The comments state the objective is to print to 2 decimal places.
There's a simple answer for Python 3:
>>> num=3.65
>>> "The number is {:.2f}".format(num)
'The number is 3.65'
or equivalently with f-strings (Python 3.6+):
>>> num = 3.65
>>> f"The number is {num:.2f}"
'The number is 3.65'
As always, the float value is an approximation:
>>> "{}".format(num)
'3.65'
>>> "{:.10f}".format(num)
'3.6500000000'
>>> "{:.20f}".format(num)
'3.64999999999999991118'
I think most use cases will want to work with floats and then only print to a specific precision.
Those that want the numbers themselves to be stored to exactly 2 decimal digits of precision, I suggest use the decimal type. More reading on floating point precision for those that are interested.
Answer from Andrew E on Stack OverflowI'm not sure if this is a problem with my IDE or if its a Python issue but when I add a floating point number, take 3.14 for example, to some numbers I get a lot of additional 0s.
y = 3.14 y1 = 3.14 + 1 print(3.14) Output: 4.140000000000001
I noticed this issue occurs when I add 1 to 4 but if I add 5 I get 8.14 instead, without all of the 0s.
Why does this only happen when adding certain numbers?
which data type should I use for most accurate calculations? float decimal or python?
How to deal with float precision?
General way to print floats without the .0 part
python - Is floating point arbitrary precision available? - Stack Overflow
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The comments state the objective is to print to 2 decimal places.
There's a simple answer for Python 3:
>>> num=3.65
>>> "The number is {:.2f}".format(num)
'The number is 3.65'
or equivalently with f-strings (Python 3.6+):
>>> num = 3.65
>>> f"The number is {num:.2f}"
'The number is 3.65'
As always, the float value is an approximation:
>>> "{}".format(num)
'3.65'
>>> "{:.10f}".format(num)
'3.6500000000'
>>> "{:.20f}".format(num)
'3.64999999999999991118'
I think most use cases will want to work with floats and then only print to a specific precision.
Those that want the numbers themselves to be stored to exactly 2 decimal digits of precision, I suggest use the decimal type. More reading on floating point precision for those that are interested.
The simple way to do this is by using the round buit-in.
round(2.6463636263,2) would be displayed as 2.65.
I am trying to write a scientific simulation in python, precision is very important, what should I use?
I know this is not a Python-specific issue. I read some posts saying it's best to cast the float as a decimal, and others say round before/after calculation. What are your recommendations? Is there a "pythonic" way to do this?
Edit: Decimal
In the standard library, the decimal module may be what you're looking for. Also, I have found mpmath to be quite helpful. The documentation has many great examples as well (unfortunately my office computer does not have mpmath installed; otherwise I would verify a few examples and post them).
One caveat about the decimal module, though. The module contains several in-built functions for simple mathematical operations (e.g. sqrt), but the results from these functions may not always match the corresponding function in math or other modules at higher precisions (although they may be more accurate). For example,
from decimal import *
import math
getcontext().prec = 30
num = Decimal(1) / Decimal(7)
print(" math.sqrt: {0}".format(Decimal(math.sqrt(num))))
print("decimal.sqrt: {0}".format(num.sqrt()))
In Python 3.2.3, this outputs the first two lines
math.sqrt: 0.37796447300922719758631274089566431939601898193359375
decimal.sqrt: 0.377964473009227227214516536234
actual value: 0.3779644730092272272145165362341800608157513118689214
which as stated, isn't exactly what you would expect, and you can see that the higher the precision, the less the results match. Note that the decimal module does have more accuracy in this example, since it more closely matches the actual value.
For this particular problem, decimal is a great way to go, because it stores the decimal digits as tuples!
>>> a = decimal.Decimal(9999999998)
>>> a.as_tuple()
DecimalTuple(sign=0, digits=(9, 9, 9, 9, 9, 9, 9, 9, 9, 8), exponent=0)
Since you're looking for a property that is most naturally expressed in decimal notation, it's a bit silly to use a binary representation. The wikipedia page you linked to didn't indicate how many "non-grafting digits" may appear before the "grafting digits" begin, so this lets you specify:
>>> def isGrafting(dec, max_offset=5):
... dec_digits = dec.as_tuple().digits
... sqrt_digits = dec.sqrt().as_tuple().digits
... windows = [sqrt_digits[o:o + len(dec_digits)] for o in range(max_offset)]
... return dec_digits in windows
...
>>> isGrafting(decimal.Decimal(9999999998))
True
>>> isGrafting(decimal.Decimal(77))
True
I think there's a good chance the result of Decimal.sqrt() will be more accurate, at least for this, than the result of math.sqrt() because of the conversion between binary representation and decimal representation. Consider the following, for example:
>>> num = decimal.Decimal(1) / decimal.Decimal(7)
>>> decimal.Decimal(math.sqrt(num) ** 2) * 7
Decimal('0.9999999999999997501998194593')
>>> decimal.Decimal(num.sqrt() ** 2) * 7
Decimal('1.000000000000000000000000000')
You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory.
With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an integer and then divide it by a power of two so 13.95 will be represented in a similar fashion to 125650429603636838/(2**53).
Double precision numbers have 53 bits (16 digits) of precision and regular floats have 24 bits (8 digits) of precision. The floating point type in Python uses double precision to store the values.
For example,
>>> 125650429603636838/(2**53)
13.949999999999999
>>> 234042163/(2**24)
13.949999988079071
>>> a = 13.946
>>> print(a)
13.946
>>> print("%.2f" % a)
13.95
>>> round(a,2)
13.949999999999999
>>> print("%.2f" % round(a, 2))
13.95
>>> print("{:.2f}".format(a))
13.95
>>> print("{:.2f}".format(round(a, 2)))
13.95
>>> print("{:.15f}".format(round(a, 2)))
13.949999999999999
If you are after only two decimal places (to display a currency value, for example), then you have a couple of better choices:
- Use integers and store values in cents, not dollars and then divide by 100 to convert to dollars.
- Or use a fixed point number like decimal.
There are new format specifications, String Format Specification Mini-Language:
You can do the same as:
"{:.2f}".format(13.949999999999999)
Note 1: the above returns a string. In order to get as float, simply wrap with float(...):
float("{:.2f}".format(13.949999999999999))
Note 2: wrapping with float() doesn't change anything:
>>> x = 13.949999999999999999
>>> x
13.95
>>> g = float("{:.2f}".format(x))
>>> g
13.95
>>> x == g
True
>>> h = round(x, 2)
>>> h
13.95
>>> x == h
True