规范化iterable类型，使输出是和为1的浮点的iterable

Question

好吧

我遇到了舍入误差问题。我有一个java程序，可以逐行读取一个列表，一些数字。程序要求这个列表是float（什么类型的float？我怀疑是单精度）数字，其和S在.999和1.001之间（即.999<；=S<；=1.001）。下面是一个例子：

from numpy import array, linalg

def Normalize(X):
    NumpyX = array(X)
    Norm = linalg.norm(NumpyX)
    Normalize = NumpyX/Norm
    return Normalize

def FileMaker(FilePointer,Vector)
    for i in Vector:
        FilePointer.write('%f\n'%i)
    return


#sum(SubstitutionPoints) does not add to 1
SubstitutionPoints =[0.00606508512067950,0.00675296642376962,0.00688999694872917,0.00580692396866418,0.00680583604896024,0.00609061670962565,0.00585881991631447,0.00577148570812953,0.00600882981888663,0.00618499536435559,0.00650767341787896,0.00670521809234427,0.00699374780209504,0.00841141135948587,0.00830145870238677,0.00879477131238090,0.00918627324146331,0.00958946761973615,0.01032404247887830,0.01093417870737930,0.01188202458790520,0.01261860720648550,0.01355451051017660,0.01473818756656830,0.01623978223562570,0.01811682034513980,0.01990010225231130,0.02154250858435480,0.02418185925226890,0.02583490296173980,0.02844212438633430,0.03194935989118780,0.03534228607419560,0.03971366519834600,0.04548573525944540,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660,0.05190994307855660]
NormedSP = Normalize(SubstitutionPoints)
SubFile = open('Sub.txt','w')
FileMaker(SubFile,NormedSP)
SubFile.close()

由于python不使用单精度浮点，我担心创建的文件会在java程序中引起问题，因为我的java程序（metasim）返回一个错误：

^{pr2}$

不幸的是，我无法调试java程序，因为它是“java可执行文件”。我唯一的选择就是给出正确的输入。在

如有任何建议/帮助，我们将不胜感激。在

编辑

根据尝试float32（）的一些建议，我修改了normalize方法如下：

def Normalize(self,X):
    Total = sum(X)
    NumpyX = array([float32(i) for i in X])
    Norm = linalg.norm(NumpyX,ord=1)
    Normalize = NumpyX/Norm
    return Normalize

我开始怀疑这是单精度浮点问题。在

我的输出是否被截断并导致错误？在

Answer 1

^{}需要ord=1关键字arg来执行您期望的操作。使用当前数据和定义。。。在

>>> Normalize(substitution_points).sum()
9.0451896403987444

您可以这样做，这样也可以正确缩放负值（linalg.norm将值的abs相加）：

^{pr2}$

0.9999999... > 0.999，因此根据您提供的规范，它应该是有效的。在

但是。。。看来还是不行。尝试使用numpy.float16并注意创建float数组的惯用方法如下：

array([ 0.,  1.,  2.,  3.,  4.], dtype=float32)

Answer 2

如果您只需要精确到小数点后三位，为什么不将值乘以1000，用整数进行所有运算，然后在输出中只添加小数点？这样就很容易保证总数是准确的。在

Answer 3

Normalize方法不会将列表的和强制为1：除以norm会将norm设置为1。若要将总计设置为1，需要除以当前总计：

def normalize(X):
   total = sum(X)
   return [x/total for x in X]