# How do you round UP a number?

## Question:

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
```

## Answers:

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)))
```

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)
```

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.

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.

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
```

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
```

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

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)))
```

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
```

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
```

You can use floor devision and add 1 to it.

2.3 // 2 + 1

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.

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)
```

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`

)

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!!

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`

.

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)
```

Try this:

```
a = 211.0
print(int(a) + ((int(a) - a) != 0))
```

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
```

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*

```
>>> def roundup(number):
... return round(number+.5)
>>> roundup(2.3)
3
>>> roundup(19.00000000001)
20
```

This function requires no modules.

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
```

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
```

You could use round like this:

```
cost_per_person = round(150 / 2, 2)
```

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.

`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.

Here is a way using `modulo`

and `bool`

```
n = 2.3
int(n) + bool(n%1)
```

Output:

```
3
```

This should work.

```
a=16
b= int(input("Please enter a number greater than 0 n"))
if b==0:
print ( "Wrong input")
elif a%b != 0:
c=a/b
d= int(c)+1
print (c)
print (d)
else:
c=a/b
d=c
print (c)
print (d)
```

Here is pretty straightforward answer, based on use of default round()

```
def round_up(arg):
if arg > round(arg):
return round(arg) + 1
else:
return round(arg)
```

It does

```
1 to 1 -1 to -1 1.0 to 1 1.00000001 to 2 -1.00000001 to -1
```