In Python, are either
n**0.5 # or
math.sqrt(n)
recognized when a number is a perfect square? Specifically, should I worry that when I use
int(n**0.5) # instead of
int(n**0.5 + 0.000000001)
I might accidentally end up with the number one less than the actual square root due to precision error?
As several answers have suggested integer arithmetic, I'll recommend the gmpy2 library. It provides functions for checking if a number is a perfect power, calculating integer square roots, and integer square root with remainder.
>>> import gmpy2
>>> gmpy2.is_power(9)
True
>>> gmpy2.is_power(10)
False
>>> gmpy2.isqrt(10)
mpz(3)
>>> gmpy2.isqrt_rem(10)
(mpz(3), mpz(1))
Disclaimer: I maintain gmpy2.
Yes, you should worry:
In [11]: int((100000000000000000000000000000000000**2) ** 0.5)
Out[11]: 99999999999999996863366107917975552L
In [12]: int(math.sqrt(100000000000000000000000000000000000**2))
Out[12]: 99999999999999996863366107917975552L
obviously adding the 0.000000001 doesn't help here either...
As #DSM points out, you can use the decimal library:
In [21]: from decimal import Decimal
In [22]: x = Decimal('100000000000000000000000000000000000')
In [23]: (x ** 2).sqrt() == x
Out[23]: True
for numbers over 10**999999999, provided you keep a check on the precision (configurable), it'll throw an error rather than an incorrect answer...
Both **0.5 and math.sqrt() perform the calculation using floating point arithmetic. The input is converted to float before the square root is calculated.
Do these calculations recognize when the input value is a perfect square?
No they do not. Floating arithmetic has no concept of perfect squares.
large integers may not be representable, for values where the number has more significant digits than available in the floating point mantissa. It's easy to see therefore that for non-representable input values, n**0.5 may be innaccurate. And you proposed fix by adding a small value will not in general fix the problem.
If your input is an integer then you should consider performing your calculation using integer arithmetic. That ultimately is the right way to deal with this.
You can use the round(number, significant_figures) before converting to an int, I cannot recall if python truncs or rounds when doing a float-to-integer conversion.
In any case, since python uses floating point arithmetic, all the pitfalls apply. See:
http://docs.python.org/2/tutorial/floatingpoint.html
Perfect-square values will have no fractional components, so your main worry would be very large values, and for such values a difference of 1 or 2 being significant means you're going to want a specific numerical library that supports such high precision (as DSM mentions, the Decimal library, standard since Python 2.4, should be able to do what you want as it supports arbitrary precision.
http://docs.python.org/library/decimal.html
sqrt is one of the easier math library functions to implement, and any math library of reasonable quality will implement it with faithful rounding (sub-ULP accuracy). If the input is a perfect square, its square root is representable (in a reasonable floating-point format). In this case, faithful rounding guarantees the result is exact.
This addresses only the value actually passed to sqrt. Whether a number can be converted without error from another format to the floating-point input for sqrt is a separate issue.
Related
I work daily with Python 2.4 at my company. I used the versatile logarithm function 'log' from the standard math library, and when I entered log(2**31, 2) it returned 31.000000000000004, which struck me as a bit odd.
I did the same thing with other powers of 2, and it worked perfectly. I ran 'log10(2**31) / log10(2)' and I got a round 31.0
I tried running the same original function in Python 3.0.1, assuming that it was fixed in a more advanced version.
Why does this happen? Is it possible that there are some inaccuracies in mathematical functions in Python?
This is to be expected with computer arithmetic. It is following particular rules, such as IEEE 754, that probably don't match the math you learned in school.
If this actually matters, use Python's decimal type.
Example:
from decimal import Decimal, Context
ctx = Context(prec=20)
two = Decimal(2)
ctx.divide(ctx.power(two, Decimal(31)).ln(ctx), two.ln(ctx))
You should read "What Every Computer Scientist Should Know About Floating-Point Arithmetic".
http://docs.sun.com/source/806-3568/ncg_goldberg.html
Always assume that floating point operations will have some error in them and check for equality taking that error into account (either a percentage value like 0.00001% or a fixed value like 0.00000000001). This inaccuracy is a given since not all decimal numbers can be represented in binary with a fixed number of bits precision.
Your particular case is not one of them if Python uses IEEE754 since 31 should be easily representable with even single precision. It's possible however that it loses precision in one of the many steps it takes to calculate log2231, simply because it doesn't have code to detect special cases like a direct power of two.
floating-point operations are never exact. They return a result which has an acceptable relative error, for the language/hardware infrastructure.
