Using Python 3, how does the following return True ?
a = 2/3
b = 4/6
print(a == b)
I have an algorithm that requires sorting a list of numbers which are each of the form x/y where x and y are integers. (y != 0).
I was concerned that the numerical precision of the division would result in instability and arbitrary ordering of cases such as above. This being an example of relevant comments.But, as per the example and for larger integers as well, it does not seem to be an issue.
Does Python remove the common factor of 2 from the numerator and denominator of b, and retain information that a and b are not just floats?
Python follows the IEEE 754 floating point specification.* (64-bit) IEEE floats are essentially a form of base 2 scientific notation, broken down as follows:
One bit for the sign (positive or negative)
53 bits for the mantissa or significand, including the implied leading one.
11 bits for the exponent.
Multiplying or dividing a floating point value by two, or any power of two, only affects the exponent, and not the mantissa.** As a result, it is normally a fairly "stable" operation by itself, so 2/3 should yield the same result as 4/6. However, IEEE floats still have the following problems:
Most operations are not associative (e.g. (a * b) * c != a * (b * c) in the general case).
More complicated operations are not required to be correctly rounded (however, as Tim Peters points out, division certainly is not a "more complicated" operation and will be correctly rounded).***
Intermediate results are always rounded to 53 bits.
You should be prepared to handle these issues and assume that most mathematically-equivalent floating point expressions will not result in identical values. In Python specifically, you can use math.isclose() to estimate whether two floats are "close enough" to be "probably the same value."
* Actually, this is a lie. Python follows C's double, which nearly always follows IEEE 754 in some fashion, but might deviate from it on sufficiently exotic architectures. In such cases the C standard provides few or no guarantees, so you will have to look to your architecture or compiler's floating point documentation.
** Provided the exponent does not overflow or underflow. If it does, then you will typically land on an appropriately-signed infinity or zero, respectively, or you might underflow to a denormal number depending on architecture and/or how Python was compiled.
*** The exact set of "more complicated" operations varies somewhat because IEEE 754 made a lot of operations optional while still demanding precision. As a result, it is seldom obvious whether a given operation conforms to IEEE 754 or only conforms to the notoriously lax C standard. In some cases, an operation may conform to no standard whatsoever.
Just noting that so long as integers x and y are exactly representable as Python floats, x / y is - on all current machines - the correctly rounded value of the infinitely precise quotient. That's what the IEEE 754 floating-point standard requires, and all current machines support that.
So the important part in your specific example isn't that the numerator and denominator in b = 4/6 have a factor of (specifically!) 2 in common, it's that (a) they have some factor in common; and, (b) 4 and 6 are both exactly representable as Python floats.
So, for example, it's guaranteed that
(2 * 9892837) / (3 * 9892837) == 2 / 3
is also true. Because the infinitely precise value of (2 * 9892837) / (3 * 9892837) is the same as the infinitely precisely value of 2/3, and IEEE 754 division acts as if the infinitely precise quotient were computed. And you can replace 9892837 with any other non-zero integer in that, provided that the products remain exactly representable as Python floats.
2/3 is the same as 4/6. (2/3)*(2/2) = 2/2 = 1, the identity element. The response is correct.
Related
I was under the impression that Python uses double precision arithmetic, with "nearest to" rounding. However, consider the following:
In the double precision system, the next number after 1.0 is 1.00...01 = 1 + 2**(-52). Now, if Python uses the "nearest to" rounding, the number 1 + 2**(-53) should round to 1 + 2**(-52). However, it turns out that 1 + 2**(-53) == 1. This would make sense if Python uses the "chop" rounding rule, but I was under the impression that nobody uses that since it biases calculations towards lower results.
The Python documentation is not strict about how floating-point arithmetic is handled. Some Python implementations use IEEE 754 with round-to-nearest-ties-to-even.
In the IEEE-754 binary64 format, also known as the “double precision” format, 1+2−53 is the midpoint between 1 and the next representable number, 1+2−52. So this is a tie, and the round-to-even rule applies. The significand for 1 is 1.000…0002, and the significand for 1+2−52 is 1.000…0012. The former is even, so the tie-breaking rule chooses it, and the result is 1.
