numpy:查找包含浮点值的两个文件之间的差异 [英] Numpy:Finding difference between two files containing float values

问题描述

我正在尝试使用Python计算两个文本文件之间的差异，并打印第一个值和它们开始发散的位置.

我不确定如何使用loadtxt:

import numpy as np
a = np.loadtxt("path/to/file", float)
b = np.loadtxt("path/to/file2", float)    
while np.absolute(a - b) !=0:


1
2
3
...

不确定如何完成此操作?开始正确吗?

解决方案

您可以使用

idx = np.where(np.abs(a-b) > 1e-6)[0]
firstidx = idx[0]

查找第一个索引，其中a和b中的值相差超过某些标称值，例如1e-6:

import numpy as np

a = np.loadtxt("path/to/file", float)
b = np.loadtxt("path/to/file2", float)

idx = np.where(np.abs(a-b) > 1e-6)[0]
firstidx = idx[0]
print(firstidx, a[firstidx], b[firstidx])

---

请注意，在处理浮点数时，很少会想与相等进行比较，例如

np.abs(a-b) == 0

或相反，

np.abs(a-b) != 0

因为浮点表示的不准确性可能会导致a和b略有不同，即使它们的值应精确地表示为 时也是如此.

所以使用类似的东西

np.abs(a-b) > 1e-6

相反. (请注意，您必须选择容忍度，例如1e-6).

---

这是一个简单示例使用相等性比较浮点数的陷阱:

In [10]: 1.2-1.0 == 0.2
Out[10]: False

I am trying to use Python to compute the difference between two text files and print the first value and location where they start to diverge.

I am not sure how to use loadtxt:

import numpy as np
a = np.loadtxt("path/to/file", float)
b = np.loadtxt("path/to/file2", float)    
while np.absolute(a - b) !=0:


1
2
3
...

Not sure how to finish this? Is the start correct?

解决方案

You could use

idx = np.where(np.abs(a-b) > 1e-6)[0]
firstidx = idx[0]

to find the first index where the values in a and b differ by more than some nominal amount like 1e-6:

import numpy as np

a = np.loadtxt("path/to/file", float)
b = np.loadtxt("path/to/file2", float)

idx = np.where(np.abs(a-b) > 1e-6)[0]
firstidx = idx[0]
print(firstidx, a[firstidx], b[firstidx])

---

Note that when dealing with floats, you rarely if ever want to compare with equality, such as with

np.abs(a-b) == 0

or the converse,

np.abs(a-b) != 0

because the inaccuracy of floating point representations can cause a and b to be slightly different even when their values should be exactly the same if their values were represented with infinite precision.

So use something like

np.abs(a-b) > 1e-6

instead. (Note that you have to choose a level of tolerance, e.g. 1e-6).

---

Here is a simple example demonstrating the pitfall of comparing floats using equality:

In [10]: 1.2-1.0 == 0.2
Out[10]: False

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