浮点运算具有任意精度吗? [英] Is floating point arbitrary precision available?

问题描述

只是为了好玩，并且因为它真的很简单，所以我编写了一个简短的程序来生成嫁接数字，但是由于浮点精度问题，因此找不到一些较大的示例.

Just for fun and because it was really easy, I've written a short program to generate Grafting numbers, but because of floating point precision issues it's not finding some of the larger examples.

def isGrafting(a):
  for i in xrange(1, int(ceil(log10(a))) + 2):
    if a == floor((sqrt(a) * 10**(i-1)) % 10**int(ceil(log10(a)))):
      return 1

a = 0
while(1):
  if (isGrafting(a)):
    print "%d %.15f" % (a, sqrt(a))
  a += 1

此代码缺少至少一个已知的嫁接号码. 9999999998 => 99999.99998999999999949999999994999999999374999999912...乘以10**5后，似乎会降低精度.

This code misses at least one known Grafting number. 9999999998 => 99999.99998999999999949999999994999999999374999999912... It seems to drop extra precision after multiplying by 10**5.

>>> a = 9999999998
>>> sqrt(a)
99999.99999
>>> a == floor((sqrt(a) * 10**(5)) % 10**int(ceil(log10(a))))
False
>>> floor((sqrt(a) * 10**(5)) % 10**int(ceil(log10(a))))
9999999999.0
>>> print "%.15f" % sqrt(a)
99999.999989999996615
>>> print "%.15f" % (sqrt(a) * 10**5)
9999999999.000000000000000

所以我写了一个简短的C ++程序，看看它是不是我的CPU截断了浮点数或python.

So I wrote a short C++ program to see if it was my CPU truncating the floating point number or python somehow.

#include <cstdio>
#include <cmath>
#include <stdint.h>

int main()
{
  uint64_t a = 9999999998;
  printf("%ld %.15f %.15f %.15f %.15f\n", a, sqrt((double)a), sqrt((double)a)*1e4, sqrt((double)a)*1e5, sqrt((double)a)*1e6);
  a = 999999999998;
  printf("%ld %.15f %.15f %.15f %.15f\n", a, sqrt((double)a), sqrt((double)a)*1e5, sqrt((double)a)*1e6, sqrt((double)a)*1e7);
  a = 99999999999998;
  printf("%ld %.15f %.15f %.15f %.15f\n", a, sqrt((double)a), sqrt((double)a)*1e6, sqrt((double)a)*1e7, sqrt((double)a)*1e8);
  return 0;
}

哪个输出:

9999999998 99999.999989999996615 999999999.899999976158142 9999999999.000000000000000 99999999990.000000000000000
999999999998 999999.999998999992386 99999999999.899993896484375 999999999999.000000000000000 9999999999990.000000000000000
99999999999998 9999999.999999899417162 9999999999999.900390625000000 99999999999999.000000000000000 999999999999990.000000000000000

因此，我似乎在努力克服浮点精度的极限，并且CPU斩波了剩余的位，因为它认为剩余的差异是浮点错误.有没有办法在Python下解决此问题?还是我需要转到C并使用GMP之类的东西?

So it looks like I'm running up hard against the limits of floating point precision and the CPU is chopping off the remaining bits because it thinks that the remaining difference is floating point error. Is there a way to work around this under Python? Or do I need to move to C and use GMP or something?

推荐答案

在标准库中， decimal 模块可能正是您所需要的.另外，我发现 mpmath 很有帮助. 文档也有很多很好的例子(很遗憾，我的办公室计算机没有mpmath已安装；否则，我将验证一些示例并将其发布).

In the standard library, the decimal module may be what you're looking for. Also, I have found mpmath to be quite helpful. The documentation has many great examples as well (unfortunately my office computer does not have mpmath installed; otherwise I would verify a few examples and post them).

不过，有关 decimal 模块的一个警告.该模块包含几个用于简单数学运算的内置函数(例如sqrt)，但是这些函数的结果可能并不总是与math或其他模块中的相应函数匹配，但精度更高(尽管它们可能更准确). .例如

One caveat about the decimal module, though. The module contains several in-built functions for simple mathematical operations (e.g. sqrt), but the results from these functions may not always match the corresponding function in math or other modules at higher precisions (although they may be more accurate). For example,

from decimal import *
import math

getcontext().prec = 30
num = Decimal(1) / Decimal(7)

print("   math.sqrt: {0}".format(Decimal(math.sqrt(num))))
print("decimal.sqrt: {0}".format(num.sqrt()))

在Python 3.2.3中，这将输出前两行

In Python 3.2.3, this outputs the first two lines

   math.sqrt: 0.37796447300922719758631274089566431939601898193359375
decimal.sqrt: 0.377964473009227227214516536234
actual value: 0.3779644730092272272145165362341800608157513118689214

如前所述，

并不完全符合您的期望，并且您可以看到精度越高，结果匹配就越少.请注意，在此示例中，decimal模块确实具有更高的准确性，因为它与实际值.

which as stated, isn't exactly what you would expect, and you can see that the higher the precision, the less the results match. Note that the decimal module does have more accuracy in this example, since it more closely matches the actual value.

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