In general, it's quite wrong to assume that floating-point operations are precise, especially with single-precision. "Accuracy problems" section from Wikipedia Floating point article :)
IEEE double floating point numbers have 52 bits of precision. Since 10^15 < 2^52 < 10^16, a double has between 15 and 16 significant figures. The result 31.000000000000004 is correct to 16 figures, so it is as good as you can expect.
This is normal. I would expect log10 to be more accurate then log(x, y), since it knows exactly what the base of the logarithm is, also there may be some hardware support for calculating base-10 logarithms.
float are imprecise
I don't buy that argument, because exact power of two are represented exactly on most platforms (with underlying IEEE 754 floating point).
So if we really want that log2 of an exact power of 2 be exact, we can.
I'll demonstrate it in Squeak Smalltalk, because it is easy to change the base system in that language, but the language does not really matter, floating point computation are universal, and Python object model is not that far from Smalltalk.
For taking log in base n, there is the log: function defined in Number, which naively use the Neperian logarithm ln:
log: aNumber
"Answer the log base aNumber of the receiver."
^self ln / aNumber ln
self ln (take the neperian logarithm of receiver) , aNumber ln and / are three operations that will round there result to nearest Float, and these rounding error can cumulate... So the naive implementation is subject to the rounding error you observe, and I guess that Python implementation of log function is not much different.
((2 raisedTo: 31) log: 2) = 31.000000000000004
But if I change the definition like this:
log: aNumber
"Answer the log base aNumber of the receiver."
aNumber = 2 ifTrue: [^self log2].
^self ln / aNumber ln
provide a generic log2 in Number class:
log2
"Answer the base-2 log of the receiver."
^self asFloat log2
and this refinment in Float class:
log2
"Answer the base 2 logarithm of the receiver.
Care to answer exact result for exact power of two."
^self significand ln / Ln2 + self exponent asFloat
where Ln2 is a constant (2 ln), then I effectively get an exact log2 for exact power of two, because significand of such number = 1.0 (including subnormal for Squeak exponent/significand definition), and 1.0 ln = 0.0.
The implementation is quite trivial, and should translate without difficulty in Python (probably in the VM); the runtime cost is very cheap, so it's just a matter of how important we think this feature is, or is not.
As I always say, the fact that floating point operations results are rounded to nearest (or whatever rounding direction) representable value is not a license to waste ulp. Exactness has a cost, both in term of runtime penalty and implementation complexity, so it's trade-offs driven.
The representation (float.__repr__) of a number in python tries to return a string of digits as close to the real value as possible when converted back, given that IEEE-754 arithmetic is precise up to a limit. In any case, if you printed the result, you wouldn't notice:
>>> from math import log
>>> log(2**31,2)
31.000000000000004
>>> print log(2**31,2)
31.0
print converts its arguments to strings (in this case, through the float.__str__ method), which caters for the inaccuracy by displaying less digits:
>>> log(1000000,2)
19.931568569324174
>>> print log(1000000,2)
19.9315685693
>>> 1.0/10
0.10000000000000001
>>> print 1.0/10
0.1
usuallyuseless' answer is very useful, actually :)
If you wish to calculate the highest power of 'k' in a number 'n'. Then the code below might be helpful:
import math
answer = math.ceil(math.log(n,k))
while k**answer>n:
answer-=1
NOTE: You shouldn't use 'if' instead of 'while' because that will give wrong results in some cases like n=2**51-1 and k=2. In this example with 'if' the answer is 51 whereas with 'while' the answer is 50, which is correct.
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
Is there a standard value for (or method for obtaining) epsilon in Python? I need to compare floating point values and want to compare against the smallest possible difference.
In C++ there's a function provided numeric_limits::epsilon( ) which gives the epsilon value for any given data type. Is there an equivalent in Python?
The information is available in sys.float_info, which corresponds to float.h in C99.
>>> import sys
>>> sys.float_info.epsilon
2.220446049250313e-16
As strcat posted, there is sys.float_info.epsilon.
But don't forget the pitfalls of using it as an absolute error margin for floating point comparisons. E.g. for large numbers, rounding error could exceed epsilon.
If you think you need a refresher, the standard reference is David Goldberg's What Every Computer Scientist Should Know About Floating-Point Arithmetic, or for a simpler review you can check out The Floating Point Guide.
If you cannot find a function to do that, remember that the algorithm to calculate the machine epsilon is very easy (you can test with your favourite programming language).E.g, for python:
eps = 1.0
while eps + 1 > 1:
eps /= 2
eps *= 2
print("The machine epsilon is:", eps)
In my case, I got:
The machine epsilon is: 2.220446049250313e-16
Surprised nobody mentioned this here; I think many people would use numpy.finfo( type(variable) ).eps instead. Or .resolution if it is to assess precision.