Consider instead 1+3•2−54. This is three-quarters of the way from 1 to 1+2−52. So rounding it to the nearest representable value will produce 1+2−52. For print(1 == 1+3*2**-54), your Python implementation will likely print “False”.
I'm new to programming and studying the basic now. I'm wondering that does the float type in python only represent approximations to real number? I know float uses binary fractions but are the floats 0.5, 0.25, 0.125, etc still the approximations? I tried:
sum([0.1] * 10) == 1
it returned False.
But
sum([0.5] * 10) == 5
It returned True.
Finally I tried:
for i in range(1, 8):
answer = sum([1 / 2 ** i] * 10)
print(answer == 1 / 2 ** i * 10)
The answer is all True.
It's that means some floats in python are exactly the real number not the approximations?
Each floating-point object represents one number (or special value such as NaN) exactly. Floating-point objects do not represent approximations.
The correct way to think about floating-point is that floating-point values are exact numbers, but floating-point operations approximate real arithmetic.
Python does not specify floating-point arithmetic precisely; each Python implementation may use the underlying arithmetic of the platform it is implemented on. Commonly, IEEE 754 formats are used, although the operations may not conform to IEEE 754 completely. To illustrate what is happening with your code, I will use IEEE-754 basic 64-bit binary floating-point.
When the source text 0.5 is processed, it is converted to floating-point. Note that conversion is an operation, just as addition or multiplication are operations. The characters are interpreted as a decimal numeral, and the conversion produces the floating-point number that is closest to the number represented by the decimal numeral. In this case, 0.5 represents one-half, and that is exactly representable in binary floating-point, so the result is exactly 0.5.
Then [0.5] * 10 produces a list containing ten copies of 0.5, and sum adds those. All of the additions performed in this summation are exact, because the floating-point format can exactly represent 0.5, 1, 1.5, 2, and so on. So the result is 5, exactly, and comparing this to 5 produces true.
On the other hand, when the source text 0.1 is processed, that decimal numeral represents one-tenth, which cannot be represented exactly. The conversion produces the nearest representable value, which is 0.1000000000000000055511151231257827021181583404541015625.
When sum adds the ten copies of this, the addition cannot always be performed exactly. Adding the first two is exact, adding 0.1000000000000000055511151231257827021181583404541015625 to 0.1000000000000000055511151231257827021181583404541015625 produces 0.200000000000000011102230246251565404236316680908203125. However, when 0.200000000000000011102230246251565404236316680908203125 is added to 0.1000000000000000055511151231257827021181583404541015625, the result is 0.3000000000000000444089209850062616169452667236328125. During this addition, the bits in the addition carried to a new position (the operands are under ¼, but the result is over ¼—the addition carried into the ¼ position. Since the floating-point format has only a fixed number of bits (53) available for the value, the operation had to discard the low bit. In doing so, it changed the result slightly. So this addition is only approximate.
As these additions go on, the final value is 0.99999999999999988897769753748434595763683319091796875. When this is compared to 1, the result is false.
Python represents floating point numbers as binary fractions. Therefore numbers like 0.5 can be represented accurately, whereas 0.1 for example can not.
Floating point numbers in Python are just approximations, if they can not be exactly represented using binary fractions.
If you need more accuracy when dealing with floating point arithmetic, I would suggest taking a look at decimals: https://docs.python.org/3/library/decimal.html
Additionally, a good resource on floating point numbers in Python can be found here: https://docs.python.org/3/tutorial/floatingpoint.html
I'm pretty new to python, and I've made a table which calculates T=1+2^-n-1 and C=2^n, which both give the same values from n=40 to n=52, but for n=52 to n=61 I get 0.0 for T, whereas C gives me progressively smaller decimals each time - why is this?
