How does Python calculate trigonometric functions?
I try to calculate using
x = ((0.1-0.001)/2)*math.sin(((1/20)*math.pi*20)+(0.5*math.pi*1))+((0.1-0.001)/2)+0.001
I'm getting
x = 0.1
why is that? in a usual calculator (radian) i'm getting 0.001
In Python 2, / is integer division,you need to import __future__ .division for floating division :
>>> from __future__ import division
>>> import math
>>> x = ((0.1-0.001)/2)*math.sin(((1/20)*math.pi*20)+(0.5*math.pi*1))+((0.1-0.001)/2)+0.001
>>> x
0.001
In python2.x, python takes the floor of integer division. Thus, you need to import division from the __future__ library at the top of your program.
from __future__ import division
x = ((0.1-0.001)/2)*math.sin(((1/20)*math.pi*20)+(0.5*math.pi*1))+((0.1-0.001)/2)+0.001
print x
Just make your integers such as 2 float 2.0, else Python 2.x uses integer division, also known as floor division (rounding towards minus infinity e.g. -9/8 gives -2, 9/8 gives 1), when dividing integers by other integers (whether plain or long):
x = ((0.1-0.001)/2.0)*math.sin(((1/20.0)*math.pi*20)+(0.5*math.pi*1))+((0.1-0.001)/2.0)+0.001
and:
print x
0.001
I want to convert 1/2 in python so that when i say print x (where x = 1/2) it returns 0.5
I am looking for the most basic way of doing this, without using any split functions, loops or maps
I have tried float(1/2) but I get 0...
can someone explain me why and how to fix it?
Is it possible to do this without modifying the variable x= 1/2 ?
In python 3.x any division returns a float;
>>> 1/2
0.5
To achieve that in python 2.x, you have to force float conversion:
>>> 1.0/2
0.5
Or to import the division from the "future"
>>> from __future__ import division
>>> 1/2
0.5
An extra: there is no built-in fraction type, but there is one in Python's standard library:
>>> from fractions import Fraction
>>> a = Fraction(1, 2) #or Fraction('1/2')
>>> a
Fraction(1, 2)
>>> print a
1/2
>>> float(a)
0.5
and so on...
If the input is a string,then you could use Fraction directly on the input:
from fractions import Fraction
x='1/2'
x=Fraction(x) #change the type of x from string to Fraction
x=float(x) #change the type of x from Fraction to float
print x
You're probably using Python 2. You can "fix" division by using:
from __future__ import division
at the start of your script (before any other imports). By default in Python 2, the / operator performs integer division when using integer operands, which discards fractional parts of the result.
This has been changed in Python 3 so that / is always floating point division. The new // operator performs integer division.
Alternatively, you can force floating point division by specifying a decimal or by multiplying by 1.0. For instance (from inside the python interpreter):
>>> print 1/2
0
>>> print 1./2
0.5
>>> x = 1/2
>>> print x
0
>>> x = 1./2
>>> print x
0.5
>>> x = 1.0 * 1/2
>>> print x
0.5
EDIT: Looks like I was beaten to the punch in the time it took to type up my response :)
There is no quantity 1/2 anywhere. Python does not represent rational numbers with a built-in type - just integers and floating-point numbers. 1 is divided by 2 - following the integer division rules - resulting in 0. float(0) is 0.
I have two variables : count, which is a number of my filtered objects, and constant value per_page. I want to divide count by per_page and get integer value but I no matter what I try - I'm getting 0 or 0.0 :
>>> count = friends.count()
>>> print count
1
>>> per_page = 2
>>> print per_page
2
>>> pages = math.ceil(count/per_pages)
>>> print pages
0.0
>>> pages = float(count/per_pages)
>>> print pages
0.0
What am I doing wrong, and why math.ceil gives float number instead of int ?
Python does integer division when both operands are integers, meaning that 1 / 2 is basically "how many times does 2 go into 1", which is of course 0 times. To do what you want, convert one operand to a float: 1 / float(2) == 0.5, as you're expecting. And, of course, math.ceil(1 / float(2)) will yield 1, as you expect.
(I think this division behavior changes in Python 3.)
