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How to make calculations extremely large numbers in python

I am trying to use the pow method on 256 bits number as the base and 1 bit to 4 thousand bits number as the exponent and then use modulo "n" on that number (n is a four thousand bits number).
meaning that the bigest number for the base is: 115792089237316195423570985008687907853269984665640564039457584007913129639935.and the bigest number for the exponent is: 13182040934309431001038897942365913631840191610932727690928034502417569281128344551079752123172122033140940756480716823038446817694240581281731062452512184038544674444386888956328970642771993930036586552924249514488832183389415832375620009284922608946111038578754077913265440918583125586050431647284603636490823850007826811672468900210689104488089485347192152708820119765006125944858397761874669301278745233504796586994514054435217053803732703240283400815926169348364799472716094576894007243168662568886603065832486830606125017643356469732407252874567217733694824236675323341755681839221954693820456072020253884371226826844858636194212875139566587445390068014747975813971748114770439248826688667129237954128555841874460665729630492658600179338272579110020881228767361200603478973120168893997574353727653998969223092798255701666067972698906236921628764772837915526086464389161570534616956703744840502975279094087587298968423516531626090898389351449020056851221079048966718878943309232071978575639877208621237040940126912767610658141079378758043403611425454744180577150855204937163460902512732551260539639221457005977247266676344018155647509515396711351487546062479444592779055555421362722504575706910949375.
But I am getting an error that says that those two numbers are to larg to convert to float.
OverflowError: int too large to convert to float
Can someone help me perform this calculation with those values?
this is the tricky part of my code:
pre_key = pow(prime_g, private_number)
pre_key = pre_key % n

As #Tim Roberts pointed out, I also don't think you should brute-force the program. That being said, you might be able to use the mpmath library, if you really NEED TO brute force your problem. I don't know, how well it handles numbers of that extreme size, but in principle you can allow for arbitrary precision.

What is a float, have I got the right idea?

I was programming a calculator and I came across float. I was told to use floats then my calculator would be able to calculate calculations including numbers like 16.4 and 4.5 and not just whole numbers.
Now, I sort of got thinking and wondered if someone could just verify I'm on the right tracks.
I understand an int is a solid number, a double is basically 2 numbers with a dot in the middle, a decimal? And now comes the tricky one, the float data type.
I just need somebody to verify I'm on the right track. I think a float is a data type that can either be a whole number (this is what I'm not sure about) and a double/decimal.
I feel like float is the safe data type kinda thing, where you're not sure and want to accept whole and decimal numbers, am I right? Is it for accepting both whole and decimals?
For anyone that struggles to understand, heres the code to my calculator, it might help.
while True:
calculation = input("Calculation: ")
if (calculation.__contains__("+")):
print(float(calculation[0] ) + float(calculation[2]))
elif (calculation.__contains__("-")):
print(float(calculation[0] ) - float(calculation[2]))
elif (calculation.__contains__("*")):
print(float(calculation[0] ) * float(calculation[2]))
elif (calculation.__contains__("/")):
print(float(calculation[0] ) / float(calculation[2]))

A float is a floating-point number (more or less your "basically 2 numbers with a dot in the middle").
The term double is short for "double precision floating-point number": a similar kind of number but typically using more bits to store it, allowing for more precision.
In Python, the type float is used to refer to all floating-point numbers, regardless of precision.

You mention Python's float and also double. These are exactly the same thing, because what Python calls float (pedantically, in most implementations of Python) is what everyone else calls double. And what C and C++ call float does not exist in Python.