Note that finfo is only for floating point types, and that it also works with Python's own float type (i.e. not restricted to numpy's types). The equivalent for integer types is iinfo, though it does not contain precision information (because, well, why would it?).
The following worked for me as well:
>>> import math
>>> math.ulp(1.0)
2.220446049250313e-16
I work daily with Python 2.4 at my company. I used the versatile logarithm function 'log' from the standard math library, and when I entered log(2**31, 2) it returned 31.000000000000004, which struck me as a bit odd.
I did the same thing with other powers of 2, and it worked perfectly. I ran 'log10(2**31) / log10(2)' and I got a round 31.0
I tried running the same original function in Python 3.0.1, assuming that it was fixed in a more advanced version.
Why does this happen? Is it possible that there are some inaccuracies in mathematical functions in Python?
This is to be expected with computer arithmetic. It is following particular rules, such as IEEE 754, that probably don't match the math you learned in school.
If this actually matters, use Python's decimal type.
Example:
from decimal import Decimal, Context
ctx = Context(prec=20)
two = Decimal(2)
ctx.divide(ctx.power(two, Decimal(31)).ln(ctx), two.ln(ctx))
You should read "What Every Computer Scientist Should Know About Floating-Point Arithmetic".
http://docs.sun.com/source/806-3568/ncg_goldberg.html
Always assume that floating point operations will have some error in them and check for equality taking that error into account (either a percentage value like 0.00001% or a fixed value like 0.00000000001). This inaccuracy is a given since not all decimal numbers can be represented in binary with a fixed number of bits precision.
Your particular case is not one of them if Python uses IEEE754 since 31 should be easily representable with even single precision. It's possible however that it loses precision in one of the many steps it takes to calculate log2231, simply because it doesn't have code to detect special cases like a direct power of two.
floating-point operations are never exact. They return a result which has an acceptable relative error, for the language/hardware infrastructure.
In general, it's quite wrong to assume that floating-point operations are precise, especially with single-precision. "Accuracy problems" section from Wikipedia Floating point article :)
IEEE double floating point numbers have 52 bits of precision. Since 10^15 < 2^52 < 10^16, a double has between 15 and 16 significant figures. The result 31.000000000000004 is correct to 16 figures, so it is as good as you can expect.
This is normal. I would expect log10 to be more accurate then log(x, y), since it knows exactly what the base of the logarithm is, also there may be some hardware support for calculating base-10 logarithms.
float are imprecise
I don't buy that argument, because exact power of two are represented exactly on most platforms (with underlying IEEE 754 floating point).
So if we really want that log2 of an exact power of 2 be exact, we can.
I'll demonstrate it in Squeak Smalltalk, because it is easy to change the base system in that language, but the language does not really matter, floating point computation are universal, and Python object model is not that far from Smalltalk.
For taking log in base n, there is the log: function defined in Number, which naively use the Neperian logarithm ln:
log: aNumber
"Answer the log base aNumber of the receiver."
^self ln / aNumber ln
self ln (take the neperian logarithm of receiver) , aNumber ln and / are three operations that will round there result to nearest Float, and these rounding error can cumulate... So the naive implementation is subject to the rounding error you observe, and I guess that Python implementation of log function is not much different.
((2 raisedTo: 31) log: 2) = 31.000000000000004
But if I change the definition like this:
log: aNumber
"Answer the log base aNumber of the receiver."
aNumber = 2 ifTrue: [^self log2].
^self ln / aNumber ln
provide a generic log2 in Number class:
log2
"Answer the base-2 log of the receiver."
^self asFloat log2
and this refinment in Float class:
log2
"Answer the base 2 logarithm of the receiver.
Care to answer exact result for exact power of two."
^self significand ln / Ln2 + self exponent asFloat
where Ln2 is a constant (2 ln), then I effectively get an exact log2 for exact power of two, because significand of such number = 1.0 (including subnormal for Squeak exponent/significand definition), and 1.0 ln = 0.0.
The implementation is quite trivial, and should translate without difficulty in Python (probably in the VM); the runtime cost is very cheap, so it's just a matter of how important we think this feature is, or is not.
As I always say, the fact that floating point operations results are rounded to nearest (or whatever rounding direction) representable value is not a license to waste ulp. Exactness has a cost, both in term of runtime penalty and implementation complexity, so it's trade-offs driven.
