I want to take a floating-point number and round it down to the nearest integer. However, if it's not a whole, I always want to round down the variable, regardless of how close it is to the next integer up. Is there a way to do this?
Simple
int(x)
will work as well.
One of these should work:
import math
math.trunc(1.5)
> 1
math.trunc(-1.5)
> -1
math.floor(1.5)
> 1
math.floor(-1.5)
> -2
x//1
The // operator returns the floor of the division. Since dividing by 1 doesn't change your number, this is equivalent to floor but no import is needed.
Notes:
This returns a float
This rounds towards -∞
To get floating point result simply use:
round(x-0.5)
It works for negative numbers as well.
I think you need a floor function :
math.floor(x)
a lot of people say to use int(x), and this works ok for most cases, but there is a little problem. If OP's result is:
x = 1.9999999999999999
it will round to
x = 2
after the 16th 9 it will round. This is not a big deal if you are sure you will never come across such thing. But it's something to keep in mind.
If you don't want to import math, you could use:
int(round(x))
Here's a piece of documentation:
>>> help(round)
Help on built-in function round in module __builtin__:
round(...)
round(number[, ndigits]) -> floating point number
Round a number to a given precision in decimal digits (default 0 digits).
This always returns a floating point number. Precision may be negative.
If you working with numpy, you can use the following solution which also works with negative numbers (it's also working on arrays)
import numpy as np
def round_down(num):
if num < 0:
return -np.ceil(abs(num))
else:
return np.int32(num)
round_down = np.vectorize(round_down)
round_down([-1.1, -1.5, -1.6, 0, 1.1, 1.5, 1.6])
> array([-2., -2., -2., 0., 1., 1., 1.])
I think it will also work if you just use the math module instead of numpy module.
Just make round(x-0.5) this will always return the next rounded down Integer value of your Float. You can also easily round up by do round(x+0.5)
Don't know if you solved this, but I just stumble upon this question. If you want to get rid of decimal points, you could use int(x) and it will eliminate all decimal digits. Theres no need to use round(x).
It may be very simple, but couldn't you just round it up then minus 1?
For example:
number=1.5
round(number)-1
> 1
I used this code where you subtract 0.5 from the number and when you round it, it is the original number rounded down.
round(a-0.5)
Related
This question already has answers here:
How to properly round-up half float numbers?
(21 answers)
Closed 7 months ago.
In Python 3, I'm trying to round the value 4800.5, so I was expecting it to 4801
but it's giving me 4800. I'm not able to track why this is happening.
Any help will be appreciated.
That's by design.
If you have a look at round function documentation (https://docs.python.org/3/library/functions.html#round) you will find that:
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
(so, for example, both round(0.5) and round(-0.5) are 0, and
round(1.5) is 2).
In simple words, 0.5 is a special case which is always rounded toward an even number.
But there're more interesting things. Please have a look at that example:
The behavior of round() for floats can be surprising: for
example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This
is not a bug: it’s a result of the fact that most decimal fractions
can’t be represented exactly as a float.
What you might want to do is to use Decimal for more conventional rounding logic: https://docs.python.org/3/library/decimal.html
For example:
>>> Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_DOWN)
Decimal('7.32')
>>> Decimal('7.325').quantize(Decimal('.01'), rounding=ROUND_UP)
Decimal('7.33')
There are lot of ways to round a number. round() behaves according to a particular rounding strategy which may or may not be the one you need for a given situation (see the first comment to your question).
If you want to round your number to the upper, you can try this:
import math
n = 4800.5
print(math.ceil(n))
you can do something like this:
from decimal import Decimal, ROUND_HALF_UP
def round_half_up(decimal_number, places=0):
if places == 0:
exp = Decimal('1')
else:
exp_str = '0' * places
exp_str = exp_str[:-1] + '1'
exp = Decimal('.{}'.format(exp_str))
return Decimal(decimal_number).quantize(exp, rounding=ROUND_HALF_UP)
print(round_half_up(4800.5)) -> 4801
print(round_half_up(4800.555, 2)) -> 4800.56
Round() function will round up to next value, if decimal is >.5
upto .5 it would round up to just the integer part.
Result of a simple division -34.11/2 returns -17.055 Is there a python function that would return round up decimal? I am expecting to see -17.06
And yes I checked these answers, but neither can answer this question:
Round up to Second Decimal Place in Python
How to round to nearest decimal in Python
Increment a Python floating point value by the smallest possible amount
Python round to nearest 0.25
Any help would be greatly appreciated.
