In Python, how can one print a number that might be an integer or real type, when the latter case would require me to limit my printout to a certain amount of digits?
Long story short, say we have the following example:
print("{0:.3f}".format(num)) # I cannot do print("{}".format(num))
# because I don't want all the decimals
Is there a "Pythy" way to ensure e.g. in case num == 1 that I print 1 instead of 1.000 (I mean other than cluttering my code with if statements)
With Python 3*, you can just use round() because in addition to rounding floats, when applied to an integer it will always return an int:
>>> num = 1.2345
>>> round(num,3)
1.234
>>> num = 1
>>> round(num,3)
1
This behavior is documented in help(float.__round__):
Help on method_descriptor:
__round__(...)
Return the Integral closest to x, rounding half toward even.
When an argument is passed, work like built-in round(x, ndigits).
And help(int.__round__):
Help on method_descriptor:
__round__(...)
Rounding an Integral returns itself.
Rounding with an ndigits argument also returns an integer.
* With Python 2, round() always returns a float.
If you need to maintain a fixed-width for float values, you could use the printf-style formatting, like this:
>>> num = 1
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1
>>> num = 1.2345
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.234
>>> num = 1.2
>>> print('%0.*f' % (isinstance(num, float) * 3, num))
1.200
If you use a fix number of floating point, you could just use a replace to remove the extra 0. For instance this would do the trick:
print("{:.3f}".format(1).replace(".000", ""))
For fix number of decimal point:
>>> num = 0.2
>>> print('%.04*f' % num)
0.2000
>>> num = 3.102
>>> print('%.02*f' % num)
3.10
This question already has answers here:
How to round a floating point number up to a certain decimal place?
(12 answers)
Closed 2 years ago.
What I want is:
if 1700 / 1000 = 1.7 = int(1) # I want this to be True
lst.append("T")
My original code was:
if 1700 / 1000 == int(1) # This is False and I want it to be True
lst.append("T")
The if statement is False because the answer is 1.7 and not 1. I would like this to be True. So I want the 1.7 to round down to 1 using int so that the if statement will be True.
// always provide integer value. If you want to get always integer then use it otherwise follow int(a/b)
if 1700 // 1000 == int(1): # I want this to be True
lst.append("T")
You either need to put the int on the other side:
if int(1700 / 1000) == 1
lst.append("T")
i.e. round the 1700/1000 to an integer before comparing it,
or use //, which is integer division and discards the fraction part:
if 1700 // 1000 == 1
lst.append("T")
You can either use the // operator (integer division) or use the floor function from the math module:
>>> from math import floor
>>> floor(1.7/1)
1
>>> floor(1.7/1) == int(1)
True
>>> 1.7 // 1
1.0
>>> 1.7 // 1 == 1
True
You can try
if int(1700/1000) == 1 # I want this to be True
lst.append("T")
int(1700/1000) will convert the 1.7 into 1 by ignoring the decimal portion of the number
Everything brilliant is simple
if int(1700 / 1000) == int(1):
lst.append("T")
You should take a look at the floor and ceil functions. The floor function rounds to the last number, whereas ceil to the next. In your case, you need to do it like this:
import math
if math.floor(1700 / 1000) == int(1):
print("TRUE")
else:
print("FALSE")
You could round up or down in Python by 'floor' and 'ceil' methods in the 'math' library.
from math import floor, ceil
print (floor(1.7), ' , ', ceil(1.7))
The result will be
1 , 2
This will work in either Python 2.x or Python 3.x
In python3, integer division works differently than in python 2.7.3. Is there a way to ensure that a number that doesn't have a remainder after division is returned as an int, while a number that does have a remainder after division is returned as a float?
I'd like to be able to check:
if (instanceof(x/n, int)):
# do something
The following occurs in python3:
>>> 4 / 2
2.0
>>> 5 / 2
2.5
Is there some way to make division act like this?
>>> 4 / 2
2
>>> 5 / 2
2.5
You'd have to implement it yourself. Obvious approach would be:
def divspecial(n, d):
q, r = divmod(n, d) # Get quotient and remainder of division (both int)
if not r:
return q # If no remainder, return quotient
return n / d # Otherwise, compute float result as accurately as possible
Of course, if you're just trying to check if division would be exact or not, don't use nonsense like the function above to check:
if isinstance(divspecial(x, n), int):
Just test the remainder directly:
if x % n == 0: # If remainder is 0, division was exact
I don't think there is a way to make it automatic, but you could always do a quick check afterwards to convert it:
r = 4/2
if r%1==0:
r=int(r)
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.
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