I think I understand why T becomes 0.0, because of python using binary floating point and because of the machine epsilon value - but I'm slightly confused as to why C doesn't also become 0.0.
import numpy as np
import math
t=np.zeros(21)
c=np.zeros(21)
for n in range(40,61):
m=n-40
t[m]=1+2**(-n)-1
c[m]=2**(-n)
print (n,t[m],c[m])
The "floating" in floating point means that values are represented by storing a fixed number of leading digits and a scale factor, rather than assuming a fixed scale (which would be fixed point).
2**-53 only takes one (binary) digit to represent (not including the scale), but 1+2**-53 would take 54 to represent exactly. Python floats only have 53 binary digits of precision; 2**-53 can be represented exactly, but 1+2**-53 gets rounded to exactly 1, and subtracting 1 from that gives exactly 0. Thus, we have
>>> 2**-53
1.1102230246251565e-16
>>> 1+(2**-53)-1
0.0
Postscript: you might wonder why 2**-53 displays as a value not equal to the exact mathematical value when I said it was exact. That's due to the float->string conversion logic, which only keeps enough decimal digits to reconstruct the original float (instead of printing a bunch of digits at the end that are usually just noise).
The difference between both is indeed due to floating-point representation. Indeed, if you perform 1 + X where X is a very very small number, then the floating-point representation sets its exponent value to 0 and the precision is ensured by the mantissa, which is 52-bit on a 64-bit computer. Therefore, 1 + 2^(-X) if X > 52 is equal to 1. However, even 2^-100 can be represented in double-precision floating-point, so you can see C decrease for a larger number of samples.
Why do some numbers lose accuracy when stored as floating point numbers?
For example, the decimal number 9.2 can be expressed exactly as a ratio of two decimal integers (92/10), both of which can be expressed exactly in binary (0b1011100/0b1010). However, the same ratio stored as a floating point number is never exactly equal to 9.2:
32-bit "single precision" float: 9.19999980926513671875
64-bit "double precision" float: 9.199999999999999289457264239899814128875732421875
How can such an apparently simple number be "too big" to express in 64 bits of memory?
In most programming languages, floating point numbers are represented a lot like scientific notation: with an exponent and a mantissa (also called the significand). A very simple number, say 9.2, is actually this fraction:
5179139571476070 * 2 -49
Where the exponent is -49 and the mantissa is 5179139571476070. The reason it is impossible to represent some decimal numbers this way is that both the exponent and the mantissa must be integers. In other words, all floats must be an integer multiplied by an integer power of 2.
9.2 may be simply 92/10, but 10 cannot be expressed as 2n if n is limited to integer values.
Seeing the Data
First, a few functions to see the components that make a 32- and 64-bit float. Gloss over these if you only care about the output (example in Python):
def float_to_bin_parts(number, bits=64):
if bits == 32: # single precision
int_pack = 'I'
float_pack = 'f'
exponent_bits = 8
mantissa_bits = 23
exponent_bias = 127
elif bits == 64: # double precision. all python floats are this
int_pack = 'Q'
float_pack = 'd'
exponent_bits = 11
mantissa_bits = 52
exponent_bias = 1023
else:
raise ValueError, 'bits argument must be 32 or 64'
bin_iter = iter(bin(struct.unpack(int_pack, struct.pack(float_pack, number))[0])[2:].rjust(bits, '0'))
return [''.join(islice(bin_iter, x)) for x in (1, exponent_bits, mantissa_bits)]
There's a lot of complexity behind that function, and it'd be quite the tangent to explain, but if you're interested, the important resource for our purposes is the struct module.
Python's float is a 64-bit, double-precision number. In other languages such as C, C++, Java and C#, double-precision has a separate type double, which is often implemented as 64 bits.
When we call that function with our example, 9.2, here's what we get:
>>> float_to_bin_parts(9.2)
['0', '10000000010', '0010011001100110011001100110011001100110011001100110']
Interpreting the Data
You'll see I've split the return value into three components. These components are:
Sign
Exponent
Mantissa (also called Significand, or Fraction)
Sign
The sign is stored in the first component as a single bit. It's easy to explain: 0 means the float is a positive number; 1 means it's negative. Because 9.2 is positive, our sign value is 0.