Thanks!
This is a floating point limitation; get around it by using decimal.Decimal.
>>> from decimal import Decimal
>>> Decimal(-34.11/2) # See? Not quite the value you thought.
Decimal('-17.05499999999999971578290569595992565155029296875')
>>> Decimal(-34.11) / Decimal(2) # This doesn't quite work either, because -34.11 is already a little bit off.
Decimal('-17.05499999999999971578290570')
>>> Decimal("-34.11") / Decimal("2")
Decimal('-17.055')
>>> round(Decimal("-34.11") / Decimal("2"), 2)
Decimal('-17.06')
How to check if a float value is within a range (0.50,150.00) and has 2 decimal digits?
For example, 15.22366 should be false (too many decimal digits). But 15.22 should be true.
I tried something like:
data= input()
if data in range(0.50,150.00):
return True
Is that you are looking for?
def check(value):
if 0.50 <= value <= 150 and round(value,2)==value:
return True
return False
Given your comment:
i input 15.22366 it is going to return true; that is why i specified the range; it should accept 15.22
Simply said, floating point values are imprecise. Many values don't have a precise representation. Say for example 1.40. It might be displayed "as it":
>>> f = 1.40
>>> print f
1.4
But this is an illusion. Python has rounded that value in order to nicely display it. The real value as referenced by the variable f is quite different:
>>> from decimal import Decimal
>>> Decimal(f)
Decimal('1.399999999999999911182158029987476766109466552734375')
According to your rule of having only 2 decimals, should f reference a valid value or not?
The easiest way to fix that issue is probably to use round(...,2) as I suggested in the code above. But this in only an heuristic -- only able to reject "largely wrong" values. See my point here:
>>> for v in [ 1.40,
... 1.405,
... 1.399999999999999911182158029987476766109466552734375,
... 1.39999999999999991118,
... 1.3999999999999991118]:
... print check(v), v
...
True 1.4
False 1.405
True 1.4
True 1.4
False 1.4
Notice how the last few results might seems surprising at first. I hope my above explanations put some light on this.
As a final advice, for your needs as I guess them from your question, you should definitively consider using "decimal arithmetic". Python provides the decimal module for that purpose.
float is the wrong data type to use for your case, Use Decimal instead.
Check python docs for issues and limitations. To quote from there (I've generalised the text in Italics)
Floating-point numbers are represented in computer hardware as base 2 (binary) fractions.
no matter how many base 2 digits you’re willing to use, some decimal value (like 0.1) cannot be represented exactly as a base 2 fraction.
Stop at any finite number of bits, and you get an approximation
On a typical machine running Python, there are 53 bits of precision available for a Python float, so the value stored internally when you enter a decimal number is the binary fraction which is close to, but not exactly equal to it.
The documentation for the built-in round() function says that it rounds to the nearest value, rounding ties away from zero.
And finally, it recommends
If you’re in a situation where you care which way your decimal halfway-cases are rounded, you should consider using the decimal module.
And this will hold for your case as well, as you are looking for a precision of 2 digits after decimal points, which float just can't guarantee.
EDIT Note: The answer below corresponds to original question related to random float generation
Seeing that you need 2 digits of sure shot precision, I would suggest generating integer random numbers in range [50, 15000] and dividing them by 100 to convert them to float yourself.
import random
random.randint(50, 15000)/100.0
Why don't you just use round?
round(random.uniform(0.5, 150.0), 2)
Probably what you want to do is not to change the value itself. As said by Cyber in the comment, even if your round a floating point number, it will always store the same precision. If you need to change the way it is printed:
n = random.uniform(0.5, 150)
print '%.2f' % n # 58.03
The easiest way is to first convert the decimal to string and split with '.' and check if the length of the character. If it is >2 then pass on. i.e. Convert use input number to check if it is in a given range.
a=15.22366
if len(str(a).split('.')[1])>2:
if 0.50 <= value <= 150:
<do your stuff>>
I found this interesting question when I was doing homework
we know, 47.36/1.6**2 == 18.5
but when I try to run the following code, it gives me a False(should be true)
print 47.36/1.6**2 == 18.5
Do anyone know what's going on?
You're probably getting an answer like 18.49999999999, which is not exactly equal to 18.5.
As always, the relevant reference for this is What Every Computer Scientist Should Know About Floating-Point Arithmetic.
Short answer: IEEE 754 floating point can't exactly represent fractions where the denominator isn't a power of two, like 1/4, 1/16, 1/256, etc. You can get awfully close, given enough digits, but never quite exactly there.
