For some reasons the code below doesn't round the value to 3 decimal places but only 1. Any ideas?
for y in range(len(candidates_list)):
analysis.write(f"\n{candidates_list[y]}: {round(percentages_list[y],3)}% ({candidates_votes[y]})")
Thanks!
Due you are using f-strings (and thats very cool instead of old .format style ), this is not a rounding representation of the number because the previous answer gives you the right way to do it, this is a truncated representation of the number on the f strings, but this answer may help to others.
Instead of printing
f'{round(percentages_list[y],3)}'
you can do:
f'{percentages_list[y]:.3f}'
Instead of using the round() function, use the appropriate formatting operations. This will including trailing zero digits after the decimal.
And use zip() to iterate through multiple lists in parallel, rather than list indexing.
for candidate, percentage, vote in zip(candidates_list, percentages_list, candidates_votes):
analysis.write(f"\n{candidate}: {percentage:.3f}% ({vote}")
Related
I want to convert some floats to Decimal retaining 5 digits after decimal place regardless of how many digits before the decimal place. Is using string formatting the most efficient way to do this?
I see in the docs:
The significance of a new Decimal is determined solely by the number of digits input. Context precision and rounding only come into play during arithmetic operations.
So that means I need to add 0 to force it to use the specified prec but the prec is total digits not after decimal so it doesn't actually help.
The best thing I can come up with is
a=[1.132434, 22.2334,99.33999434]
[Decimal("%.5f" % round(x,5)) for x in a]
to get [Decimal('1.13243'), Decimal('22.23340'), Decimal('99.33999')]
Is there a better way? It feels like turning floats into strings just to convert them back to a number format isn't very good although I can't articulate why.
Do all the formatting on the way out from your code, inside the print and write statements. There is no reason I can think of to lose precision (and convert the numbers to some fixed format) while doing numeric calculations inside the code.
Is the algorithm used for rounding a float in Python to a specified number of digits specified in any Python documentation? The semantics of round with zero fractional digits (i.e. rounding to an integer) are simple to understand, but it's not clear to me how the case where the number of digits is nonzero is implemented.
The most straightforward implementation of the function that I can think of (given the existence of round to zero fractional digits) would be:
def round_impl(x, ndigits):
return (10 ** -ndigits) * round(x * (10 ** ndigits))
I'm trying to write some C++ code that mimics the behavior of Python's round() function for all values of ndigits, and the above agrees with Python for the most part, when translated to equivalent C++ calls. However, there are some cases where it differs, e.g.:
>>> round(0.493125, 5)
0.49312
>>> round_impl(0.493125, 5)
0.49313
There is clearly a difference that occurs when the value to be rounded is at or very near the exact midpoint between two potential output values. Therefore, it seems important that I try to use the same technique if I want similar results.
Is the specific means for performing the rounding specified by Python? I'm using CPython 2.7.15 in my tests, but I'm specifically targeting v2.7+.
Also refer to What Every Programmer Should Know About Floating-Point Arithmetic, which has more detailed explanations for why this is happening as it is.
This is a mess. First of all, as far as float is concerned, there is no such number as 0.493125, when you write 0.493125 what you actually get is:
0.493124999999999980015985556747182272374629974365234375
So this number is not exactly between two decimals, it's actually closer to 0.49312 than it is to 0.49313, so it should definitely round to 0.49312, that much is clear.
The problem is that when you multiply by 105, you get the exact number 49312.5. So what happened here is the multiplication gave you an inexact result which by coincidence canceled out the rounding error in the original number. Two rounding errors canceled each other out, yay! But the problem is that when you do this, the rounding is actually incorrect... at least if you want to round up at midpoints, but Python 3 and Python 2 behave differently. Python 2 rounds away from 0, and Python 3 rounds towards even least-significant digits.
Python 2
if two multiples are equally close, rounding is done away from 0
Python 3
...if two multiples are equally close, rounding is done toward the even choice...