Exponent
The exponent is stored in the middle component as 11 bits. In our case, 0b10000000010. In decimal, that represents the value 1026. A quirk of this component is that you must subtract a number equal to 2(# of bits) - 1 - 1 to get the true exponent; in our case, that means subtracting 0b1111111111 (decimal number 1023) to get the true exponent, 0b00000000011 (decimal number 3).
Mantissa
The mantissa is stored in the third component as 52 bits. However, there's a quirk to this component as well. To understand this quirk, consider a number in scientific notation, like this:
6.0221413x1023
The mantissa would be the 6.0221413. Recall that the mantissa in scientific notation always begins with a single non-zero digit. The same holds true for binary, except that binary only has two digits: 0 and 1. So the binary mantissa always starts with 1! When a float is stored, the 1 at the front of the binary mantissa is omitted to save space; we have to place it back at the front of our third element to get the true mantissa:
1.0010011001100110011001100110011001100110011001100110
This involves more than just a simple addition, because the bits stored in our third component actually represent the fractional part of the mantissa, to the right of the radix point.
When dealing with decimal numbers, we "move the decimal point" by multiplying or dividing by powers of 10. In binary, we can do the same thing by multiplying or dividing by powers of 2. Since our third element has 52 bits, we divide it by 252 to move it 52 places to the right:
0.0010011001100110011001100110011001100110011001100110
In decimal notation, that's the same as dividing 675539944105574 by 4503599627370496 to get 0.1499999999999999. (This is one example of a ratio that can be expressed exactly in binary, but only approximately in decimal; for more detail, see: 675539944105574 / 4503599627370496.)
Now that we've transformed the third component into a fractional number, adding 1 gives the true mantissa.
Recapping the Components
Sign (first component): 0 for positive, 1 for negative
Exponent (middle component): Subtract 2(# of bits) - 1 - 1 to get the true exponent
Mantissa (last component): Divide by 2(# of bits) and add 1 to get the true mantissa
Calculating the Number
Putting all three parts together, we're given this binary number:
1.0010011001100110011001100110011001100110011001100110 x 1011
Which we can then convert from binary to decimal:
1.1499999999999999 x 23 (inexact!)
And multiply to reveal the final representation of the number we started with (9.2) after being stored as a floating point value:
9.1999999999999993
Representing as a Fraction
9.2
Now that we've built the number, it's possible to reconstruct it into a simple fraction:
1.0010011001100110011001100110011001100110011001100110 x 1011
Shift mantissa to a whole number:
10010011001100110011001100110011001100110011001100110 x 1011-110100
Convert to decimal:
5179139571476070 x 23-52
Subtract the exponent:
5179139571476070 x 2-49
Turn negative exponent into division:
5179139571476070 / 249
Multiply exponent:
5179139571476070 / 562949953421312
Which equals:
9.1999999999999993
9.5
>>> float_to_bin_parts(9.5)
['0', '10000000010', '0011000000000000000000000000000000000000000000000000']
Already you can see the mantissa is only 4 digits followed by a whole lot of zeroes. But let's go through the paces.
Assemble the binary scientific notation:
1.0011 x 1011
Shift the decimal point:
10011 x 1011-100
Subtract the exponent:
10011 x 10-1
Binary to decimal:
19 x 2-1
Negative exponent to division:
19 / 21
Multiply exponent:
19 / 2
Equals:
9.5
Further reading
The Floating-Point Guide: What Every Programmer Should Know About Floating-Point Arithmetic, or, Why don’t my numbers add up? (floating-point-gui.de)
What Every Computer Scientist Should Know About Floating-Point Arithmetic (Goldberg 1991)
IEEE Double-precision floating-point format (Wikipedia)
Floating Point Arithmetic: Issues and Limitations (docs.python.org)
Floating Point Binary
This isn't a full answer (mhlester already covered a lot of good ground I won't duplicate), but I would like to stress how much the representation of a number depends on the base you are working in.
Consider the fraction 2/3
In good-ol' base 10, we typically write it out as something like
0.666...