You compare floating point numbers by defining "equals" as "within a certain delta". You could write something like:
def almost_equals(a, b, delta=0.0005):
return abs(a - b) <= delta
and then test for "probably equal" with:
>>> almost_equals(47.36/1.6**2, 18.5)
True
I would avoid checking for exact equality when comparing two floats. Instead take the difference and see if it is smaller than a value you consider close to zero.
(47.36/1.6**2 - 18.5) < 0.00000000001
will be
True
>>> 47.36/1.6**2
18.499999999999996
See this page on Floating Point Arithmetic: Issues and Limitations.
Here is how you can calculate this to exactly 18.5 without using any rounding or "close enough" behavior by using the decimal module:
>>> from decimal import Decimal
>>> Decimal('47.36') / Decimal('1.6')**2 == Decimal('18.5')
True
>>> float(Decimal('47.36') / Decimal('1.6')**2) == 18.5
True
As others have said:
>>> 47.36/1.6**2
18.499999999999996
But, this is NOT due to a floating-point arithmetic problem as far as I can tell. Even if you use decimal math by wrapping the operands in Decimal() (after from decimal import Decimal) you will still get Decimal('18.49999999999999772404279952') as the answer.
It's possible I'm using Decimal() wrong here and my result also has some sort of floating point error; however, if I'm correct, that expression flat out does not equal 18.5, no matter what kind of math you use.
Edit: As Greg points out in the comments, the problem with my approach here is that Decimal(1.6) will just convert the float representation of 1.6, inaccuracies intact, into a Decimal. This gives the correct answer:
>>> Decimal('47.36') / Decimal('1.6')**2
Decimal('18.5')
Better still would be to use the fractions module as suggested by Kirk.
47.36/1.6*2 return integer. So 47.36/1.6*2 would be 18, which is not equal to 18.5.
Edit
Sorry about that, actually it is being stored as 18.499999.
You should do this
import numpy as np
print np.around((47.36/1.6**2), decimals=1) == 18.5
This would return True.
I have been asked to test a library provided by a 3rd party. The library is known to be accurate to n significant figures. Any less-significant errors can safely be ignored. I want to write a function to help me compare the results:
def nearlyequal( a, b, sigfig=5 ):
The purpose of this function is to determine if two floating-point numbers (a and b) are approximately equal. The function will return True if a==b (exact match) or if a and b have the same value when rounded to sigfig significant-figures when written in decimal.
Can anybody suggest a good implementation? I've written a mini unit-test. Unless you can see a bug in my tests then a good implementation should pass the following:
assert nearlyequal(1, 1, 5)
assert nearlyequal(1.0, 1.0, 5)
assert nearlyequal(1.0, 1.0, 5)
assert nearlyequal(-1e-9, 1e-9, 5)
assert nearlyequal(1e9, 1e9 + 1 , 5)
assert not nearlyequal( 1e4, 1e4 + 1, 5)
assert nearlyequal( 0.0, 1e-15, 5 )
assert not nearlyequal( 0.0, 1e-4, 6 )
Additional notes:
Values a and b might be of type int, float or numpy.float64. Values a and b will always be of the same type. It's vital that conversion does not introduce additional error into the function.
Lets keep this numerical, so functions that convert to strings or use non-mathematical tricks are not ideal. This program will be audited by somebody who is a mathematician who will want to be able to prove that the function does what it is supposed to do.
Speed... I've got to compare a lot of numbers so the faster the better.
I've got numpy, scipy and the standard-library. Anything else will be hard for me to get, especially for such a small part of the project.
As of Python 3.5, the standard way to do this (using the standard library) is with the math.isclose function.
It has the following signature:
isclose(a, b, rel_tol=1e-9, abs_tol=0.0)
An example of usage with absolute error tolerance:
from math import isclose
a = 1.0
b = 1.00000001
assert isclose(a, b, abs_tol=1e-8)
If you want it with precision of n significant digits, simply replace the last line with:
assert isclose(a, b, abs_tol=10**-n)
There is a function assert_approx_equal in numpy.testing (source here) which may be a good starting point.
def assert_approx_equal(actual,desired,significant=7,err_msg='',verbose=True):
"""
Raise an assertion if two items are not equal up to significant digits.
.. note:: It is recommended to use one of `assert_allclose`,
`assert_array_almost_equal_nulp` or `assert_array_max_ulp`
instead of this function for more consistent floating point
comparisons.
