Int too large to convert to float питон ошибка

Answer by Johnny Woodward

But I got OverflowError: long int too large to convert to float,Thanks for contributing an answer to Stack Overflow!,

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,Note the L; the factorial of 170 is still convertable to a float:

Factorials get large real fast:

>>> math.factorial(170)
7257415615307998967396728211129263114716991681296451376543577798900561843401706157852350749242617459511490991237838520776666022565442753025328900773207510902400430280058295603966612599658257104398558294257568966313439612262571094946806711205568880457193340212661452800000000000000000000000000000000000000000L

Note the L; the factorial of 170 is still convertable to a float:

>>> float(math.factorial(170))
7.257415615307999e+306

but the next factorial is too large:

>>> float(math.factorial(171))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: long int too large to convert to float

You could use the decimal module; calculations will be slower, but the Decimal() class can handle factorials this size:

>>> from decimal import Decimal
>>> Decimal(math.factorial(171))
Decimal('1241018070217667823424840524103103992616605577501693185388951803611996075221691752992751978120487585576464959501670387052809889858690710767331242032218484364310473577889968548278290754541561964852153468318044293239598173696899657235903947616152278558180061176365108428800000000000000000000000000000000000000000')

You’ll have to use Decimal() values throughout:

from decimal import *

with localcontext() as ctx:
    ctx.prec = 32  # desired precision
    p = ctx.power(3, idx)
    depart = ctx.exp(-3) * p 
    depart /= math.factorial(idx)

Answer by Jaden O’Neill

OverflowError: int too large to convert to float,The problem is you’re computing enormous factorials and then trying to convert them to floats. One way around it is by not computing the power on top of the fraction and the factorial separately, so it doesn’t blow up:,This is the code I’m using to try to find the approximation of sin of an angle x. I want to increase the variable NumberOfTerms to increase the precision but when I do i get this error,We used getcontext().prec = # to try and control how many decimal places were returned. No matter what, we either got the same number (if we used something like 4) as not using Decimal, or we got 50.

import math

def factorial(n):
    if n == 1:
        return n
    else:
        return n*factorial(n-1)


def sin(x):
    OddNumber = 3
    counter = 1
    NumberOfTerms = 100
    while counter < NumberOfTerms:
        x -= (x**OddNumber)/(factorial(OddNumber))
        OddNumber += 2
        counter += 1
        x += (x**OddNumber)/(factorial(OddNumber))
        OddNumber += 2
        counter += 1
    return x


theta = float(input("input angle in degrees: "))
theta = float((theta*math.pi/180))
print(sin(theta))

Answer by Alden Briggs

OverflowError: Python int too large to convert to C long.,How to resolve OverflowError: Python int too large to convert to C long?,What is OverflowError: Python int too large to convert to C long?,OverflowError: Python int too large to convert to C long in PandasResolving the problem

import math
e = 250000
m = math.exp(e)
print(m)

Answer by Whitley Carroll

When I try to decrypt the ciphertext by using priv.decrypt(n1_r1_enc) and n1_r1_enc is the sum of n1 and r1, it raised the following error.,It is weird because it can decrypt correctly when n1 and r1 are the following values.,The error occurred when n1 and r1 are the following values:,how you encrypted the numbers

…roblem.

Answer by Alena Barr

It will raise «OverflowError: long int too large to convert to float».
But without converting to float,you would get 0.How to deal with large numbers like this in python?
Are there any better ways to calculate cumulative hypergeometric distribution in python?,Everything should be self-explanitory.,even that won’t work when N is very large … You need to use a log-hypercdf,nice, last time I checked scipy.stats would return overflow errors on large-N hypergeocdfs … they must’ve updated things since then.

Here are the codes:

from math import factorial

def Binomial(n,k):
    if n > k:
        return factorial(n) / (factorial(k)*factorial(n-k))
    elif n == k:
        return 1

def Hypergeometric(x,m,n,N):
    return 1.0*Binomial(m,x)*Binomial(N-m,n-x)/Binomial(N,n)

def HypergeometricCDF(x,m,n,N):
    cdf=0
    for i in range(x,min(m,n)+1):
        cdf+=Hypergeometric(i,m,n,N)
    return cdf

print HypergeometricCDF(5,40,50,500)
print HypergeometricCDF(50,400,500,5000)

Answer by Mackenzie Bennett

Recently, while I was working with the panda’s module recently and I discovered an OverflowError: Python int too large to convert to C long. I was running a python script, where I have to convert a string column from a pandas df to int, using the astype(int) method. However, I got the error. The code was as follows:,TypeError: coercing to Unicode: need string or buffer, NoneType found in Python,TypeError: exceptions must derive from BaseException in Python,TypeError: can’t use a string pattern on a bytes-like object in Python

 import pandas as pd
 
df = pd.DataFrame({'t': ['123456789985', '178965423698']})
df['int'] = df['test'].astype('int')
 
print(df['int'])

Answer by Vada Hahn

Im running an example from here: https://ntguardian.wordpress.com/2017/06/12/getting-started-with-backtrader/ and I ran into this error:,Looks like your connection to Backtrader Community was lost, please wait while we try to reconnect.,Seems like a simple error- self.price could easily be huge. Wondering if anyone else has encountered it, and if not what the next steps are (Im VERY new to github),

B

backtrader

administrators

last edited by

It’s a limitation of floating point storage capabilities (64 bits and you have to account for sign, mantissa and exponent). What is what you want to do?

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Im running an example from here: https://ntguardian.wordpress.com/2017/06/12/getting-started-with-backtrader/ and I ran into this error:

~/anaconda3/lib/python3.6/site-packages/backtrader/position.py in update(self, size, price, dt)
    191             if size < 0:  # increased position
    192                 opened, closed = size, 0
--> 193                 self.price = (self.price * oldsize + size * price) / self.size
    194 
    195             elif self.size < 0:  # reduced position

OverflowError: int too large to convert to float