0.666
0.667
When we look at those representations, we tend to associate each of them with the fraction 2/3, even though only the first representation is mathematically equal to the fraction. The second and third representations/approximations have an error on the order of 0.001, which is actually much worse than the error between 9.2 and 9.1999999999999993. In fact, the second representation isn't even rounded correctly! Nevertheless, we don't have a problem with 0.666 as an approximation of the number 2/3, so we shouldn't really have a problem with how 9.2 is approximated in most programs. (Yes, in some programs it matters.)
Number bases
So here's where number bases are crucial. If we were trying to represent 2/3 in base 3, then
(2/3)10 = 0.23
In other words, we have an exact, finite representation for the same number by switching bases! The take-away is that even though you can convert any number to any base, all rational numbers have exact finite representations in some bases but not in others.
To drive this point home, let's look at 1/2. It might surprise you that even though this perfectly simple number has an exact representation in base 10 and 2, it requires a repeating representation in base 3.
(1/2)10 = 0.510 = 0.12 = 0.1111...3
Why are floating point numbers inaccurate?
Because often-times, they are approximating rationals that cannot be represented finitely in base 2 (the digits repeat), and in general they are approximating real (possibly irrational) numbers which may not be representable in finitely many digits in any base.
While all of the other answers are good there is still one thing missing:
It is impossible to represent irrational numbers (e.g. π, sqrt(2), log(3), etc.) precisely!
And that actually is why they are called irrational. No amount of bit storage in the world would be enough to hold even one of them. Only symbolic arithmetic is able to preserve their precision.
Although if you would limit your math needs to rational numbers only the problem of precision becomes manageable. You would need to store a pair of (possibly very big) integers a and b to hold the number represented by the fraction a/b. All your arithmetic would have to be done on fractions just like in highschool math (e.g. a/b * c/d = ac/bd).
But of course you would still run into the same kind of trouble when pi, sqrt, log, sin, etc. are involved.
TL;DR
For hardware accelerated arithmetic only a limited amount of rational numbers can be represented. Every not-representable number is approximated. Some numbers (i.e. irrational) can never be represented no matter the system.
There are infinitely many real numbers (so many that you can't enumerate them), and there are infinitely many rational numbers (it is possible to enumerate them).
The floating-point representation is a finite one (like anything in a computer) so unavoidably many many many numbers are impossible to represent. In particular, 64 bits only allow you to distinguish among only 18,446,744,073,709,551,616 different values (which is nothing compared to infinity). With the standard convention, 9.2 is not one of them. Those that can are of the form m.2^e for some integers m and e.
You might come up with a different numeration system, 10 based for instance, where 9.2 would have an exact representation. But other numbers, say 1/3, would still be impossible to represent.
Also note that double-precision floating-points numbers are extremely accurate. They can represent any number in a very wide range with as much as 15 exact digits. For daily life computations, 4 or 5 digits are more than enough. You will never really need those 15, unless you want to count every millisecond of your lifetime.
Why can we not represent 9.2 in binary floating point?
Floating point numbers are (simplifying slightly) a positional numbering system with a restricted number of digits and a movable radix point.
A fraction can only be expressed exactly using a finite number of digits in a positional numbering system if the prime factors of the denominator (when the fraction is expressed in it's lowest terms) are factors of the base.
The prime factors of 10 are 5 and 2, so in base 10 we can represent any fraction of the form a/(2b5c).
On the other hand the only prime factor of 2 is 2, so in base 2 we can only represent fractions of the form a/(2b)
Why do computers use this representation?
Because it's a simple format to work with and it is sufficiently accurate for most purposes. Basically the same reason scientists use "scientific notation" and round their results to a reasonable number of digits at each step.
It would certainly be possible to define a fraction format, with (for example) a 32-bit numerator and a 32-bit denominator. It would be able to represent numbers that IEEE double precision floating point could not, but equally there would be many numbers that can be represented in double precision floating point that could not be represented in such a fixed-size fraction format.
However the big problem is that such a format is a pain to do calculations on. For two reasons.
If you want to have exactly one representation of each number then after each calculation you need to reduce the fraction to it's lowest terms. That means that for every operation you basically need to do a greatest common divisor calculation.
