arrays as this matches the post processing to be done. The test can be done like:

>>> from numpy.random import normal as randn
>>> fs, N = 1.E3, 2**22
>>> f = fs/N*arange(N).reshape((N,1))
>>> x = randn(size=(N,1))
>>> w1 = testA(x,f,200.,550.)
>>> z1 = testB(x,f,200.,550.)

在我的笔记本电脑的种皮需要18.7秒和 TESTB 需要19.3 - 既为N = 2 ** 22。在 TESTB 我也试过，包括Z = [无] * LEN（六），以preallocate在另一个线程建议，但并不真正使任何区别。

On my laptop testA takes 18.7 seconds and testB takes 19.3 - both for N=2**22. In testB I also tried to include "z = [None]*len(f)" to preallocate as suggested in another thread but this doesn't really make any difference.

我有两个问题，我希望能有相同的答案：

I have two questions, which I hope to have the same answer:

1. 什么是正确的Python解决这个问题？


2. 有什么我能做得到的答案快？

我特意不使用任何时候都使用Python的编译 - 例如我想先有一定的工作code。也希望一些东西，这是很好的Python的风格。我希望能够得到执行时间为N = 2 ** 22两秒钟所以下面。所以执行时间事执行此特定的操作将被使用很多次。

I have deliberately not used any time at all using compiled Python for example - I wanted to have some working code first. Hopefully also something, which is good Python style. I hope to be able to get the execution time for N=2**22 below two seconds or so. This particular operation will be used many times so the execution time does matter.

我提前道歉，如果问题是愚蠢的 - 我一直没能找到一个答案并不总是方便Python文档的压倒性数量或在另一个线程

I apologize in advance if the question is stupid - I haven't been able to find an answer in the overwhelming amount of not always easily accessible Python documentation or in another thread.

推荐答案

使用布尔数组访问量阵列Y元素：

use bool array to access elements in array y:

def testC(y, f, FC1, FC2):
    f2 = abs(f)
    idx = (f2>=FC1) & (f2<=FC2)
    y[~idx] = 0
    return y

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