Numpy to weak to calculate a precise mean value

This question is very similar to this post - but not exactly
I have some data in a .csv file. The data has precision to the 4th digit (#.####).
Calculating the mean in Excel or SAS gives a result with precision to 5th digit (#.#####) but using numpy gives:
import numpy as np
data = np.recfromcsv(path2file, delimiter=';', names=['measurements'], dtype=np.float64)
rawD = data['measurements']
print np.average(rawD)
gives a number like this
#.#####999999999994
Clearly something is wrong..
using
from math import fsum
print fsum(rawD.ravel())/rawD.size
gives
#.#####
Is there anything in the np.average that I set wrong _______?
BONUS info:
I'm only working with 200 data points in the array
UPDATE
I thought I should make my case more clear.
I have numbers like 4.2730 in my csv (giving a 4 decimal precision - even though the 4th always is zero [not part of the subject so don't mind that])
Calculating an average/mean by numpy gives me this
4.2516499999999994
Which gives a print by
>>>print "%.4f" % np.average(rawD)
4.2516
During the same thing in Excel or SAS gives me this:
4.2517
Which I actually believe as being the true average value because it finds it to be 4.25165.
This code also illustrate it:
answer = 0
for number in rawD:
answer += int(number*1000)
print answer/2
425165
So how do I tell np.average() to calculate this value ___?
I'm a bit surprised that numpy did this to me... I thought that I only needed to worry if I was dealing with 16 digits numbers. Didn't expect a round off on the 4 decimal place would be influenced by this..
I know I could use
fsum(rawD.ravel())/rawD.size
But I also have other things (like std) I want to calculate with the same precision
UPDATE 2
I thought I could make a temp solution by
>>>print "%.4f" % np.float64("%.5f" % np.mean(rawD))
4.2416
Which did not solve the case. Then I tried
>>>print "%.4f" % float("4.24165")
4.2416
AHA! There is a bug in the formatter: Issue 5118
To be honest I don't care if python stores 4.24165 as 4.241649999... It's still a round off error - NO MATTER WHAT.
If the interpeter can figure out how to display the number
>>>print float("4.24165")
4.24165
Then should the formatter as well and deal with that number when rounding..
It still doesn't change the fact that I have a round off problem (now both with the formatter and numpy)
In case you need some numbers to help me out then I have made this modified .csv file:
Download it from here
(I'm aware that this file does not have the number of digits I explained earlier and that the average gives ..9988 at the end instead of ..9994 - it's modified)
Guess my qeustion boils down to how do I get a string output like the one excel gives me if I use =average()
and have it round off correctly if I choose to show only 4 digits
I know that this might seem strange for some.. But I have my reasons for wanting to reproduce the behavior of Excel.
Any help would be appreciated, thank you.

To get exact decimal numbers, you need to use decimal arithmetic instead of binary. Python provides the decimal module for this.
If you want to continue to use numpy for the calculations and simply round the result, you can still do this with decimal. You do it in two steps, rounding to a large number of digits to eliminate the accumulated error, then rounding to the desired precision. The quantize method is used for rounding.
from decimal import Decimal,ROUND_HALF_UP
ten_places = Decimal('0.0000000001')
four_places = Decimal('0.0001')
mean = 4.2516499999999994
print Decimal(mean).quantize(ten_places).quantize(four_places, rounding=ROUND_HALF_UP)
4.2517

The result value of average is a double. When you print out a double, by default all digits are printed. What you see here is the result of limited digital precision, which is not a problem of numpy, but a general computing problem. When you care of the presentation of your float value, use "%.4f" % avg_val. There is also a package for rational numbers, to avoid representing fractions as real numbers, but I guess that's not what you're looking for.
For your second statement, summarizing all the values by hand and then dividing it, I suppose you're using python 2.7 and all your input values are integer. In that way, you would have an integer division, which truncates everything after the dot, resulting in another integer value.

Why do Python's math.ceil() and math.floor() operations return floats instead of integers?

Can someone explain this (straight from the docs- emphasis mine):
math.ceil(x) Return the ceiling of x as a float, the smallest integer value greater than or equal to x.
math.floor(x) Return the floor of x as a float, the largest integer value less than or equal to x.
Why would .ceil and .floor return floats when they are by definition supposed to calculate integers?
EDIT:
Well this got some very good arguments as to why they should return floats, and I was just getting used to the idea, when #jcollado pointed out that they in fact do return ints in Python 3...

As pointed out by other answers, in python they return floats probably because of historical reasons to prevent overflow problems. However, they return integers in python 3.
>>> import math
>>> type(math.floor(3.1))
<class 'int'>
>>> type(math.ceil(3.1))
<class 'int'>
You can find more information in PEP 3141.

The range of floating point numbers usually exceeds the range of integers. By returning a floating point value, the functions can return a sensible value for input values that lie outside the representable range of integers.
Consider: If floor() returned an integer, what should floor(1.0e30) return?
Now, while Python's integers are now arbitrary precision, it wasn't always this way. The standard library functions are thin wrappers around the equivalent C library functions.