Summary
In Python 2,
>>> round(49312.5)
49313.0
>>> round(0.493125, 5)
0.49312
In Python 3,
>>> round(49312.5)
49312
>>> round(0.493125, 5)
0.49312
And in both cases, 0.493125 is really just a short way of writing 0.493124999999999980015985556747182272374629974365234375.
So, how does it work?
I see two plausible ways for round() to actually behave.
Choose the closest decimal number with the specified number of digits, and then round that decimal number to float precision. This is hard to implement, because it requires doing calculations with more precision than you can get from a float.
Take the two closest decimal numbers with the specified number of digits, round them both to float precision, and return whichever is closer. This will give incorrect results, because it rounds numbers twice.
And Python chooses... option #1! The exactly correct, but much harder to implement version. Refer to Objects/floatobject.c:927 double_round(). It uses the following process:
Write the floating-point number to a string in decimal format, using the requested precision.
Parse the string back in as a float.
This uses code based on David Gay's dtoa library. If you want C++ code that gets the actual correct result like Python does, this is a good start. Fortunately you can just include dtoa.c in your program and call it, since its licensing is very permissive.
The Python documentation for and 2.7 specifies the behaviour:
Values are rounded to the closest multiple of 10 to the power minus
ndigits; if two multiples are equally close, rounding is done away
from 0.
For 3.7:
For the built-in types supporting round(), values are rounded to the
closest multiple of 10 to the power minus ndigits; if two multiples
are equally close, rounding is done toward the even choice
Update:
The (cpython) implementation can be found floatobjcet.c in the function float___round___impl, which calls round if ndigits is not given, but double_round if it is.
double_round has two implementations.
One converts the double to a string (aka decimal) and back to a double.
The other one does some floating point calculations, calls to pow and at its core calls round. It seems to have more potential problems with overflows, since it actually multiplies the input by 10**-ndigits.
For the precise algorithm, look at the linked source file.
So I have a list of tuples of two floats each. Each tuple represents a range. I am going through another list of floats which represent values to be fit into the ranges. All of these floats are < 1 but positive, so precision matter. One of my tests to determine if a value fits into a range is failing when it should pass. If I print the value and the range that is causing problems I can tell this much:
curValue = 0.00145000000671
range = (0.0014500000067055225, 0.0020968749796738849)
The conditional that is failing is:
if curValue > range[0] and ... blah :
# do some stuff
From the values given by curValue and range, the test should clearly pass (don't worry about what is in the conditional). Now, if I print explicitly what the value of range[0] is I get:
range[0] = 0.00145000000671
Which would explain why the test is failing. So my question then, is why is the float changing when it is accessed. It has decimal values available up to a certain precision when part of a tuple, and a different precision when accessed. Why would this be? What can I do to ensure my data maintains a consistent amount of precision across my calculations?
The float doesn't change. The built-in numberic types are all immutable. The cause for what you're observing is that:
print range[0] uses str on the float, which (up until very recent versions of Python) printed less digits of a float.
Printing a tuple (be it with repr or str) uses repr on the individual items, which gives a much more accurate representation (again, this isn't true anymore in recent releases which use a better algorithm for both).
As for why the condition doesn't work out the way you expect, it's propably the usual culprit, the limited precision of floats. Try print repr(curVal), repr(range[0]) to see if what Python decided was the closest representation of your float literal possible.
In modern day PC's floats aren't that precise. So even if you enter pi as a constant to 100 decimals, it's only getting a few of them accurate. The same is happening to you. This is because in 32-bit floats you only get 24 bits of mantissa, which limits your precision (and in unexpected ways because it's in base2).
Please note, 0.00145000000671 isn't the exact value as stored by Python. Python only diplays a few decimals of the complete stored float if you use print. If you want to see exactly how python stores the float use repr.
If you want better precision use the decimal module.
It isn't changing per se. Python is doing its best to store the data as a float, but that number is too precise for float, so Python modifies it before it is even accessed (in the very process of storing it). Funny how something so small is such a big pain.
You need to use a arbitrary fixed point module like Simple Python Fixed Point or the decimal module.