Integer division is the default of the / operator in Python < 3.0. This has behaviour that seems a little weird. It returns the dividend without a remainder.
>>> 10 / 3
3
If you're running Python 2.6+, try:
from __future__ import division
>>> 10 / 3
3.3333333333333335
If you're running a lower version of Python than this, you will need to convert at least one of the numerator or denominator to a float:
>>> 10 / float(3)
3.3333333333333335
Also, math.ceil always returns a float...
>>> import math
>>> help(math.ceil)
ceil(...)
ceil(x)
Return the ceiling of x as a float.
This is the smallest integral value >= x.
From Python documentation (math module):
math.ceil(x)
Return the ceiling of x as a float, the smallest integer value greater than or equal to x.
They're integers, so count/per_pages is zero before the functions ever get to do anything beyond that. I'm not a Python programmer really but I know that (count * 1.0) / pages will do what you want. There's probably a right way to do that however.
edit — yes see #mipadi's answer and float(x)
its because how you have it set up is performing the operation and then converting it to a float try
count = friends.count()
print count
per_page = float(2)
print per_page
pages = math.ceil(count/per_pages)
print pages
pages = count/per_pages
By converting either count or per_page to a float all of its future operations should be able to do divisions and end up with non whole numbers
>>> 10 / float(3)
3.3333333333333335
>>> #Or
>>> 10 / 3.0
3.3333333333333335
>>> #Python make any decimal number to float
>>> a = 3
>>> type(a)
<type 'int'>
>>> b = 3.0
>>> type(b)
<type 'float'>
>>>
The best solution maybe is to use from __future__ import division
Is there a way to get the ceil of a high precision Decimal in python?
>>> import decimal;
>>> decimal.Decimal(800000000000000000001)/100000000000000000000
Decimal('8.00000000000000000001')
>>> math.ceil(decimal.Decimal(800000000000000000001)/100000000000000000000)
8.0
math rounds the value and returns non precise value
The most direct way to take the ceiling of a Decimal instance x is to use x.to_integral_exact(rounding=ROUND_CEILING). There's no need to mess with the context here. Note that this sets the Inexact and Rounded flags where appropriate; if you don't want the flags touched, use x.to_integral_value(rounding=ROUND_CEILING) instead. Example:
>>> from decimal import Decimal, ROUND_CEILING
>>> x = Decimal('-123.456')
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')
Unlike most of the Decimal methods, the to_integral_exact and to_integral_value methods aren't affected by the precision of the current context, so you don't have to worry about changing precision:
>>> from decimal import getcontext
>>> getcontext().prec = 2
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')
By the way, in Python 3.x, math.ceil works exactly as you want it to, except that it returns an int rather than a Decimal instance. That works because math.ceil is overloadable for custom types in Python 3. In Python 2, math.ceil simply converts the Decimal instance to a float first, potentially losing information in the process, so you can end up with incorrect results.
x = decimal.Decimal('8.00000000000000000000001')
with decimal.localcontext() as ctx:
ctx.prec=100000000000000000
ctx.rounding=decimal.ROUND_CEILING
y = x.to_integral_exact()
You can do this using the precision and rounding mode option of the Context constructor.
ctx = decimal.Context(prec=1, rounding=decimal.ROUND_CEILING)
ctx.divide(decimal.Decimal(800000000000000000001), decimal.Decimal(100000000000000000000))
EDIT: You should consider changing the accepted answer.. Although the prec can be increased as needed, to_integral_exact is a simpler solution.
>>> decimal.Context(rounding=decimal.ROUND_CEILING).quantize(
... decimal.Decimal(800000000000000000001)/100000000000000000000, 0)
Decimal('9')
def decimal_ceil(x):
int_x = int(x)
if x - int_x == 0:
return int_x
return int_x + 1
Just use potency to make this.
import math
def lo_ceil(num, potency=0): # Use 0 for multiples of 1, 1 for multiples of 10, 2 for 100 ...
n = num / (10.0 ** potency)
c = math.ceil(n)
return c * (10.0 ** potency)
lo_ceil(8.0000001, 1) # return 10
How does one round a number UP in Python?
