Python 溢出错误：（34，'结果太大'）
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OverflowError: (34, 'Result too large')
提问by user3033766
I'am getting an overflow error(OverflowError: (34, 'Result too large')
I want to calculate pi to 100 decimals here's my code:
我收到溢出错误（OverflowError: (34, 'Result too large'）
我想将 pi 计算为 100 位小数，这是我的代码：
def pi():
pi = 0
for k in range(350):
pi += (4./(8.*k+1.) - 2./(8.*k+4.) - 1./(8.*k+5.) - 1./(8.*k+6.)) / 16.**k
return pi
print(pi())
采纳答案by Peter DeGlopper
Python floats are neither arbitary precision nor of unlimited size. When k = 349, 16.**kis much too large - that's almost 2^1400. Fortunately, the decimallibrary allows arbitrary precision and can handle the size:
Python 浮点数既不是任意精度也不是无限大小。当 k = 349 时，16.**k太大了 - 几乎是 2^1400。幸运的是，该decimal库允许任意精度并且可以处理大小：
import decimal
decimal.getcontext().prec = 100
def pi():
pi = decimal.Decimal(0)
for k in range(350):
pi += (decimal.Decimal(4)/(decimal.Decimal(8)*decimal.Decimal(k+1))...)
回答by Martijn Pieters
You reached the limits of your platform's floatsupport, probably after k = 256:
您达到了平台float支持的极限，可能是在k = 256以下情况之后：
>>> k = 256
>>> (4./(8.*k+1.) - 2./(8.*k+4.) - 1./(8.*k+5.) - 1./(8.*k+6.)) / 16.**k
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: (34, 'Result too large')
>>> k = 255
>>> (4./(8.*k+1.) - 2./(8.*k+4.) - 1./(8.*k+5.) - 1./(8.*k+6.)) / 16.**k
3.19870064997e-313
See sys.float_infofor the exact limitations, but you are unlikely to run into a current CPU and OS combination that'll give you 100 significant digits in any case; my MacBook Pro with 64-bit OS X will only support 15.
查看sys.float_info确切的限制，但您不太可能遇到在任何情况下都会为您提供 100 位有效数字的当前 CPU 和操作系统组合；我的 64 位 OS X MacBook Pro 只支持 15。
Use the decimalmoduleto go beyond your hardware limitations.
使用该decimal模块超越您的硬件限制。
from decimal import Decimal, localcontext
def pi():
with localcontext() as ctx:
ctx.prec = 100 # 100 digits precision
pi = Decimal(0)
for k in range(350):
pi += (Decimal(4)/(Decimal(8)*k+1) - Decimal(2)/(Decimal(8)*k+4) - Decimal(1)/(Decimal(8)*k+5) - Decimal(1)/(Decimal(8)*k+6)) / Decimal(16)**k
return pi
回答by Bence
16.**256 is too large to be stored in double precision float. I suggest that you run your cycle for less, like range(250), because larger k values will not contribute to the first hundred digits anyway.
16.**256 太大，无法存储在双精度浮点数中。我建议你少运行你的循环，比如 range(250)，因为更大的 k 值无论如何都不会影响前一百位数字。
Another thing you might try is to multiply by 16.*(-k) instead of dividing by 16.*k. This number will be rounded to zero for large k, therefore will not give you runtime errors.
您可能会尝试的另一件事是乘以 16.* (-k) 而不是除以 16.*k。对于大 k，此数字将四舍五入为零，因此不会给您带来运行时错误。
I suggest that you use numpy.power instead of **, it handles overflows better. For example, in your code numpy.power(16.,256) would evaluate to inf, and dividing a finite number by inf gives zero, which avoids runtime errors just like the method suggested in the previous paragraph.
我建议你使用 numpy.power 而不是 **，它可以更好地处理溢出。例如，在您的代码中 numpy.power(16.,256) 将评估为 inf，并且将有限数除以 inf 得到零，这避免了运行时错误，就像上一段中建议的方法一样。
回答by u7420336
I use python3.6 AMD64,I also meet this problem,this is because python built-in floatis double-precision-float,it's 64 bit,in most progamming task,64 bit is enough,but in some extra task,it's not enough(like scitific computing,big data compute)
我使用python3.6 AMD64，我也遇到这个问题，这是因为python内置float是双精度浮点数，它是64位，在大多数编程任务中，64位就足够了，但在一些额外的任务中，这还不够（如科学计算、大数据计算）