The representation (float.__repr__) of a number in python tries to return a string of digits as close to the real value as possible when converted back, given that IEEE-754 arithmetic is precise up to a limit. In any case, if you printed the result, you wouldn't notice:
>>> from math import log
>>> log(2**31,2)
31.000000000000004
>>> print log(2**31,2)
31.0
print converts its arguments to strings (in this case, through the float.__str__ method), which caters for the inaccuracy by displaying less digits:
>>> log(1000000,2)
19.931568569324174
>>> print log(1000000,2)
19.9315685693
>>> 1.0/10
0.10000000000000001
>>> print 1.0/10
0.1
usuallyuseless' answer is very useful, actually :)
If you wish to calculate the highest power of 'k' in a number 'n'. Then the code below might be helpful:
import math
answer = math.ceil(math.log(n,k))
while k**answer>n:
answer-=1
NOTE: You shouldn't use 'if' instead of 'while' because that will give wrong results in some cases like n=2**51-1 and k=2. In this example with 'if' the answer is 51 whereas with 'while' the answer is 50, which is correct.
So I've decided to try to solve my physics homework by writing some python scripts to solve problems for me. One problem that I'm running into is that significant figures don't always seem to come out properly. For example this handles significant figures properly:
from decimal import Decimal
>>> Decimal('1.0') + Decimal('2.0')
Decimal("3.0")
But this doesn't:
>>> Decimal('1.00') / Decimal('3.00')
Decimal("0.3333333333333333333333333333")
So two questions:
Am I right that this isn't the expected amount of significant digits, or do I need to brush up on significant digit math?
Is there any way to do this without having to set the decimal precision manually? Granted, I'm sure I can use numpy to do this, but I just want to know if there's a way to do this with the decimal module out of curiosity.
Changing the decimal working precision to 2 digits is not a good idea, unless you absolutely only are going to perform a single operation.
You should always perform calculations at higher precision than the level of significance, and only round the final result. If you perform a long sequence of calculations and round to the number of significant digits at each step, errors will accumulate. The decimal module doesn't know whether any particular operation is one in a long sequence, or the final result, so it assumes that it shouldn't round more than necessary. Ideally it would use infinite precision, but that is too expensive so the Python developers settled for 28 digits.
Once you've arrived at the final result, what you probably want is quantize:
>>> (Decimal('1.00') / Decimal('3.00')).quantize(Decimal("0.001"))
Decimal("0.333")
You have to keep track of significance manually. If you want automatic significance tracking, you should use interval arithmetic. There are some libraries available for Python, including pyinterval and mpmath (which supports arbitrary precision). It is also straightforward to implement interval arithmetic with the decimal library, since it supports directed rounding.
You may also want to read the Decimal Arithmetic FAQ: Is the decimal arithmetic ‘significance’ arithmetic?
Decimals won't throw away decimal places like that. If you really want to limit precision to 2 d.p. then try
decimal.getcontext().prec=2
EDIT: You can alternatively call quantize() every time you multiply or divide (addition and subtraction will preserve the 2 dps).
Just out of curiosity...is it necessary to use the decimal module? Why not floating point with a significant-figures rounding of numbers when you are ready to see them? Or are you trying to keep track of the significant figures of the computation (like when you have to do an error analysis of a result, calculating the computed error as a function of the uncertainties that went into the calculation)? If you want a rounding function that rounds from the left of the number instead of the right, try:
def lround(x,leadingDigits=0):
"""Return x either as 'print' would show it (the default)
or rounded to the specified digit as counted from the leftmost
non-zero digit of the number, e.g. lround(0.00326,2) --> 0.0033
"""
assert leadingDigits>=0
if leadingDigits==0:
return float(str(x)) #just give it back like 'print' would give it
return float('%.*e' % (int(leadingDigits),x)) #give it back as rounded by the %e format
The numbers will look right when you print them or convert them to strings, but if you are working at the prompt and don't explicitly print them they may look a bit strange:
>>> lround(1./3.,2),str(lround(1./3.,2)),str(lround(1./3.,4))
(0.33000000000000002, '0.33', '0.3333')
Decimal defaults to 28 places of precision.
The only way to limit the number of digits it returns is by altering the precision.
What's wrong with floating point?
>>> "%8.2e"% ( 1.0/3.0 )
'3.33e-01'
It was designed for scientific-style calculations with a limited number of significant digits.
If I undertand Decimal correctly, the "precision" is the number of digits after the decimal point in decimal notation.
You seem to want something else: the number of significant digits. That is one more than the number of digits after the decimal point in scientific notation.
I would be interested in learning about a Python module that does significant-digits-aware floating point point computations.