I tried round(number) but it rounds the number down. Example:
round(2.3) = 2.0
and not 3, as I would like.
The I tried int(number + .5) but it round the number down again! Example:
int(2.3 + .5) = 2
The math.ceil (ceiling) function returns the smallest integer higher or equal to x.
For Python 3:
import math
print(math.ceil(4.2))
For Python 2:
import math
print(int(math.ceil(4.2)))
I know this answer is for a question from a while back, but if you don't want to import math and you just want to round up, this works for me.
>>> int(21 / 5)
4
>>> int(21 / 5) + (21 % 5 > 0)
5
The first part becomes 4 and the second part evaluates to "True" if there is a remainder, which in addition True = 1; False = 0. So if there is no remainder, then it stays the same integer, but if there is a remainder it adds 1.
If working with integers, one way of rounding up is to take advantage of the fact that // rounds down: Just do the division on the negative number, then negate the answer. No import, floating point, or conditional needed.
rounded_up = -(-numerator // denominator)
For example:
>>> print(-(-101 // 5))
21
Interesting Python 2.x issue to keep in mind:
>>> import math
>>> math.ceil(4500/1000)
4.0
>>> math.ceil(4500/1000.0)
5.0
The problem is that dividing two ints in python produces another int and that's truncated before the ceiling call. You have to make one value a float (or cast) to get a correct result.
In javascript, the exact same code produces a different result:
console.log(Math.ceil(4500/1000));
5
You might also like numpy:
>>> import numpy as np
>>> np.ceil(2.3)
3.0
I'm not saying it's better than math, but if you were already using numpy for other purposes, you can keep your code consistent.
Anyway, just a detail I came across. I use numpy a lot and was surprised it didn't get mentioned, but of course the accepted answer works perfectly fine.
Use math.ceil to round up:
>>> import math
>>> math.ceil(5.4)
6.0
NOTE: The input should be float.
If you need an integer, call int to convert it:
>>> int(math.ceil(5.4))
6
BTW, use math.floor to round down and round to round to nearest integer.
>>> math.floor(4.4), math.floor(4.5), math.floor(5.4), math.floor(5.5)
(4.0, 4.0, 5.0, 5.0)
>>> round(4.4), round(4.5), round(5.4), round(5.5)
(4.0, 5.0, 5.0, 6.0)
>>> math.ceil(4.4), math.ceil(4.5), math.ceil(5.4), math.ceil(5.5)
(5.0, 5.0, 6.0, 6.0)
I am surprised nobody suggested
(numerator + denominator - 1) // denominator
for integer division with rounding up. Used to be the common way for C/C++/CUDA (cf. divup)
The syntax may not be as pythonic as one might like, but it is a powerful library.
https://docs.python.org/2/library/decimal.html
from decimal import *
print(int(Decimal(2.3).quantize(Decimal('1.'), rounding=ROUND_UP)))
For those who want to round up a / b and get integer:
Another variant using integer division is
def int_ceil(a, b):
return (a - 1) // b + 1
>>> int_ceil(19, 5)
4
>>> int_ceil(20, 5)
4
>>> int_ceil(21, 5)
5
Note: a and b must be non-negative integers
Here is a way using modulo and bool
n = 2.3
int(n) + bool(n%1)
Output:
3
Try this:
a = 211.0
print(int(a) + ((int(a) - a) != 0))
Be shure rounded value should be float
a = 8
b = 21
print math.ceil(a / b)
>>> 0
but
print math.ceil(float(a) / b)
>>> 1.0
The above answers are correct, however, importing the math module just for this one function usually feels like a bit of an overkill for me. Luckily, there is another way to do it:
g = 7/5
g = int(g) + (not g.is_integer())
True and False are interpreted as 1 and 0 in a statement involving numbers in python. g.is_interger() basically translates to g.has_no_decimal() or g == int(g). So the last statement in English reads round g down and add one if g has decimal.
