为什么这些浮点数不被评估为相等?

发布于 2024-10-28 18:35:54 字数 594 浏览 2 评论 0原文

在下面的代码中,如果 x0 == x1 且 y0 == y1,那么它应该告诉我这一点。在我使用的示例中,它们是。然而,Python 只表示 x 值相等,而不表示 y 值相等。

这是代码:

print "arc2x0 = ", arc2x0, ", arc2x1 = ", arc2x1
print "arc2y0 = ", arc2y0, ", arc2y1 = ", arc2y1

if arc2x0 == arc2x1:
    print "x0 == x1"
else:
    print "x0 != x1"

if arc2y0 == arc2y1:
    print "y0 == y1"
else:
    print "y0 != y1"

这是输出:

arc2x0 =  5 , arc2x1 =  5.0
arc2y0 =  -4.16026900507 , arc2y1 =  -4.16026900507
x0 == x1
y0 != y1

知道为什么 y 值测试相等吗?这些值的计算方式相同。也许还有比我所看到的更精确的不相等的精度?有什么方法可以打印该精度或任何其他关于如何调试它的想法?

In the following code, if x0 == x1 and y0 == y1, then it should tell me so. In the example I'm using, they are. However, Python is only saying the x values are equal, not the y.

Here's the code:

print "arc2x0 = ", arc2x0, ", arc2x1 = ", arc2x1
print "arc2y0 = ", arc2y0, ", arc2y1 = ", arc2y1

if arc2x0 == arc2x1:
    print "x0 == x1"
else:
    print "x0 != x1"

if arc2y0 == arc2y1:
    print "y0 == y1"
else:
    print "y0 != y1"

And here's the output:

arc2x0 =  5 , arc2x1 =  5.0
arc2y0 =  -4.16026900507 , arc2y1 =  -4.16026900507
x0 == x1
y0 != y1

Any idea why the y values are testing as equal? These values are calculated in the same way. Maybe there's more precision beyond what I'm seeing that isn't equal? Any way to print that precision or any other ideas on how to debug this?

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(2

乄_柒ぐ汐 2024-11-04 18:35:54

是的,比您所看到的更精确。问题是 == 运算符通常不应该与浮点数一起使用。浮点数的精度有限,会累积舍入误差,并且由于无法精确存储所有十进制值而导致不精确。例如,

>>> print '%.18f' % 0.1
0.100000000000000006

>>> 0.4 - 0.3 == 0.1
False

比较浮点数是否相等的最佳方法是比较它们是否几乎相等:

def nearly_equal(x, y, epsilon=1e-7):
    return abs(x - y) < epsilon

通过比较浮点数之间的差异是否小于 epsilon 值,该值指示多接近才被认为足够接近相等。

您始终可以使用decimal.Decimal类型来避免这些问题。

Yes there is more precision beyond what you're seeing. The problem is that the == operator should not usually be used with floats. Floating point numbers have limited precision and accumulate rounding error, and imprecision from not being able to store all decimal values precisely. e.g.

>>> print '%.18f' % 0.1
0.100000000000000006

>>> 0.4 - 0.3 == 0.1
False

The best way to compare floats for equality is to compare if they are nearly equal:

def nearly_equal(x, y, epsilon=1e-7):
    return abs(x - y) < epsilon

By comparing if the difference between floats is less than an epsilon value, which indicates how close is considered close enough to be equal.

You can always use the decimal.Decimal type to avoid these problems.

半世晨晓 2024-11-04 18:35:54

这是由于浮点的工作原理所致,您应该手动截断数字以将精度强制达到您想要的数字:

a=-4.16026900507

figures =9
a=round(a*(10**figures))/(10**figures) # truncates the digits
print a

但我认为最好的方法是使用十进制:

from decimal import *
a = Decimal ('your number as a string')

This is due to how floating point works,you should manually truncate the numbers to force the precision to the figures you want:

a=-4.16026900507

figures =9
a=round(a*(10**figures))/(10**figures) # truncates the digits
print a

But i think the best way would be to use decimal:

from decimal import *
a = Decimal ('your number as a string')
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文