Because python's math library is a thin wrapper around the C math library which returns floats.

The source of your confusion is evident in your comment:
The whole point of ceil/floor operations is to convert floats to integers!
The point of the ceil and floor operations is to round floating-point data to integral values. Not to do a type conversion. Users who need to get integer values can do an explicit conversion following the operation.
Note that it would not be possible to implement a round to integral value as trivially if all you had available were a ceil or float operation that returned an integer. You would need to first check that the input is within the representable integer range, then call the function; you would need to handle NaN and infinities in a separate code path.
Additionally, you must have versions of ceil and floor which return floating-point numbers if you want to conform to IEEE 754.

Before Python 2.4, an integer couldn't hold the full range of truncated real numbers.
http://docs.python.org/whatsnew/2.4.html#pep-237-unifying-long-integers-and-integers

Because the range for floats is greater than that of integers -- returning an integer could overflow

This is a very interesting question! As a float requires some bits to store the exponent (=bits_for_exponent) any floating point number greater than 2**(float_size - bits_for_exponent) will always be an integral value! At the other extreme a float with a negative exponent will give one of 1, 0 or -1. This makes the discussion of integer range versus float range moot because these functions will simply return the original number whenever the number is outside the range of the integer type. The python functions are wrappers of the C function and so this is really a deficiency of the C functions where they should have returned an integer and forced the programer to do the range/NaN/Inf check before calling ceil/floor.
Thus the logical answer is the only time these functions are useful they would return a value within integer range and so the fact they return a float is a mistake and you are very smart for realizing this!

Maybe because other languages do this as well, so it is generally-accepted behavior. (For good reasons, as shown in the other answers)

This totally caught me off guard recently. This is because I've programmed in C since the 1970's and I'm only now learning the fine details of Python. Like this curious behavior of math.floor().
The math library of Python is how you access the C standard math library. And the C standard math library is a collection of floating point numerical functions, like sin(), and cos(), sqrt(). The floor() function in the context of numerical calculations has ALWAYS returned a float. For 50 YEARS now. It's part of the standards for numerical computation. For those of us familiar with the math library of C, we don't understand it to be just "math functions". We understand it to be a collection of floating-point algorithms. It would be better named something like NFPAL - Numerical Floating Point Algorithms Libary. :)
Those of us that understand the history instantly see the python math module as just a wrapper for the long-established C floating-point library. So we expect without a second thought, that math.floor() is the same function as the C standard library floor() which takes a float argument and returns a float value.
The use of floor() as a numerical math concept goes back to 1798 per the Wikipedia page on the subject: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Notation
It never has been a computer science covert floating-point to integer storage format function even though logically it's a similar concept.
The floor() function in this context has always been a floating-point numerical calculation as all(most) the functions in the math library. Floating-point goes beyond what integers can do. They include the special values of +inf, -inf, and Nan (not a number) which are all well defined as to how they propagate through floating-point numerical calculations. Floor() has always CORRECTLY preserved values like Nan and +inf and -inf in numerical calculations. If Floor returns an int, it totally breaks the entire concept of what the numerical floor() function was meant to do. math.floor(float("nan")) must return "nan" if it is to be a true floating-point numerical floor() function.
When I recently saw a Python education video telling us to use:
i = math.floor(12.34/3)
to get an integer I laughed to myself at how clueless the instructor was. But before writing a snarkish comment, I did some testing and to my shock, I found the numerical algorithms library in Python was returning an int. And even stranger, what I thought was the obvious answer to getting an int from a divide, was to use:
i = 12.34 // 3
Why not use the built-in integer divide to get the integer you are looking for! From my C background, it was the obvious right answer. But low and behold, integer divide in Python returns a FLOAT in this case! Wow! What a strange upside-down world Python can be.
A better answer in Python is that if you really NEED an int type, you should just be explicit and ask for int in python:
i = int(12.34/3)
Keeping in mind however that floor() rounds towards negative infinity and int() rounds towards zero so they give different answers for negative numbers. So if negative values are possible, you must use the function that gives the results you need for your application.
Python however is a different beast for good reasons. It's trying to address a different problem set than C. The static typing of Python is great for fast prototyping and development, but it can create some very complex and hard to find bugs when code that was tested with one type of objects, like floats, fails in subtle and hard to find ways when passed an int argument. And because of this, a lot of interesting choices were made for Python that put the need to minimize surprise errors above other historic norms.
Changing the divide to always return a float (or some form of non int) was a move in the right direction for this. And in this same light, it's logical to make // be a floor(a/b) function, and not an "int divide".
Making float divide by zero a fatal error instead of returning float("inf") is likewise wise because, in MOST python code, a divide by zero is not a numerical calculation but a programming bug where the math is wrong or there is an off by one error. It's more important for average Python code to catch that bug when it happens, instead of propagating a hidden error in the form of an "inf" which causes a blow-up miles away from the actual bug.
And as long as the rest of the language is doing a good job of casting ints to floats when needed, such as in divide, or math.sqrt(), it's logical to have math.floor() return an int, because if it is needed as a float later, it will be converted correctly back to a float. And if the programmer needed an int, well then the function gave them what they needed. math.floor(a/b) and a//b should act the same way, but the fact that they don't I guess is just a matter of history not yet adjusted for consistency. And maybe too hard to "fix" due to backward compatibility issues. And maybe not that important???
In Python, if you want to write hard-core numerical algorithms, the correct answer is to use NumPy and SciPy, not the built-in Python math module.
import numpy as np
nan = np.float64(0.0) / 0.0 # gives a warning and returns float64 nan
nan = np.floor(nan) # returns float64 nan
Python is different, for good reasons, and it takes a bit of time to understand it. And we can see in this case, the OP, who didn't understand the history of the numerical floor() function, needed and expected it to return an int from their thinking about mathematical integers and reals. Now Python is doing what our mathematical (vs computer science) training implies. Which makes it more likely to do what a beginner expects it to do while still covering all the more complex needs of advanced numerical algorithms with NumPy and SciPy. I'm constantly impressed with how Python has evolved, even if at times I'm totally caught off guard.