Given two numbers, check that they are approximately equal.
Approximately equal is defined as the number of significant digits
that agree.
Here's a take.
def nearly_equal(a,b,sig_fig=5):
return ( a==b or
int(a*10**sig_fig) == int(b*10**sig_fig)
)
I believe your question is not defined well enough, and the unit-tests you present prove it:
If by 'round to N sig-fig decimal places' you mean 'N decimal places to the right of the decimal point', then the test assert nearlyequal(1e9, 1e9 + 1 , 5) should fail, because even when you round 1000000000 and 1000000001 to 0.00001 accuracy, they are still different.
And if by 'round to N sig-fig decimal places' you mean 'The N most significant digits, regardless of the decimal point', then the test assert nearlyequal(-1e-9, 1e-9, 5) should fail, because 0.000000001 and -0.000000001 are totally different when viewed this way.
If you meant the first definition, then the first answer on this page (by Triptych) is good.
If you meant the second definition, please say it, I promise to think about it :-)
There are already plenty of great answers, but here's a think:
def closeness(a, b):
"""Returns measure of equality (for two floats), in unit
of decimal significant figures."""
if a == b:
return float("infinity")
difference = abs(a - b)
avg = (a + b)/2
return math.log10( avg / difference )
if closeness(1000, 1000.1) > 3:
print "Joy!"
This is a fairly common issue with floating point numbers. I solve it based on the discussion in Section 1.5 of Demmel[1]. (1) Calculate the roundoff error. (2) Check that the roundoff error is less than some epsilon. I haven't used python in some time and only have version 2.4.3, but I'll try to get this correct.
Step 1. Roundoff error
def roundoff_error(exact, approximate):
return abs(approximate/exact - 1.0)
Step 2. Floating point equality
def float_equal(float1, float2, epsilon=2.0e-9):
return (roundoff_error(float1, float2) < epsilon)
There are a couple obvious deficiencies with this code.
Division by zero error if the exact value is Zero.
Does not verify that the arguments are floating point values.
Revision 1.
def roundoff_error(exact, approximate):
if (exact == 0.0 or approximate == 0.0):
return abs(exact + approximate)
else:
return abs(approximate/exact - 1.0)
def float_equal(float1, float2, epsilon=2.0e-9):
if not isinstance(float1,float):
raise TypeError,"First argument is not a float."
elif not isinstance(float2,float):
raise TypeError,"Second argument is not a float."
else:
return (roundoff_error(float1, float2) < epsilon)
That's a little better. If either the exact or the approximate value is zero, than the error is equal to the value of the other. If something besides a floating point value is provided, a TypeError is raised.
At this point, the only difficult thing is setting the correct value for epsilon. I noticed in the documentation for version 2.6.1 that there is an epsilon attribute in sys.float_info, so I would use twice that value as the default epsilon. But the correct value depends on both your application and your algorithm.
[1] James W. Demmel, Applied Numerical Linear Algebra, SIAM, 1997.
"Significant figures" in decimal is a matter of adjusting the decimal point and truncating to an integer.
>>> int(3.1415926 * 10**3)
3141
>>> int(1234567 * 10**-3)
1234
>>>
Oren Shemesh got part of the problem with the problem as stated but there's more:
assert nearlyequal( 0.0, 1e-15, 5 )
also fails the second definition (and that's the definition I learned in school.)
No matter how many digits you are looking at, 0 will not equal a not-zero. This could prove to be a headache for such tests if you have a case whose correct answer is zero.
There is a interesting solution to this by B. Dawson (with C++ code)
at "Comparing Floating Point Numbers". His approach relies on strict IEEE representation of two numbers and the enforced lexicographical ordering when said numbers are represented as unsigned integers.
I have been asked to test a library provided by a 3rd party
If you are using the default Python unittest framework, you can use assertAlmostEqual
self.assertAlmostEqual(a, b, places=5)
There are lots of ways of comparing two numbers to see if they agree to N significant digits. Roughly speaking you just want to make sure that their difference is less than 10^-N times the largest of the two numbers being compared. That's easy enough.
But, what if one of the numbers is zero? The whole concept of relative-differences or significant-digits falls down when comparing against zero. To handle that case you need to have an absolute-difference as well, which should be specified differently from the relative-difference.
I discuss the problems of comparing floating-point numbers -- including a specific case of handling zero -- in this blog post:
http://randomascii.wordpress.com/2012/02/25/comparing-floating-point-numbers-2012-edition/