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Question

好，

我遇到了舍入错误问题。我有一个java程序，读取列表，逐行，一些数字。该程序要求该列表是浮点数（什么类型的浮点数？我怀疑单精度）数字，其和S在.999和1.001之间（即.999 <= S <= 1.001）。这是一个例子：

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不使用单精度浮点数，我担心创建的文件会导致java程序出现问题，因为我的java程序（metasim）返回错误：

Simulating files:
[BorreliaBurgdorferiB31_CP32-3.fasta]
java.lang.Exception: Substitution rates don't sum to 1.0.
!!! FAILED !!!

不幸的是，我无法调试java程序，因为它是'java executable'。我唯一的选择是给出正确的输入。

感谢任何建议/帮助。

修改

从尝试float32（）的一些建议我修改了我的normalize方法：

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

我开始怀疑它是单精度浮点问题。

我的输出是否被截断并导致错误？

Answer 1

如果你只需要精确到三位小数，为什么不将这些值乘以1000，用整数进行所有数学计算，只在输出中添加小数点？这样就很容易确保总和是准确的。

Answer 2

numpy.linalg.norm要求ord=1关键字arg执行您期望的操作。使用您当前的数据＆amp;定义...

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

您可能会这样做 - 这也会正确地缩放负值（linalg.norm对值的abs求和）：

>>> def normalize(x):
...     a = numpy.array(x)
...     return a / a.sum()
... 
>>> normalize(substitution_points).sum()
0.99999999999999789

0.9999999... > 0.999，根据您提供的规范，应工作。

但......似乎它仍然没有用。尝试使用numpy.float16 - 并注意创建浮点数组的更惯用的方法是这样的：

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

Answer 3

您的Normalize方法不强制列表的总和为1：除以norm将 norm 设置为1.要将总数设置为1，您需要除以当前值总：

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

规范化可迭代类型，使得输出是可迭代的浮点数，其总和为1

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3 个答案: