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formating e notation and division in python

I have
"%f"%(9584629447823472134871239847192/2)
when I run this in idle the output I get is
4792314723911735847389621125120.000000
As we all know that the output is not correct. I am new in this. Kindly help how to get the correct output.
The output should be this
4792314723911736067435619923596

You are using floating point arithmetic, which isn’t precise.
Use integer arithmetic if appropriate in your situation and you want precise results. Integer division in Python is done using the // operator.
For example,
"%d"%(9584629447823472134871239847192//2)

Mitigating Floating Point Approximation Issues with Numpy

My code is quite simple, and only 1 line is causing an issue:
np.tan(np.radians(rotation))
Instead of my expected output for rotation = 45 as 1, I get 0.9999999999999999. I understand that 0 and a ton of 9's is 1. In my use case, however, it seems like the type of thing that will definitely build up over iterations.
What is causing the floating point error: np.tan or np.radians, and how do I get the problem function to come out correctly regardless of floating point inaccuracies?
Edit:
I should clarify that I am familiar with floating point inaccuracies. My concern is that as that number gets multiplied, added, and compared, the 1e-6 error suddenly becomes a tangible issue. I've normally been able to safely ignore floating point issues, but now I am far more concerned about the build up of error. I would like to reduce the possibility of such an error.
Edit 2:
My current solution is to just round to 8 decimal places because that's most likely enough. It's sort of a temporary solution because I'd much prefer a way to get around the IEEE decimal representations.

What is causing the floating point error: np.tan or np.radians, and how do I get the problem function to come out correctly regardless of floating point inaccuracies?
Both functions incur rounding error, since in neither case is the exact result representable in floating point.
My current solution is to just round to 8 decimal places because that's most likely enough. It's sort of a temporary solution because I'd much prefer a way to get around the IEEE decimal representations.
The problem has nothing to do with decimal representation, and this will give worse results outside of the exact case you mention above, e.g.
>>> np.tan(np.radians(60))
1.7320508075688767
>>> round(np.tan(np.radians(60)), 8)
1.73205081
>>> np.sqrt(3) # sqrt is correctly rounded, so this is the closest float to the true result
1.7320508075688772
If you absolutely need higher accuracy than the 15 decimal digits you would get from code above, then you can use an arbitrary precision library like gmpy2.

Take a look here: https://docs.scipy.org/doc/numpy/user/basics.types.html .
Standard dtypes in numpy do not go beyond 64 bits precision. From the docs:
Be warned that even if np.longdouble offers more precision than python
float, it is easy to lose that extra precision, since python often
forces values to pass through float. For example, the % formatting
operator requires its arguments to be converted to standard python
types, and it is therefore impossible to preserve extended precision
even if many decimal places are requested. It can be useful to test
your code with the value 1 + np.finfo(np.longdouble).eps.
You can increase precision with np.longdouble, but this is platform dependent
In spyder (windows):
np.finfo(np.longdouble).eps #same precision as float
>> 2.220446049250313e-16
np.finfo(np.longdouble).precision
>> 15
In google colab:
np.finfo(np.longdouble).eps #larger precision
>> 1.084202172485504434e-19
np.finfo(np.longdouble).precision
>> 18
print(np.tan(np.radians(45, dtype=np.float), dtype=np.float) - 1)
print(np.tan(np.radians(45, dtype=np.longfloat), dtype=np.longfloat) - 1)
>> -1.1102230246251565e-16
0.0

Python : arbitrary precision with floats

I tried to compute math.exp(9500) but encountered an OverflowError: math range error (it's roughly 6.3e4125). From this question it seems like it's due to a too large float, the accepted answer says "(...) is slightly outside of the range of a double, so it causes an overflow".
I know that Python can deal with arbitrarily large integers (long type), is there a way to deal with arbitrarily large floats in the same manner ?
Edit : my original question was about using integers for calculating exp(n) but as Eric Duminil said, the simplest way to do that would be 3**n which doesn't provide any useful result. I know realize this question might be similar to this one.

