We Keep Coding

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

calculating decimal power of 10

I need some help with this calculation :
10**7342345.54334 % m
I am calculating it in python but that says Result too large
pow(10,7342345.54334,m)
Can not be used because all parameters must be integers. If I convert this decimal value to integer then the answer changes.

Hint: Make use of the equivalence of the following expressions
and
For further details, see Modular exponentiation in Wikipedia.

107342345.54334 is equal to 107342345 times 100.54334. Now can you do it?

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.

Getting decimals from python math function?

I'm a bit new to python and can't seem to figure out what I'm doing wrong.
a = 9
b = 13
print ((a-b)/a)
-1
But on my calculator, the correct answer is -0.444444444 (meaning 'a' is about 45% lower than 'b').
How can I get a few decimals to show up?
I tried
print Decimal((a-b)/a)
print float((a-b)/a)
both with the same result. it works if I make a = 9.0 but I wanted to see if there was anything I can do without changing the variables.
I'm sure this is super easy but I'm not sure what to try. Any suggestions?
Thanks

Try converting one (or both) of the arguments to a float, rather than the result:
print ((a-b)/float(a))
Or just upgrade to Python 3 where this behaviour has been fixed:
>>> a = 9
>>> b = 13
>>> print ((a-b)/a)
-0.4444444444444444
By the way, if you want integer division in Python 3.x you can use // instead of /. See the PEP about this change if you are interested.

You need to specify that you want to operate on floating point numbers.
For example:
3.0/4.0 or just 3.0/4 will give you floating point. Right now it's just performing integer operations.
EDIT: you could use float(3)/4 too

You are performing division on two integer operands and in Python 2.x this means integer division. That is, the result is an integer.
You need what is known as floating point division. To force this you just need at least one of the operands to the division to be a float. For example:
print ((a-b)/float(a))
or
print (float(a-b)/a)

Python float - str - float weirdness

>>> float(str(0.65000000000000002))
0.65000000000000002
>>> float(str(0.47000000000000003))
0.46999999999999997 ???
What is going on here?
How do I convert 0.47000000000000003 to string and the resultant value back to float?
I am using Python 2.5.4 on Windows.

str(0.47000000000000003) give '0.47' and float('0.47') can be 0.46999999999999997.
This is due to the way floating point number are represented (see this wikipedia article)
Note: float(repr(0.47000000000000003)) or eval(repr(0.47000000000000003)) will give you the expected result, but you should use Decimal if you need precision.

float (and double) do not have infinite precision. Naturally, rounding errors occur when you operate on them.

This is a Python FAQ
The same question comes up quite regularly in comp.lang.python also.
I think reason it is a FAQ is that because python is perfect in all other respects ;-), we expect it to perform arithmetic perfectly - just like we were taught at school. However, as anyone who has done a numerical methods course will tell you, floating point numbers are a very long way from perfect.
Decimal is a good alternative and if you want more speed and more options gmpy is great too.

by this example
I think this is an error in Python when you devide
>>> print(int(((48/5.0)-9)*5))
2
the easy way, I solve this problem by this
>>> print(int(round(((48/5.0)-9)*5,2)))
3