为什么在使用python和numpy时sin(180)不为零? [英] Why is sin(180) not zero when using python and numpy?

问题描述

有人知道为什么下限值不等于0吗?

Does anyone know why the below doesn't equal 0?

import numpy as np
np.sin(np.radians(180))

或:

np.sin(np.pi)

当我将其输入python时，它的值为1.22e-16.

When I enter it into python it gives me 1.22e-16.

推荐答案

数字π不能完全表示为浮点数.因此，np.radians(180)不会给您π，它会给您3.1415926535897931.

The number π cannot be represented exactly as a floating-point number. So, np.radians(180) doesn't give you π, it gives you 3.1415926535897931.

sin(3.1415926535897931)实际上与1.22e-16类似.

那么，您如何处理呢?

您必须计算出或至少猜测出适当的绝对和/或相对误差范围，然后您编写:代替x == y:

You have to work out, or at least guess at, appropriate absolute and/or relative error bounds, and then instead of x == y, you write:

abs(y - x) < abs_bounds and abs(y-x) < rel_bounds * y

(这也意味着您必须组织计算，以使相对于y的相对误差大于相对于x的误差.在您的情况下，因为y是常数0，所以这很简单-只是向后做.)

(This also means that you have to organize your computation so that the relative error is larger relative to y than to x. In your case, because y is the constant 0, that's trivial—just do it backward.)

Numpy提供了一个在整个数组中为您执行此操作的功能， allclose :

Numpy provides a function that does this for you across a whole array, allclose:

np.allclose(x, y, rel_bounds, abs_bounds)

(这实际上检查了abs(y - x) < abs_ bounds + rel_bounds * y)，但这几乎总是足够的，并且在不满足要求的情况下，您可以轻松地重新组织代码.)

(This actually checks abs(y - x) < abs_ bounds + rel_bounds * y), but that's almost always sufficient, and you can easily reorganize your code when it's not.)

在您的情况下:

np.allclose(0, np.sin(np.radians(180)), rel_bounds, abs_bounds)

那么，您如何知道正确的界限呢?在SO答案中，没有任何方法可以教给您足够的错误分析. Wikipedia上的不确定性的传播提供了一个高层次的概述.如果您真的不知道，可以使用默认值，相对值1e-5和绝对值1e-8

So, how do you know what the right bounds are? There's no way to teach you enough error analysis in an SO answer. Propagation of uncertainty at Wikipedia gives a high-level overview. If you really have no clue, you can use the defaults, which are 1e-5 relative and 1e-8 absolute.

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