floats inside tuples changing values when accessed

So I have a list of tuples of two floats each. Each tuple represents a range. I am going through another list of floats which represent values to be fit into the ranges. All of these floats are < 1 but positive, so precision matter. One of my tests to determine if a value fits into a range is failing when it should pass. If I print the value and the range that is causing problems I can tell this much:
curValue = 0.00145000000671
range = (0.0014500000067055225, 0.0020968749796738849)
The conditional that is failing is:
if curValue > range[0] and ... blah :
# do some stuff
From the values given by curValue and range, the test should clearly pass (don't worry about what is in the conditional). Now, if I print explicitly what the value of range[0] is I get:
range[0] = 0.00145000000671
Which would explain why the test is failing. So my question then, is why is the float changing when it is accessed. It has decimal values available up to a certain precision when part of a tuple, and a different precision when accessed. Why would this be? What can I do to ensure my data maintains a consistent amount of precision across my calculations?

The float doesn't change. The built-in numberic types are all immutable. The cause for what you're observing is that:
print range[0] uses str on the float, which (up until very recent versions of Python) printed less digits of a float.
Printing a tuple (be it with repr or str) uses repr on the individual items, which gives a much more accurate representation (again, this isn't true anymore in recent releases which use a better algorithm for both).
As for why the condition doesn't work out the way you expect, it's propably the usual culprit, the limited precision of floats. Try print repr(curVal), repr(range[0]) to see if what Python decided was the closest representation of your float literal possible.

In modern day PC's floats aren't that precise. So even if you enter pi as a constant to 100 decimals, it's only getting a few of them accurate. The same is happening to you. This is because in 32-bit floats you only get 24 bits of mantissa, which limits your precision (and in unexpected ways because it's in base2).
Please note, 0.00145000000671 isn't the exact value as stored by Python. Python only diplays a few decimals of the complete stored float if you use print. If you want to see exactly how python stores the float use repr.
If you want better precision use the decimal module.

It isn't changing per se. Python is doing its best to store the data as a float, but that number is too precise for float, so Python modifies it before it is even accessed (in the very process of storing it). Funny how something so small is such a big pain.
You need to use a arbitrary fixed point module like Simple Python Fixed Point or the decimal module.

Not sure it would work in this case, because I don't know if Python's limiting in the output or in the storage itself, but you could try doing:
if curValue - range[0] > 0 and...