I have a string:
x = "12.000"
And I want it to convert it to digits. However, I have used int, float, and others but I only get 12.0 and i want to keep all the zeroes. Please help!
I want x = 12.000 as a result.
decimal.Decimal allows you to use a specific precision.
>>> decimal.Decimal('12.000')
Decimal('12.000')
If you really want to perform calculations that take precision into account, the easiest way is to probably to use the uncertainties module. Here is an example
>>> import uncertainties
>>> x = uncertainties.ufloat('12.000')
>>> x
12.0+/-0.001
>>> print 2*x
24.0+/-0.002
The uncertainties module transparently handles uncertainties (precision) for you, whatever the complexity of the mathematical expressions involved.
The decimal module, on the other hand, does not handle uncertainties, but instead sets the number of digits after the decimal point: you can't trust all the digits given by the decimal module. Thus,
>>> 100*decimal.Decimal('12.1')
Decimal('1210.0')
whereas 100*(12.1±0.1) = 1210±10 (not 1210.0±0.1):
>>> 100*uncertainties.ufloat('12.1')
1210.0+/-10.0
Thus, the decimal module gives '1210.0' even though the precision on 100*(12.1±0.1) is 100 times larger than 0.1.
So, if you want numbers that have a fixed number of digits after the decimal point (like for accounting applications), the decimal module is good; if you instead need to perform calculations with uncertainties, then the uncertainties module is appropriate.
(Disclaimer: I'm the author of the uncertainties module.)
You may be interested by the decimal python lib.
You can set the precision with getcontext().prec.
Related
I'm using the modulus operator and I'm get some floating point errors. For example,
>>> 7.2%3
1.2000000000000002
Is my only recourse to handle this by using the round function? E.g.
>>> round(7.2%3, 1)
1.2
I don't a priori know the number of digits I'm going to need to round to, so I'm wondering if there's a better solution?
If you want arbitrary precision, use the decimal module:
>>> import decimal
>>> decimal.Decimal('7.2') % decimal.Decimal('3')
Decimal('1.2')
Please read the documentation carefully.
Notice I used a str as an argument to Decimal. Look what happens if I didn't:
>>> decimal.Decimal(7.2) % decimal.Decimal(3)
Decimal('1.200000000000000177635683940')
>>>
I need to write a simple program that calculates a mathematical formula.
The only problem here is that one of the variables can take the value 10^100.
Because of this I can not write this program in C++/C (I can't use external libraries like gmp).
Few hours ago I read that Python is capable of calculating such values.
My question is:
Why
print("%.10f"%(10.25**100))
is returning the number "118137163510621843218803309161687290343217035128100169109374848108012122824436799009169146127891562496.0000000000"
instead of
"118137163510621850716311252946961817841741635398513936935237985161753371506358048089333490072379307296.453937046171461"?
By default, Python uses a fixed precision floating-point data type to represent fractional numbers (just like double in C). You can work with precise rational numbers, though:
>>> from fractions import Fraction
>>> Fraction("10.25")
Fraction(41, 4)
>>> x = Fraction("10.25")
>>> x**100
Fraction(189839102486063226543090986563273122284619337618944664609359292215966165735102377674211649585188827411673346619890309129617784863285653302296666895356073140724001, 1606938044258990275541962092341162602522202993782792835301376)
You can also use the decimal module if you want arbitrary precision decimals (only numbers that are representable as finite decimals are supported, though):
>>> from decimal import *
>>> getcontext().prec = 150
>>> Decimal("10.25")**100
Decimal('118137163510621850716311252946961817841741635398513936935237985161753371506358048089333490072379307296.453937046171460995169093650913476028229144848989')
Python is capable of handling arbitrarily large integers, but not floating point values. They can get pretty large, but as you noticed, you lose precision in the low digits.
>>> str(1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702)
'1.41421356237'
Is there a way I can make str() record more digits of the number into the string? I don't understand why it truncates by default.
Python's floating point numbers use double precision only, which is 64 bits. They simply cannot represent (significantly) more digits than you're seeing.
If you need more, have a look at the built-in decimal module, or the mpmath package.
Try this:
>>> from decimal import *
>>> Decimal('1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702')
Decimal('1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702')
The float literal is truncated by default to fit in the space made available for it (i.e. it's not because of str):
>>> 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702
1.4142135623730951
If you need more decimal places use decimal instead.
The Python compiler is truncating; your float literal has more precision than can be represented in a C double. Express the number as a string in the first place if you need more precision.
That's because it's converting to a float. It's not the conversion to the string that's causing it.
You should use decimal.Decimal for representing such high precision numbers.
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.
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.