Not sure it would work in this case, because I don't know if Python's limiting in the output or in the storage itself, but you could try doing:
if curValue - range[0] > 0 and...
I am depending on some code that uses the Decimal class because it needs precision to a certain number of decimal places. Some of the functions allow inputs to be floats because of the way that it interfaces with other parts of the codebase. To convert them to decimal objects, it uses things like
mydec = decimal.Decimal(str(x))
where x is the float taken as input. My question is, does anyone know what the standard is for the 'str' method as applied to floats?
For example, take the number 2.1234512. It is stored internally as 2.12345119999999999 because of how floats are represented.
>>> x = 2.12345119999999999
>>> x
2.1234511999999999
>>> str(x)
'2.1234512'
Ok, str(x) in this case is doing something like '%.6f' % x. This is a problem with the way my code converts to decimals. Take the following:
>>> d = decimal.Decimal('2.12345119999999999')
>>> ds = decimal.Decimal(str(2.12345119999999999))
>>> d - ds
Decimal('-1E-17')
So if I have the float, 2.12345119999999999, and I want to pass it to Decimal, converting it to a string using str() gets me the wrong answer. I need to know what are the rules for str(x) that determine what the formatting will be, because I need to determine whether this code needs to be re-written to avoid this error (note that it might be OK, because, for example, the code might round to the 10th decimal place once we have a decimal object)
There must be some set of rules in python's docs that hopefully someone here can point me to. Thanks!
In the Python source, look in "Include/floatobject.h". The precision for the string conversion is set a few lines from the top after an comment with some explanation of the choice:
/* The str() precision PyFloat_STR_PRECISION is chosen so that in most cases,
the rounding noise created by various operations is suppressed, while
giving plenty of precision for practical use. */
#define PyFloat_STR_PRECISION 12
You have the option of rebuilding, if you need something different. Any changes will change formatting of floats and complex numbers. See ./Objects/complexobject.c and ./Objects/floatobject.c. Also, you can compare the difference between how repr and str convert doubles in these two files.
There's a couple of issues worth discussing here, but the summary is: you cannot extract information that is not stored on your system already.
If you've taken a decimal number and stored it as a floating point, you'll have lost information, since most decimal (base 10) numbers with a finite number of digits cannot be stored using a finite number of digits in base 2 (binary).
As was mentioned, str(a_float) will really call a_float.__str__(). As the documentation states, the purpose of that method is to
return a string containing a nicely printable representation of an object
There's no particular definition for the float case. My opinion is that, for your purposes, you should consider __str__'s behavior to be undefined, since there's no official documentation on it - the current implementation can change anytime.
If you don't have the original strings, there's no way to extract the missing digits of the decimal representation from the float objects. All you can do is round predictably, using string formatting (which you mention):
Decimal( "{0:.5f}".format(a_float) )
You can also remove 0s on the right with resulting_string.rstrip("0").
Again, this method does not recover the information that has been lost.
I am trying to write a method in Python 3.2 that encrypts a phrase and then decrypts it. The problem is that the numbers are so big that when Python does math with them it immediately converts it into scientific notation. Since my code requires all the numbers to function scientific notation, this is not useful.
What I have is:
coded = ((eval(input(':'))+1213633288469888484)/2)+1042
Basically, I just get a number from the user and do some math to it.
I have tried format() and a couple other things but I can't get them to work.
EDIT: I use only even integers.
In python3, '/' does real division (e.g. floating point). To get integer division, you need to use //. In other words 100/2 yields 50.0 (float) whereas 100//2 yields 50 (integer)
Your code probably needs to be changed as:
coded = ((eval(input(':'))+1213633288469888484)//2)+1042
As a cautionary tale however, you may want to consider using int instead of eval:
coded = ((int(input(':'))+1213633288469888484)//2)+1042
If you know that the floating point value is really an integer, or you don't care about dropping the fractional part, you can just convert it to an int before you print it.
>>> print 1.2e16
1.2e+16
>>> print int(1.2e16)
12000000000000000