In case anyone is looking to round up to a specific decimal place:
import math
def round_up(n, decimals=0):
multiplier = 10 ** decimals
return math.ceil(n * multiplier) / multiplier
Without importing math // using basic envionment:
a) method / class method
def ceil(fl):
return int(fl) + (1 if fl-int(fl) else 0)
def ceil(self, fl):
return int(fl) + (1 if fl-int(fl) else 0)
b) lambda:
ceil = lambda fl:int(fl)+(1 if fl-int(fl) else 0)
>>> def roundup(number):
... return round(number+.5)
>>> roundup(2.3)
3
>>> roundup(19.00000000001)
20
This function requires no modules.
x * -1 // 1 * -1
Confusing but it works: For x=7.1, you get 8.0. For x = -1.1, you get -1.0
No need to import a module.
For those who doesn't want to use import.
For a given list or any number:
x = [2, 2.1, 2.5, 3, 3.1, 3.5, 2.499,2.4999999999, 3.4999999,3.99999999999]
You must first evaluate if the number is equal to its integer, which always rounds down. If the result is True, you return the number, if is not, return the integer(number) + 1.
w = lambda x: x if x == int(x) else int(x)+1
[w(i) for i in z]
>>> [2, 3, 3, 3, 4, 4, 3, 3, 4, 4]
Math logic:
If the number has decimal part: round_up - round_down == 1, always.
If the number doens't have decimal part: round_up - round_down == 0.
So:
round_up == x + round_down
With:
x == 1 if number != round_down
x == 0 if number == round_down
You are cutting the number in 2 parts, the integer and decimal. If decimal isn't 0, you add 1.
PS:I explained this in details since some comments above asked for that and I'm still noob here, so I can't comment.
If you don't want to import anything, you can always write your own simple function as:
def RoundUP(num):
if num== int(num):
return num
return int(num + 1)
To do it without any import:
>>> round_up = lambda num: int(num + 1) if int(num) != num else int(num)
>>> round_up(2.0)
2
>>> round_up(2.1)
3
I know this is from quite a while back, but I found a quite interesting answer, so here goes:
-round(-x-0.5)
This fixes the edges cases and works for both positive and negative numbers, and doesn't require any function import
Cheers
I'm surprised I haven't seen this answer yet round(x + 0.4999), so I'm going to put it down. Note that this works with any Python version. Changes made to the Python rounding scheme has made things difficult. See this post.
Without importing, I use:
def roundUp(num):
return round(num + 0.49)
testCases = list(x*0.1 for x in range(0, 50))
print(testCases)
for test in testCases:
print("{:5.2f} -> {:5.2f}".format(test, roundUp(test)))
Why this works
From the docs
For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus n; if two multiples are equally close, rounding is done toward the even choice
Therefore 2.5 gets rounded to 2 and 3.5 gets rounded to 4. If this was not the case then rounding up could be done by adding 0.5, but we want to avoid getting to the halfway point. So, if you add 0.4999 you will get close, but with enough margin to be rounded to what you would normally expect. Of course, this will fail if the x + 0.4999 is equal to [n].5000, but that is unlikely.
You could use round like this:
cost_per_person = round(150 / 2, 2)
You can use floor devision and add 1 to it.
2.3 // 2 + 1
when you operate 4500/1000 in python, result will be 4, because for default python asume as integer the result, logically:
4500/1000 = 4.5 --> int(4.5) = 4
and ceil of 4 obviouslly is 4
using 4500/1000.0 the result will be 4.5 and ceil of 4.5 --> 5
Using javascript you will recieve 4.5 as result of 4500/1000, because javascript asume only the result as "numeric type" and return a result directly as float
Good Luck!!
I think you are confusing the working mechanisms between int() and round().
int() always truncates the decimal numbers if a floating number is given; whereas round(), in case of 2.5 where 2 and 3 are both within equal distance from 2.5, Python returns whichever that is more away from the 0 point.
round(2.5) = 3
int(2.5) = 2
My share
I have tested print(-(-101 // 5)) = 21 given example above.
Now for rounding up:
101 * 19% = 19.19
I can not use ** so I spread the multiply to division:
(-(-101 //(1/0.19))) = 20
I'm basically a beginner at Python, but if you're just trying to round up instead of down why not do:
round(integer) + 1