If after your calculation you end up with an unrepresentable result because the numerator or denominator you need to find the closest representable result. This is non-trivil.
Some Languages do offer fraction types, but usually they do it in combination with arbitary precision, this avoids needing to worry about approximating fractions but it creates it's own problem, when a number passes through a large number of calculation steps the size of the denominator and hence the storage needed for the fraction can explode.
Some languages also offer decimal floating point types, these are mainly used in scenarios where it is imporant that the results the computer gets match pre-existing rounding rules that were written with humans in mind (chiefly financial calculations). These are slightly more difficult to work with than binary floating point, but the biggest problem is that most computers don't offer hardware support for them.
The built-in Python str() function outputs some weird results when passing in floats with many decimals. This is what happens:
>>> str(19.9999999999999999)
>>> '20.0'
I'm expecting to get:
>>> '19.9999999999999999'
Does anyone know why? and maybe workaround it?
Thanks!
It's not str() that rounds, it's the fact that you're using floats in the first place. Float types are fast, but have limited precision; in other words, they are imprecise by design. This applies to all programming languages. For more details on float quirks, please read "What Every Programmer Should Know About Floating-Point Arithmetic"
If you want to store and operate on precise numbers, use the decimal module:
>>> from decimal import Decimal
>>> str(Decimal('19.9999999999999999'))
'19.9999999999999999'
A float has 32 bits (in C at least). One of those bits is allocated for the sign, a few allocated for the mantissa, and a few allocated for the exponent. You can't fit every single decimal to an infinite number of digits into 32 bits. Therefore floating point numbers are heavily based on rounding.
If you try str(19.998), it will probably give you something at least close to 19.998 because 32 bits have enough precision to estimate that, but something like 19.999999999999999 is too precise to estimate in 32 bits, so it rounds to the nearest possible value, which happens to be 20.
Please note that this is a problem of understanding floating point (fixed-length) numbers. Most languages do exactly (or very similar to) what Python does.
Python float is IEEE 754 64-bit binary floating point. It is limited to 53 bits of precision i.e. slightly less than 16 decimal digits of precision. 19.9999999999999999 contains 18 decimal digits; it cannot be represented exactly as a float. float("19.9999999999999999") produces the nearest floating point value, which happens to be the same as float("20.0").
>>> float("19.9999999999999999") == float("20.0")
True
If by "many decimals" you mean "many digits after the decimal point", please be aware that the same "weird" results happen when there are many decimal digits before the decimal point:
>>> float("199999999999999999")
2e+17
If you want the full float precision, don't use str(), use repr():
>>> x = 1. / 3.
>>> str(x)
'0.333333333333'
>>> str(x).count('3')
12
>>> repr(x)
'0.3333333333333333'
>>> repr(x).count('3')
16
>>>
Update It's interesting how often decimal is prescribed as a cure-all for float-induced astonishment. This is often accompanied by simple examples like 0.1 + 0.1 + 0.1 != 0.3. Nobody stops to point out that decimal has its share of deficiencies e.g.
>>> (1.0 / 3.0) * 3.0
1.0
>>> (Decimal('1.0') / Decimal('3.0')) * Decimal('3.0')
Decimal('0.9999999999999999999999999999')
>>>
True, float is limited to 53 binary digits of precision. By default, decimal is limited to 28 decimal digits of precision.
>>> Decimal(2) / Decimal(3)
Decimal('0.6666666666666666666666666667')
>>>
You can change the limit, but it's still limited precision. You still need to know the characteristics of the number format to use it effectively without "astonishing" results, and the extra precision is bought by slower operation (unless you use the 3rd-party cdecimal module).
For any given binary floating point number, there is an infinite set of decimal fractions that, on input, round to that number. Python's str goes to some trouble to produce the shortest decimal fraction from this set; see GLS's paper http://kurtstephens.com/files/p372-steele.pdf for the general algorithm (IIRC they use a refinement that avoids arbitrary-precision math in most cases). You happened to input a decimal fraction that rounds to a float (IEEE double) whose shortest possible decimal fraction is not the same as the one you entered.