I don't think it's possible to approximate exp() with integers. If you use 3**n instead of 2.71828182845905**n, your calculations will be completely useless.
One possible solution would be to use Sympy. According to the documentation:
There is essentially no upper precision limit
>>> from sympy import *
>>> exp(9500)
exp(9500)
>>> exp(9500).evalf()
6.27448493490172e+4125
You can also specify the desired precision:
>>> exp(9500).evalf(1000)
6.274484934901720177929867046175406311474380389941415760684209191232450360090766458256588885184199320756050569665785657269735313171886975309933254563488343491718198237894473901620914303565550450204805537225888529509352754121292701357622411614860860409639719786022989336837263283678476008817556351031696366815467221836948040042378034720460820127399855873232167818091083005170669482845098735176209372328114732133251096196535355946589133977397512846130629857604295369747597459602137604440011394793443041829253598478244189078131130488653468669559814695095974271938947640276013215753183113041899037415404445478806695965167014404297848725756879184380559837391976534521522360723388582608454995349380217499779247330557664230806254642768796486899322646423713763772064068933790640394967085887914192401473425799354391464743910233873602389444180426155866237536459654917521713769608318128404177877383203786348495822099924812081683286880293701785567962687838594752986160305764297117036426951203418854463404773701882e+4125
With exp(9500).evalf(5000), you even get the integer closest to exp(9500).

Here's another way to calculate the result with Python:
exp(9500)
is too big.
But log10(exp(9500)) isn't. You cannot calculate it this way in Python, but log10(exp(9500)) is log(exp(9500))/ln(10), which is 9500/ln(10):
>>> from math import log
>>> 9500/log(10)
4125.797578080892
>>> int(9500/log(10))
4125
>>> 10**(9500/log(10) % 1)
6.274484934896202
This way, you can calculate that exp(9500) is 6.27448493 * 10**4125 in plain Python, without any library!

try long type.
int type has been remove from python since 3.0 version.

Python Shell - "Extras" in float subtraction [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Floating Point Limitations
Using Python 2.7 here.
Can someone explain why this happens in the shell?
>>> 5.2-5.0
0.20000000000000018
Searching yielded things about different scales of numbers not producing the right results (a very small number and a very large number), but that seemed pretty general, and considering the numbers I'm using are of the same scale, I don't think that's why this happens.
EDIT: I suppose I didn't define that the "this thing happening" I meant was that it returns 0.2 ... 018 instead of simply resulting in 0.2. I get that print rounds, and removed the print part in the code snippet, as that was misleading.

You need to understand that 5.2-5.0 really is 0.20000000000000018, not 0.2. The standard explanation for this is found in What Every Computer Scientist Should Know About Floating-Point Arithmetic.
If you don't want to read all of that, just accept that 5.2, 5.0, and 0.20000000000000018 are all just approximations, as close as the computer can get to the numbers you really way.
Python has some tricks to allow you to not know what every computer scientist should know and still get away with it. The main trick is that str(f)—that is, the human-readable rendition of a floating-point number—is truncated to 12 significant digits, so str(5.2-5.0) is "0.2", not "0.20000000000000018". But sometimes you need all the precision you can get, so repr(f)—that is, the machine-readable rendition—is not truncated, so repr(5.2-5.0) is "0.20000000000000018".
Now the only thing left to understand is what the interpreter shell does. As Ashwini Chaudhary explains, just evaluating something in the shell prints out its repr, while the print statement prints out its str.

shell uses repr():
In [1]: print repr(5.2-5.0)
0.20000000000000018
In [2]: print str(5.2-5.0)
0.2
In [3]: print 5.2-5.0
0.2

The default implementation of float.__str__ limits the output to 12 digits only.
Thus, the least significant digits are dropped and what is left is the value 0.2.
To print more digits (if available), use string formatting:
print '%f' % result # prints 0.200000
That defaults to 6 digits, but you can specify more precision:
print '%.16f' % result # prints 0.2000000000000002
Alternatively, python offers a newer string formatting method too:
print '{0:.16f}'.format(result) # prints 0.2000000000000002
Why python produces the 'imprecise' result in the first place has everything to do with the imprecise nature of floating point arithmetic. Use the decimal module instead if you need more predictable precision:
>>> from decimal import *
>>> getcontext().prec = 1
>>> Decimal(5.2) - Decimal(5.0)
Decimal('0.2')

Python has two different ways of converting an object to a string, the __str__ and __repr__ methods. __str__ is meant to be a normal string output and is used by print; __repr__ is meant to be a more exact representation and is what is displayed when you don't use print, or when you print the contents of a list or dictionary. __str__ rounds floating-point values.
As for why the actual result of the subtraction is 0.20000000000000018 rather than 0.2 exactly, it has to do with the internal representation of floating point. It's impossible to represent 5.2 exactly because it's an infinitely repeating binary number. The closest that you can come is approximately 5.20000000000000018.

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.