This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 2 years ago.
I am using python 3.6.8 and tried to multiply 0.1235 with 10 and the answer is 1.2349999999999999 rather than 1.235.
After importing the decimal module, when we multiply decimal.Decimal(0.1235) with 10 we get Decimal('1.234999999999999986677323704') rather than Decimal('1.235').
So How to do precision float calculations with python?
The value 0.1235 is a decimal fraction that needs to be converted to a binary fraction in memory, resulting in not a 100 percent accuracy. please refer to Floating Point Arithmetic: Issues and Limitations
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This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 7 months ago.
It seems, from what I can tell, that python 3.10.4 math.floor() will sometimes round up instead of down. This seems to be in contrast to the purpose of the function.
Could someone please explain this?
Example:
>>> math.floor(0.9999999999999999)
0
>>> math.floor(0.99999999999999999)
1
floor() has nothing to do with this:
>>> 0.9999999999999999
0.9999999999999999
>>> 0.99999999999999999
1.0
That is, your second literal rounds up to 1.0 all on its own before math.floor() happens. Thus, math.floor() is flooring the number 1, not 0.99999999999999999. Floating-point literals are automatically converted to internal machine binary floating-point format, which has only 53 bits of precision ("IEEE 754 double" format on almost all machines).
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 2 years ago.
I run the following code:
numero = 80.12
a = int(numero)
b = abs(numero) - abs(int(numero))
print(a,b)
And as a result I get...
80, 0.12000000000000455
I don't know what could be the problem of printing excess of decimal places.
The full gory details can be found in this documentation. Floating point representations of decimal fractions are almost never precise.
This question already has answers here:
How to suppress scientific notation when printing float values?
(16 answers)
Closed 3 years ago.
When dividing 2 large numbers in Python the output was 1.5640891676957637e+308. How do I print the entire number?
This can't be done because that's all the precision the float type has. It tells you the first 16 or something digits of your number and the exponent, but the digits after the first 16 have never been calculated because there's no space for them.
If you want to work with huge numbers and have basically infinite precision, almost like with Python's integers, try the SymPy library.
number_str = str(int(1.5640891676957637e+308))
print(number_str)
Prints:
156408916769576373071379516999674270294758197183972476505692635672561429946607721482839700799900977784426920800145985096418278978450607600874550703086464871105809270941181641564392002031609107640705147719606017681794554578537463358952125037388161745430586972528713238507284919924435316681000630776819257180160
You can consider the decimal data type in the standard library https://docs.python.org/3/library/decimal.html "The decimal module provides support for fast correctly-rounded decimal floating point arithmetic. It offers several advantages over the float datatype...Decimal numbers can be represented exactly"
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 4 years ago.
Below code's output different result , while in daily mathematics it results same
d=1.11-1.10
e=2.11-2.10
print('d= ', d ,'e= ' , e)
you are using floating point mathematics, which has limited accuracy. The results are expected!
If you want accurate results, you could use decimal arithmetics:
from decimal import Decimal
Decimal('1.11') - Decimal('1.10')
Decimal('2.11') - Decimal('2.10')
This question already has answers here:
Limiting floats to two decimal points
(35 answers)
Closed 7 years ago.
So I was surprised I didn't find anything regarding this.
I have a python script which is testing a C++ program. It needs to format a float in the same way std::setprecision does. That is a float like 1.265 should be rounded UP to 1.27 (2 dp).
Now I have the following code:
"{:.2f}".format(myFloat)
The issue is that numbers like 1.265 are rounded to 1.26 and my tests fail. setprecision rounds 1.265 to 1.27.
What is the best way to fix this issue?
You can use double rounding to overcome the inability of binary arithmetic to exactly represent a decimal value.
round(round(1.265, 3) + 0.0005, 2)